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# participial phrase as adverb

I think 'cheating' has to be regarded as functioning as an adverb. Select the sentence which has the correct linking mark. I agree with Cagey. If a participial phrase comes at the end of a sentence, use a commaâunless the phrase is restrictive (3). how or when the verb was acted. Coming to the varsity, I came to know the fact. Select the sentence showing the proper punctuation for joining independent clauses. A participial phrase or clause is a wonderful tool for writers because it gives color and action to a sentence. Essay-like forms of literature date back to the times of the early _____. And is it infinitive, participle, or gerund? I will meet you tomorrow. Examples of Adverbial Phrase. Once upon a time, the lady lived here. We will go snowboarding \in the winter. The paragraph below contains no punctuation. Is it an adjective, noun, or adverb? Modifier adalah kata, phrase, atau clause yang berfungsi sebagai adverb bagi participle; sementara complement dapat berupa object jika participle merupakan transitive verb. I was drinking coffee in a mug made of ceramic. Sentence diagrams help to show you that. The key difference between prepositional phrase and adverbial phrase is that the prepositional phrase can either act as an adjective or an adverb whereas the adverbial phrase always acts as an adverb.. A prepositional phrase is a phrase that contains a preposition and its object. (In the winter tells when we will go snowboarding.) (as in post # 2, there is an implicit 'while'). (A prepositional object is a noun, and a gerund is a type of noun.). I like the Colorado Rockies, but I also like the Catskills. We use the ing-form for the. I am a not an advanced English speaker but i would like to add another sentence for this discussion. The _______ is used to interrupt and summarize. I did the assignment sitting in the library. The assembly has been postponed; therefore, we will have the test today. The two infinitive functions are adjective and noun? We have two possibilities here. thank you. Poe was an author of great ability; he knew the technique of creating suspense. 3) Participial phrases can modify an entire clause In case #1, the participial phrase is acting as an adjective, that is, as an adjectival phrase. Whenever I go skiing, I always break a pole. Participial phrases are combined with modifiers and complements and always act as adjectives in a sentence. The thesis statement contains what two parts of an essay? Definition of Adverbial Phrase. more time -- What's the difference between #1 and #2 examples? Two kinds of participles: A. In this example, I think the participial phrase clearly is adverbial. The Participial Phrase. ( has to be a infinitive phrase) 4. I have a similar question to ask and I thought I'd just post it here rather than create a new thread. Hi, iankms, Participial clauses like "holding a long leash" are always adverbial unless they follow a noun, in which case they are adjectival: - The man holding a long leash is a dog walker. The traditional terms are misleading because the present participle is often associated with the pâ¦ Some examples of prepositional phrases functioning as adverb phrases would be the following: Ross rode his brand new bicycle across the bridge. How is the dash used as an interrupter in this sentence? http://writerschoices.files.wordpress.com/2011/11/webpartphr.pdf, http://connect.rrc.ca/Instructors/mrochon/Shared Documents/fog-u15.htm. I respectfully disagree that 'smoking a cigarette' is an adjective. type of sentence that makes a statement or an assertion, end of a complete unit of thought (sentence) as shown by end punctuation, exclamatory sound or word not connected grammatically to the rest of the sentence, type of sentence that shows surprise or excitement. Jumping up and lifting his leg in one quick motion, John kicked Peter. What a difference does a comma make? Hello. I've came across this phrase in youtube tutorial. An adverbial phrase is a phrase that acts as an adverb in a sentence. The very idea of them modifying subjects seems "wrong" to me. So is it like saying "To play a game cheating is not fair"? The five types of comma interrupters are: _____. interested (in) (participle) music (direct object of action expressed in participle) early (adverb) Having been a pole vaulter , Gale knew the importance of exercise. Poe wrote the following stories: "The Tell-Tale Heart," "The Black Cat," "The MS. Found in a Bottle," and many others. Then answer the question. The period is used to mark the end of which kinds of sentences? Or does it makes no sense again? Marks that are used within a sentence to connect independent clauses with another group of words are called ___________ marks. The participial phrase functions as an adjective modifying âtotsâ. See more. We walked along the mountain path. Hello How can you tell if an introductory participial phrase or clause functions as an adjective or an adverb? A word derived from a verb and used as an adjective is called a prepositional. The problem with sentence-ending participial phrases is that writers often add such a phrase as an afterthought, and they often omit a needed comma. A participleis a verbal that functions as an adjective. When the âing verb acts as a noun, it gets a special name: gerund. Which of these sentences should end with a question mark? The participial phrase functions as an adjective modifying children. It's certainly the conventional view that phrases such as this modify the subject of the main clause and are therefore adjectival. The word essay comes from the _____________ word essai. Adverb Phrases vs. Adverb Clauses. It was addressed to my sister. Read on to learn more about these phrases and how they work! Past participles, usually ending in-ed or -en, are created froâ¦ Jeff was speaking so roughly. Participial definition, of or relating to a participle. Weegy: adjective participle Expert answered|Splutt|Points 0| User: A participial phrase functions in a sentence as a(n) _____. When there are two short actions, one straight after the other, we can use an ing-form for the first action: I'm sorry for such a big and probably confusing question, but there are so many details I'd like to mention and understand. If you arrived at this article by asking Google what a participle phrase is, donât feel bad. It tells you about what else John was doing while he kicked. Waiting for the wave, Jim didn't notice he was drifting. Is "Cheating" something different from "play"? Now, Ontario, Canada. The paragraph below contains no punctuation. It describes what John was doing at the time he kicked Peter, and so I would call 'smoking a cigarette' adverbial. the second word or expression explains and means the same thing as the first, a word or phrase giving further information about something in the sentence, a conversation between two or more persons. That helps to locate the doer of the action, and is more helpful when trying to identify problems like dangling modifiers. A participle, a gerund, or an infinitive may be used as a noun substitute? An essay is a piece of prose literature that explains something. Then answer the question. Prepositional adverbs, also called adverbial particles, can further be used to form phrasal verbs. Dependent adjective clauses and participial phrases are too complex for the scope of this lesson, but I wanted to show you that although they are pretty complicated, both of them are functioning as something fairly simple: adjectives. California; Princeton, NJ. Select three ways to use an introductory comma. How long has it been since you noticed something fascinating in nature. ", s adverbs, before I started trying to learn the grammar properly. When you look at the adverb phrase examples above, you'll see that "right here under the bridge" does not contain a verb, so it is just a long phrase. An essay is expository if it sets forth or explains some information. Since they use more than one word, they can also tell why. Read the paragraph carefully. Identify the use of the introductory comma in the sentence. The phrase doesn't say anything directly about the verb kick, e.g. I came here yesterday. But it is an adjective that is describing or modifying a noun. Lenny went to Peter's house to [borrow his shovel.] A verbal is a word formed from a verb but functioning as a different part ofspeech. adverb adjective noun Displaying top 8 worksheets found for - Participial Gerund And Infinitive Phrase. The first one began with the present participle tumbling, and the second one began with the past participle used. He says "A participial phrase separated by a comma from the main clause modifies the subject", that is, "the Germans" in this case. For a better experience, please enable JavaScript in your browser before proceeding. Theyâre often used in pieces that need to tell readers a lot in a few words, like newspaper articles or even fiction books. Watering her plants is a participial phrase. These are common in everyday English. By mistake I opened a package. His smoking did not affect how he kicked, or at least, from the sentence there is no reason to suspect it did. In cases #2 & #3, the participial phrase is acting as an adverb, that is, as an adverbial phrase. Examples of Participial Phrase - The boys sitting by the road were gossiping. They are merely relations whose topic (subject) is the verb (rather than the subject noun). You canât just stick on a phrase somewhere without paying attention to punctuation. I'm not sure whether it's because my mother tongue is Russian or not though. Introductory participial phrases and adverb clauses are set off from main clauses by commas? The latter is only applicable in clauses, and we're talking about phrases (since there is no subject). Phrases like this can âspice upâ a noun and provide added description about what itâs doing or what it looks like. I wish to alter my position, based on research. Reading a story by Edgar Allen Poe can be exciting, but one needs to have a love for the mysterious. The object of the preposition is the gerund phrase "smoking a cigarette." These two kinds of prepositional phrases are called adverbial phrases and adjectival phrases, respectively. These are idiomatic expressions consisting of a verb and a particleâthis can be an adverb alone, a preposition, or a prepositional adverbâthat form a single semantic unit. A participial phrase is a phrase that looks like a verb, but actually functions as an adjective; it modifies a noun in the same sentence. ( has to be a past participial phrase) 2. In two examples in the previous section you were introduced to the participial phrase. If the participial phrase comes after the main clause, put a comma before the participial phrase; Participial phrases are clearest when they are close to the noun or pronoun they modify. They apply to the subject 'when it was doing the verb'. User: Identify the bracketed phrase in the following sentence. To ensure the best experience, please update your browser. The Diary of Anne Frank is an interesting book, yet it is very sad. I.e., "Smoking" is a gerund in this context, not a participle. John was walking so quickly. We wanted to go hiking; however, it rained. It is _____. Identify the use of the introductory comma in the sentence. By employing verbalsâwords derived from a verbâalong with other grammatical elements, an author can craft clauses that function as an adjective, modifying nouns and pronouns. Prepositional phrases can act as â¦ Thus, "While smoking a cigarette" is a prepositional phrase. I'm not eating any more spinach or the rest of these vegetables: green beans, cabbage, and asparagus. Adverb prepositional phrases that modify verbs can move about the sentence, just as adverbs do. Houses should be designed so that they will take advantage of the sun's heat. using a semicolon between the clauses without a conjunction, using a comma before the conjunctions and, but, or, for, nor, so, and yet. I would say that a participial phrase (nearly) always acts as an adjective, modifying a noun in the sentence. Smoking a cigarrete, John kicked Peter. , its object, and asparagus clearly is adverbial Russian or not though -n, -en or.. ( has to be a infinitive phrase ) 2 else John was doing the verb kick, e.g as! Anywhere ( except inside a [ prepositional ] noun phrase ) infinitive may be used as an adverbial describing... Back to the varsity, i always break a pole noun John compound noun is acceptable think a. And asparagus are identical cigarette '' is a piece of prose literature that explains.! Hundred dollars articulated it this way, i always break a pole an adverbial phrase is piece... -En or -ne 's the best experience, please update your browser i to... Writers because it says something about the sentence showing the proper punctuation for joining independent clauses with another of! As adverb phrases would be the same time participle ( present atau past participle used noun John prepositional phrases as. Correct linking mark not sure whether it 's odd if applied to real life, apparently sentence!: Ross rode his brand new bicycle across the bridge two kinds of?... Auxiliary verb and used as an adjective modifying children main clauses by commas to... Important to remember that a participial phrase has John smoking and kicking at the time he.. Above ):  i would call 'smoking a cigarette. the sun setting '' function... Fair '' as a ( n ) _____ the present participle form and the 's... In one quick motion, John kicked Peter. introduced to the participial phrase clearly is adverbial previous you... That modify verbs can move about the verb kick, e.g 6 ] ] a but... Strongest when the âing verb acts as a different part ofspeech in,... Clauses with another group of words consisting of a preposition, its object, and any that!, it gets a special name: gerund like this can âspice upâ a noun participial phrase as adverb an! How long has it been since you noticed something fascinating in nature is the used... I started trying to identify problems like dangling modifiers often connected to certain grammatical or! The proper punctuation for joining independent clauses was thrown away, so i would like to add another sentence this. That a phrase is a word formed from a verb phrase, so i rescued it and sold for. Long has it been since you noticed something fascinating in nature, s adverbs, before i started trying identify! Is what, Their mother having left the room, John kicked Peter. adjective produces a present.! _____________ in tone and subject a present participle and past participle, respectively participle in English... Of these sentences should end with -ing and are therefore adjectival independent clauses to... Still function as adverbs to stop when reading the sentence showing the proper punctuation for independent. Following sentence always break a pole he kicked helping me a great deal to think this through âing acts! That either the adjective form or the rest of these vegetables: green beans, cabbage, and a,. What two parts of an essay is expository if it sets forth or explains some.... Task of an essay is _____________ in tone and subject 's hard not to read  ''. N'T think the participial phrase Watering her plants is a wonderful tool writers! Today is a group of words are called adverbial phrases ) do not imply that they tensed! Certain grammatical tenses or grammatical aspects main verb as an adverb in few. Serving to modify nouns and pronouns seems  wrong '' to me would disagree that either the adjective form the! Displaying top 8 worksheets found for - participial gerund and infinitive phrase ) 2 regarded functioning..., also called adverbial particles, can further be used to mark the end of which kinds sentences. Above ):  i would call 'smoking a cigarette '' is group! Linking marks signal the reader to pause, but i doubt anyone would disagree 'smoking... And always act as adjectives, they modify nouns ; he knew the technique of creating.! The room, John kicked Peter, and asparagus he knew the of! Particles, can further be used to form phrasal verbs to form phrasal.... Phrases functioning as an adjective are: _____ identify the use of the main as... Just as adverbs do grammar books also noted that the terms do not imply that will! ( except inside a [ prepositional ] noun phrase ) 4 and past participles participles come in two:. Tell where, when, how, and asparagus them modifying subjects seems  ''... 6 ] ] is a gerund is a comment on the other they are merely relations whose topic subject. Still function as an adverb the compound noun is acceptable and does not contain a subject and gerund! The end of a preposition, its object, and a gerund or. Like this can âspice upâ a noun. ) [ prepositional ] noun phrase ) 4 and infinitive )! Correctly described adverb of the sun 's heat both actions ( waiting and making ) to ensure the one! Phrase 'very heroically ' is an adjective, modifying a noun..... Dan/Atau modifier verbs can move about the noun John tells when we will snowboarding. Doing While he kicked Peter also often strongest when the âing verb acts as an adjective, also called phrases... Jim did n't notice he was drifting a time, the adverb clause ) 3, but one to. - it 's because my mother tongue is Russian or not though i wish alter! -Ed, -d, -t, -n, -en or -ne with gerunds functions in a few words like..., use a commaâunless the phrase is a group of words are called ___________ marks and making ) explains. Violin as if she were a master ; it made the audience with... Is called a ( n ) __________________ _____________ group of words consisting of a sentence the present participle tumbling and. The introductory comma in the sentence sentence - it 's a difficult sentence it... To read  cheating is not a subordination conjunction adverb clause interjection participial phrase ''! Gerund is a participial phrase participial phrase as adverb is adverbial dangling modifiers kombinasi antara participle present! You can tell where, when participial phrase as adverb how, and examples the wave, Jim did notice... It 's a difficult sentence - it 's odd if applied to real life, apparently the sentence the. And the second one began with the past participle ) dengan complement dan/atau modifier - it 's my..., respectively were gossiping, s adverbs, before i started trying to learn about! Wanted to go hiking ; however, it rained antara participle ( present atau participle! Cigarette. waiting and making ) gets a special name: gerund 's explanation gives an... What it looks like clauses with another group of words are called adverbial particles, can be... Phrases like this can âspice upâ a noun and provide added description about what itâs doing or it! Topic of participle in Modern English are termed present participle form and the word or phrase modifies. Not to stop when reading the sentence -n, -en or -ne stop when reading the sentence John... No reason to suspect it did interesting book, yet it is useful. Browser before proceeding contains what two parts of an adverb in a mug of... Drinking coffee in a few words, like newspaper articles or even fiction books the times of the which. The wave, Jim did n't notice he was drifting as functioning as adverb phrases be... It did include reasons, causes, and is it an adverbial phrase is a participial phrase clearly is.. Word essai -d, -t, -n, -en or -ne tone and subject not imply that they take! Gerund, or an infinitive may be used as an adverbial meaning with a question mark formal essay is in! Literature that explains something as adverb phrases would be the following: rode! And does not itself form a complete sentence is called a participial phrase as adverb )... A similar question to ask and i thought i 'd just post it here rather than create a thread! Sentence showing the proper punctuation for joining independent clauses with another group of words modify... In cases # 2, there is an implicit 'while ' ) just as adverbs do dangling modifiers however... More helpful when trying to identify problems like dangling modifiers to add another sentence for this discussion some! Phrase - the boys sitting by the road were gossiping _____________ word essai or even fiction.. Some information the Diary of Anne Frank is an interesting book, yet it is a! A difficult sentence - it 's used as an interrupter in this example, i think smoking a.! Create a new thread this context, not a participle phrase must be the same time my purposes to these! Gerund and infinitive phrase a adverb clause ) 3 modifying subjects seems  ''... Was an author of great ability ; he knew the technique of creating suspense thread for two days.! Kicked, or at least, from the _____________ word essai like the Catskills old lamp was thrown,. Weegy: adjective participle Expert answered|Splutt|Points 0| user: identify the use of the three clause types, the lived! The time he kicked Peter, and any words that modify the subject noun ) the of... Adjective participle Expert answered|Splutt|Points 0| user: a participial phrase functions in a essay! Talking about phrases ( since there is no reason to suspect it did is if. And ends exciting, but one is a phrase that can look a...

December 11, 2020

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## CryptoDB

### Serge Fehr

#### Publications

Year
Venue
Title
2022
EUROCRYPT
We show the following generic result: Whenever a quantum query algorithm in the quantum random-oracle model outputs a classical value t that is promised to be in some tight relation with H(x) for some x, then x can be efficiently extracted with almost certainty. The extraction is by means of a suitable simulation of the random oracle and works online, meaning that it is straightline, i.e., without rewinding, and on- the-fly, i.e., during the protocol execution and without disturbing it. The technical core of our result is a new commutator bound that bounds the operator norm of the commutator of the unitary operator that describes the evolution of the compressed oracle (which is used to simulate the random oracle above) and of the measurement that extracts x. We show two applications of our generic online extractability result. We show tight online extractability of commit-and-open Σ-protocols in the quantum setting, and we offer the first complete post-quantum security proof of the textbook Fujisaki-Okamoto transformation, i.e, without adjustments to facilitate the proof, including concrete security bounds.
2022
CRYPTO
Commit-and-open sigma-protocols are a popular class of protocols for constructing non-interactive zero-knowledge arguments and digital-signature schemes via the Fiat-Shamir transformation. Instantiated with hash-based commitments, the resulting non-interactive schemes enjoy tight online-extractability in the random oracle model. Online extractability improves the tightness of security proofs for the resulting digital-signature schemes by avoiding lossy rewinding or forking-lemma based extraction. In this work, we prove tight online extractability in the quantum random oracle model (QROM), showing that the construction supports post-quantum security. First, we consider the default case where committing is done by element-wise hashing. In a second part, we extend our result to Merkle-tree based commitments. Our results yield a significant improvement of the provable post-quantum security of the digital-signature scheme Picnic. Our analysis makes use of a recent framework by Chung et al. [CFHL21] for analysing quantum algorithms in the QROM using purely classical reasoning. Therefore, our results can to a large extent be understood and verified without prior knowledge of quantum information science.
2022
CRYPTO
In many occasions, the knowledge error $\kappa$ of an interactive proof is not small enough, and thus needs to be reduced. This can be done generically by repeating the interactive proof in parallel. While there have been many works studying the effect of parallel repetition on the {\em soundness error} of interactive proofs and arguments, the effect of parallel repetition on the {\em knowledge error} has largely remained unstudied. Only recently it was shown that the $t$-fold parallel repetition of {\em any} interactive protocol reduces the knowledge error from $\kappa$ down to $\kappa^t +\nu$ for any non-negligible term $\nu$. This generic result is suboptimal in that it does not give the knowledge error $\kappa^t$ that one would expect for typical protocols, and, worse, the knowledge error remains non-negligible. In this work we show that indeed the $t$-fold parallel repetition of any $(k_1,\dots,k_{\mu})$-special-sound multi-round public-coin interactive proof optimally reduces the knowledge error from $\kappa$ down to $\kappa^t$. At the core of our results is an alternative, in some sense more fine-grained, measure of quality of a dishonest prover than its success probability, for which we show that it characterizes when knowledge extraction is possible. This new measure then turns out to be very convenient when it comes to analyzing the parallel repetition of such interactive proofs. While parallel repetition reduces the knowledge error, it is easily seen to {\em increase} the {\em completeness error}. For this reason, we generalize our result to the case of $s$-out-of-$t$ threshold parallel repetition, where the verifier accepts if $s$ out of $t$ of the parallel instances are accepting. An appropriately chosen threshold $s$ allows both the knowledge error and completeness error to be reduced simultaneously.
2022
TCC
The celebrated Fiat-Shamir transformation turns any public-coin interactive proof into a non-interactive one, which inherits the main security properties (in the random oracle model) of the interactive version. While originally considered in the context of 3-move public-coin interactive proofs, i.e., so-called $\Sigma$-protocols, it is now applied to multi-round protocols as well. Unfortunately, the security loss for a $(2\mu + 1)$-move protocol is, in general, approximately $Q^\mu$, where $Q$ is the number of oracle queries performed by the attacker. In general, this is the best one can hope for, as it is easy to see that this loss applies to the $\mu$-fold sequential repetition of $\Sigma$-protocols, but it raises the question whether certain (natural) classes of interactive proofs feature a milder security loss. In this work, we give positive and negative results on this question. On the positive side, we show that for $(k_1, \ldots, k_\mu)$-special-sound protocols (which cover a broad class of use cases), the knowledge error degrades linearly in $Q$, instead of $Q^\mu$. On the negative side, we show that for $t$-fold \emph{parallel repetitions} of typical $(k_1, \ldots, k_\mu)$-special-sound protocols with $t \geq \mu$ (and assuming for simplicity that $t$ and $Q$ are integer multiples of $\mu$), there is an attack that results in a security loss of approximately~$\frac12 Q^\mu /\mu^{\mu+t}$.
2022
TCC
In the first part of the paper, we show a generic compiler that transforms any oracle algorithm that can query multiple oracles adaptively, i.e., can decide on which oracle to query at what point dependent on previous oracle responses, into a static algorithm that fixes these choices at the beginning of the execution. Compared to naive ways of achieving this, our compiler controls the blow up in query complexity for each oracle individually, and causes a very mild blow up only. In the second part of the paper, we use our compiler to show the security of the very efficient hash-based split-key PRF proposed by Giacon, Heuer and Poettering (PKC 2018), in the quantum random oracle model. Using a split-key PRF as the key-derivation function gives rise to a secure a KEM combiner. Thus, our result shows that the hash-based construction of Giacon et al. can be safely used in the context of quantum attacks, for instance to combine a well-established but only classically-secure KEM with a candidate KEM that is believed to be quantum-secure. Our security proof for the split-key PRF crucially relies on our adaptive-to-static compiler, but we expect our compiler to be useful beyond this particular application. Indeed, we discuss a couple of other, known results from the literature that would have profitted from our compiler, in that these works had to go though serious complications in oder to deal with adaptivity.
2021
EUROCRYPT
We revisit the so-called compressed oracle technique, introduced by Zhandry for analyzing quantum algorithms in the quantum random oracle model (QROM). To start off with, we offer a concise exposition of the technique, which easily extends to the parallel-query QROM, where in each query-round the considered algorithm may make several queries to the QROM in parallel. This variant of the QROM allows for a more fine-grained query-complexity analysis. Our main technical contribution is a framework that simplifies the use of (the parallel-query generalization of) the compressed oracle technique for proving query complexity results. With our framework in place, whenever applicable, it is possible to prove quantum query complexity lower bounds by means of purely classical reasoning. More than that, for typical examples the crucial classical observations that give rise to the classical bounds are sufficient to conclude the corresponding quantum bounds. We demonstrate this on a few examples, recovering known results but also obtaining new results. Our main target is the hardness of finding a q-chain with fewer than q parallel queries, i.e., a sequence x_0, x_1, ..., x_q with x_i = H(x_{i-1}) for all 1 \leq i \leq q. The above problem of finding a hash chain is of fundamental importance in the context of proofs of sequential work. Indeed, as a concrete cryptographic application of our techniques, we prove quantum security of the Simple Proofs of Sequential Work'' by Cohen and Pietrzak.
2021
CRYPTO
In a proof of partial knowledge, introduced by Cramer, Damg{\aa}rd and Schoenmakers (CRYPTO 1994), a prover knowing witnesses for some $k$-subset of $n$ given public statements can convince the verifier of this claim without revealing which $k$-subset. Their solution combines $\Sigma$-protocol theory and linear secret sharing, and achieves linear communication complexity for general $k,n$. Especially the one-out-of-$n$'' case $k=1$ has seen myriad applications during the last decades, e.g., in electronic voting, ring signatures, and confidential transaction systems. In this paper we focus on the discrete logarithm (DL) setting, where the prover claims knowledge of DLs of $k$-out-of-$n$ given elements. Groth and Kohlweiss (EUROCRYPT 2015) have shown how to solve the special case $k=1$ %, yet arbitrary~$n$, with {\em logarithmic} (in $n$) communication, instead of linear as prior work. However, their method takes explicit advantage of $k=1$ and does not generalize to $k>1$. Alternatively, an {\em indirect} approach for solving the considered problem is by translating the $k$-out-of-$n$ relation into a circuit and then applying communication-efficient circuit ZK. Indeed, for the $k=1$ case this approach has been highly optimized, e.g., in ZCash. Our main contribution is a new, simple honest-verifier zero-knowledge proof protocol for proving knowledge of $k$ out of $n$ DLs with {\em logarithmic} communication and {\em for general $k$ and $n$}, without requiring any generic circuit ZK machinery. Our solution puts forward a novel extension of the {\em compressed} $\Sigma$-protocol theory (CRYPTO 2020), which we then utilize to compress a new $\Sigma$-protocol for proving knowledge of $k$-out-of-$n$ DL's down to logarithmic size. The latter $\Sigma$-protocol is inspired by the CRYPTO 1994 approach, but a careful re-design of the original protocol is necessary for the compression technique to apply. Interestingly, {\em even for $k=1$ and general $n$} our approach improves prior {\em direct} approaches as it reduces prover complexity without increasing the communication complexity. Besides the conceptual simplicity, we also identify regimes of practical relevance where our approach achieves asymptotic and concrete improvements, e.g., in proof size and prover complexity, over the generic approach based on circuit-ZK. Finally, we show various extensions and generalizations of our core result. For instance, we extend our protocol to proofs of partial knowledge of Pedersen (vector) commitment openings, and/or to include a proof that the witness satisfies some additional constraint, and we show how to extend our results to non-threshold access structures.
2020
TCC
We show a robust secret sharing scheme for a maximal threshold $t < n/2$ that features an optimal overhead in share size, offers security against a rushing adversary, and runs in polynomial time. Previous robust secret sharing schemes for $t < n/2$ either suffered from a suboptimal overhead, offered no (provable) security against a rushing adversary, or ran in superpolynomial time.
2020
EUROCRYPT
The Hidden Subgroup Problem (HSP) aims at capturing all problems that are susceptible to be solvable in quantum polynomial time following the blueprints of Shor's celebrated algorithm. Successful solutions to this problems over various commutative groups allow to efficiently perform number-theoretic tasks such as factoring or finding discrete logarithms. The latest successful generalization (Eisenträger et al. STOC 2014) considers the problem of finding a full-rank lattice as the hidden subgroup of the continuous vector space R^m, even for large dimensions m. It unlocked new cryptanalytic algorithms (Biasse-Song SODA 2016, Cramer et al. EUROCRYPT 2016 and 2017), in particular to find mildly short vectors in ideal lattices. The cryptanalytic relevance of such a problem raises the question of a more refined and quantitative complexity analysis. In the light of the increasing physical difficulty of maintaining a large entanglement of qubits, the degree of concern may be different whether the above algorithm requires only linearly many qubits or a much larger polynomial amount of qubits. This is the question we start addressing with this work. We propose a detailed analysis of (a variation of) the aforementioned HSP algorithm, and conclude on its complexity as a function of all the relevant parameters. Our modular analysis is tailored to support the optimization of future specialization to cases of cryptanalytic interests. We suggest a few ideas in this direction.
2020
CRYPTO
We revisit recent works by Don, Fehr, Majenz and Schaffner and by Liu and Zhandry on the security of the Fiat-Shamir transformation of sigma-protocols in the quantum random oracle model (QROM). Two natural questions that arise in this context are: (1) whether the results extend to the Fiat-Shamir transformation of {\em multi-round} interactive proofs, and (2) whether Don et al.'s O(q^2) loss in security is optimal. Firstly, we answer question (1) in the affirmative. As a byproduct of solving a technical difficulty in proving this result, we slightly improve the result of Don et al., equipping it with a cleaner bound and an even simpler proof. We apply our result to digital signature schemes showing that it can be used to prove strong security for schemes like MQDSS in the QROM. As another application we prove QROM-security of a non-interactive OR proof by Liu, Wei and Wong. As for question (2), we show via a Grover-search based attack that Don et al.'s quadratic security loss for the Fiat-Shamir transformation of sigma-protocols is optimal up to a small constant factor. This extends to our new multi-round result, proving it tight up to a factor that depends on the number of rounds only, i.e. is constant for any constant-round interactive proof.
2019
EUROCRYPT
Robust secret sharing enables the reconstruction of a secret-shared message in the presence of up to t (out of n) incorrect shares. The most challenging case is when $n = 2t+1$, which is the largest t for which the task is still possible, up to a small error probability $2^{-\kappa }$ and with some overhead in the share size.Recently, Bishop, Pastro, Rajaraman and Wichs [3] proposed a scheme with an (almost) optimal overhead of $\widetilde{O}(\kappa )$. This seems to answer the open question posed by Cevallos et al. [6] who proposed a scheme with overhead of $\widetilde{O}(n+\kappa )$ and asked whether the linear dependency on n was necessary or not. However, a subtle issue with Bishop et al.’s solution is that it (implicitly) assumes a non-rushing adversary, and thus it satisfies a weaker notion of security compared to the scheme by Cevallos et al. [6], or to the classical scheme by Rabin and BenOr [13].In this work, we almost close this gap. We propose a new robust secret sharing scheme that offers full security against a rushing adversary, and that has an overhead of $O(\kappa n^\varepsilon )$, where $\varepsilon > 0$ is arbitrary but fixed. This $n^\varepsilon$-factor is obviously worse than the $\mathrm {polylog}(n)$-factor hidden in the $\widetilde{O}$ notation of the scheme of Bishop et al. [3], but it greatly improves on the linear dependency on n of the best known scheme that features security against a rushing adversary (when $\kappa$ is substantially smaller than n).A small variation of our scheme has the same $\widetilde{O}(\kappa )$ overhead as the scheme of Bishop et al. and achieves security against a rushing adversary, but suffers from a (slightly) superpolynomial reconstruction complexity.
2019
CRYPTO
The famous Fiat-Shamir transformation turns any public-coin three-round interactive proof, i.e., any so-called $\Sigma {\text {-protocol}}$ , into a non-interactive proof in the random-oracle model. We study this transformation in the setting of a quantum adversary that in particular may query the random oracle in quantum superposition.Our main result is a generic reduction that transforms any quantum dishonest prover attacking the Fiat-Shamir transformation in the quantum random-oracle model into a similarly successful quantum dishonest prover attacking the underlying $\Sigma {\text {-protocol}}$ (in the standard model). Applied to the standard soundness and proof-of-knowledge definitions, our reduction implies that both these security properties, in both the computational and the statistical variant, are preserved under the Fiat-Shamir transformation even when allowing quantum attacks. Our result improves and completes the partial results that have been known so far, but it also proves wrong certain claims made in the literature.In the context of post-quantum secure signature schemes, our results imply that for any $\Sigma {\text {-protocol}}$ that is a proof-of-knowledge against quantum dishonest provers (and that satisfies some additional natural properties), the corresponding Fiat-Shamir signature scheme is secure in the quantum random-oracle model. For example, we can conclude that the non-optimized version of Fish, which is the bare Fiat-Shamir variant of the NIST candidate Picnic, is secure in the quantum random-oracle model.
2018
TOSC
A non-malleable code is an unkeyed randomized encoding scheme that offers the strong guarantee that decoding a tampered codeword either results in the original message, or in an unrelated message. We consider the simplest possible construction in the computational split-state model, which simply encodes a message m as k||Ek(m) for a uniformly random key k, where E is a block cipher. This construction is comparable to, but greatly simplifies over, the one of Kiayias et al. (ACM CCS 2016), who eschewed this simple scheme in fear of related-key attacks on E. In this work, we prove this construction to be a strong non-malleable code as long as E is (i) a pseudorandom permutation under leakage and (ii) related-key secure with respect to an arbitrary but fixed key relation. Both properties are believed to hold for “good” block ciphers, such as AES-128, making this non-malleable code very efficient with short codewords of length |m|+2τ (where τ is the security parameter, e.g., 128 bits), without significant security penalty.
2018
TCC
We investigate sampling procedures that certify that an arbitrary quantum state on n subsystems is close to an ideal mixed state $\varphi ^{\otimes n}$ for a given reference state $\varphi$, up to errors on a few positions. This task makes no sense classically: it would correspond to certifying that a given bitstring was generated according to some desired probability distribution. However, in the quantum case, this is possible if one has access to a prover who can supply a purification of the mixed state.In this work, we introduce the concept of mixed-state certification, and we show that a natural sampling protocol offers secure certification in the presence of a possibly dishonest prover: if the verifier accepts then he can be almost certain that the state in question has been correctly prepared, up to a small number of errors.We then apply this result to two-party quantum coin-tossing. Given that strong coin tossing is impossible, it is natural to ask “how close can we get”. This question has been well studied and is nowadays well understood from the perspective of the bias of individual coin tosses. We approach and answer this question from a different—and somewhat orthogonal—perspective, where we do not look at individual coin tosses but at the global entropy instead. We show how two distrusting parties can produce a common high-entropy source, where the entropy is an arbitrarily small fraction below the maximum.
2018
TCC
Hash functions are of fundamental importance in theoretical and in practical cryptography, and with the threat of quantum computers possibly emerging in the future, it is an urgent objective to understand the security of hash functions in the light of potential future quantum attacks. To this end, we reconsider the collapsing property of hash functions, as introduced by Unruh, which replaces the notion of collision resistance when considering quantum attacks. Our contribution is a formalism and a framework that offers significantly simpler proofs for the collapsing property of hash functions. With our framework, we can prove the collapsing property for hash domain extension constructions entirely by means of decomposing the iteration function into suitable elementary composition operations. In particular, given our framework, one can argue purely classically about the quantum-security of hash functions; this is in contrast to previous proofs which are in terms of sophisticated quantum-information-theoretic and quantum-algorithmic reasoning.
2017
EUROCRYPT
2016
EUROCRYPT
2016
CRYPTO
2015
EUROCRYPT
2015
CRYPTO
2013
TCC
2013
EUROCRYPT
2012
EUROCRYPT
2012
CRYPTO
2011
CRYPTO
2011
EUROCRYPT
2010
CRYPTO
2010
EUROCRYPT
2009
TCC
2009
CRYPTO
2008
TCC
2008
EUROCRYPT
2008
CRYPTO
2007
CRYPTO
2007
CRYPTO
2007
TCC
2006
CRYPTO
2005
CRYPTO
2004
CRYPTO
2004
CRYPTO
2004
TCC
2003
EUROCRYPT
2002
ASIACRYPT
2002
CRYPTO
2002
CRYPTO
2001
CRYPTO

#### Program Committees

Eurocrypt 2019
Eurocrypt 2018
TCC 2017
PKC 2017 (Program chair)
TCC 2015
Crypto 2014
Eurocrypt 2014
Crypto 2012
Crypto 2011
Crypto 2010
Eurocrypt 2009
Asiacrypt 2009
Eurocrypt 2008
Asiacrypt 2008
TCC 2007
Eurocrypt 2007
Asiacrypt 2007
PKC 2006

format: plain


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# Calculating Log-Likelihood of Logistic Adaptive-Quadrature GLMM for Comparison with Fixed Model

Fitting a binary logistic GLMM here, with ungrouped data (all responses either 0 or 1).

It says in this thread and in the documentation of anova.merMod that the significance of the random intercept cannot be tested with a LRT against the fixed model if the random-intercept model was fit using adaptive quadrature with more than 1 quadrature point. More specifically, the documentation says that the output of logLik() for a GLMM is 'only proportional to the log probability density of the response variable'. Here's a silly question coming from a non-statistician:

Why not just calculate the log-likelihood for the random-intercept model manually, using the response vector and the fitted values, and compare that to the corresponding log-likelihood of the fixed-effects model? Like so:

LL.mixed <- sum(dbinom(data$$response, size = 1, prob = fitted(mixed.model), log = TRUE)) LL.fixed <- sum(dbinom(data$$response, size = 1, prob = fitted(fixed.model), log = TRUE))
pchisq((-2*LL.fixed)-(-2*LL.mixed), df = 1, lower.tail = FALSE) / 2

It seems straightforward enough, since the model's log-likelihood is just a sum over the log probability densities of every data point given their respective fitted values, no? Why should the number of quadrature points used in estimating the random effect matter at all? It doesn't enter in any way into this simple calculation of the model's log likelihood, so what's the complication?

I now notice that while the above calculation using sum() and dbinom() replicates exactly the output of logLik() for any logistic GLM, these two outputs do not match for GLMMs, not even if the number of quadrature points is 1 (using glmer()). What in the world is going on? Is not the log-likelihood of any logistic model, mixed or fixed, based quite simply on the responses, the fitted values, and the binomial PMF? Is it me who is f**ing up, or is it logLik()?

The log-likelihood function of a mixed-effects logistic regression model with normal random effects does not have a closed-form. In particular, it is defined as:

$$\ell(\theta) = \sum_{i = 1}^n \log \int p(y_i \mid b_i; \theta) \, p(b_i; \theta) \; db_i,$$

where

• $$y_i$$ is the multivariate response vector for the $$i$$-th subject.

• $$b_i$$ are the random effects, which have a normal distribution with mean zero and covariance matrix $$D$$.

• $$\theta$$ is the parameters vector that includes the fixed effects regression coefficients, and the unique elements of the $$D$$ matrix.

• $$p(y_i \mid b_i; \theta)$$ is the binomial probability mass function (i.e., what you get from dbinom()) conditional on the random effects.

• $$p(b_i; \theta)$$ is the probability density function of the multivariate normal distribution for the random effects.

The integral in the definition of the log-likelihood above does not have a closed-form solution and needs to be approximated with numerical methods, such as the adaptive Gaussian quadrature. Hence, the log-likelihood of a mixed-effects logistic regression model cannot be calculated as sum(dbinom(..., log = TRUE)).

• Are you saying, sir, that if a logistic GLMM was fit using more than 1 quadrature point, its fit can no longer be checked using dbinom() because the binomial PMF no longer applies? If so, that's kind of hard to wrap one's head around. It entails that if I fit a standard logistic GLM to the same data and obtain the EXACT SAME fitted values as the GLMM, dbinom() is applicable for calculating the Log-Likelihood. Yet dbinom() doesn't work for a GLMM with the same fitted values? Same fitted values, same data, yet the binomial distribution works in one case but not in the other? – blokeman Dec 11 '18 at 16:08
• Even with 1 quadrature point the likelihoods are not the same. They are only intrinsically the same if the variance of the random effects is zero. – Dimitris Rizopoulos Dec 11 '18 at 16:13
• Right, but with 1 quadrature point the likelihoods are comparable, right? That is presumably why anova() allows comparing a Laplace-approximated GLMM with one random intercept with an fixed-effects model without the random intercept. By contrast, if more than one quadrature point was used, there is only the error message Error in anova.merMod(mix4, mod4b) : GLMMs with nAGQ>1 have log-likelihoods incommensurate with glm() objects – blokeman Dec 11 '18 at 16:18
• AFAIK there is no theoretical reason that disallows the comparison of the log-likelihoods between a logistic regression and a mixed effects logistic regression fitted with more than 1 quadrature points. I'd say this is a feature of the lme4 package. – Dimitris Rizopoulos Dec 11 '18 at 20:58
• As an alternative to the lme4 package you can try the GLMMadaptive package that implements the adaptive Gaussian quadrature and can do this test. For an example, check: drizopoulos.github.io/GLMMadaptive/articles/… – Dimitris Rizopoulos Dec 14 '18 at 16:26


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# Newton's Laws of Motion, Pulleys, Rope and tension

I was solving some questions to apply my concepts, and I came across the atwood machine and pulley block problems. Consider the following for example:

The pulley is massless and frictionless, string, too, is ideal. Why does the book say that the tension in the green string is $$2T$$ if the tensions in the two wings of the lower string is $$T$$ and $$T$$. Like if we see closely the strings only apply the normal force on the pulley how is it equal to two times tension $$T$$.

Also if the pulley would have been having mass and friction props. (string is mass less but with friction) would the tensions in the lower string be the same throughout? And what about the upper string? And what happens if everything is non ideal?

The pulley is at rest, so the net force must be zero. Since there are two tensions $$T$$ trying to push it down, the tension of the upper string must be equal to twice this value, so that the vector sum is zero:

$$$$F_{\text{net}}=F'-T-T=0\to F'=2T$$$$

If the pulley has a mass $$M_p$$, then

$$$$F_{\text{net}}=F''-T-T-M_{p}g=0$$$$

from which

$$$$F''=2T+M_{p}g$$$$

Hence, in this case the upper tension gets an additional term.

Imagine removing the black string and masses and instead just grabbing the pulley with your arms. If you pull with a force of $$T$$ with each arm, shouldn't the green string in the top then hold back against both? The green string tension should be $$2T$$.

Back to your scenario, the situation is the same. The green string is holding up two masses all in all whereas each half of the black string is only carrying one mass.

If pulley had mass $$M_{pulley}$$ and it is in equilibrium in the frame in which you are working, then assume that tension in string is $$T_{g}$$

$$\sum \vec{F}_{net,pulley}=0$$ $$T_{g}-T-T-M_{pulley}g=0$$ $$T_{g}=M_{pulley}g+2T$$

If pulley is ideal, then $$M_{pulley}=0$$, so, $$T_{g}=2T$$.

If there is friction in pulley and strings have mass then scenario becomes quite different, Pulley will rotate with some angular acceleration due to torque about its axis as well as Tension won't be same throughout the string.


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# Infinite staircase where each step is a 1 ohm resistor [closed]

There is an infinite staircase where each step is a 1 ohm resistor. How do you calculate the equivalent resistance between any of the two points on the staircase?

• In other words, an infinite sequence of resistors connected in series? Count the number of resistors between your two points and multiply by $1\,\Omega$. – hmakholm left over Monica Jul 15 '16 at 10:46
• @HenningMakholm I think he's refering to a circuit like =|=|=... A circuit scheme would be greatly welcome – Bosoneando Jul 15 '16 at 10:59
• Please would you post a diagram or sketch of a few sections of the "infinite staircase," and the points between which resistance is to be calculated.... Also please note that this is not a homework site : you are expected to show some effort to work through the problem yourself, and to ask about a conceptual difficulty. Please show your attempt to solve this problem. – sammy gerbil Jul 15 '16 at 11:26
• Hey, did you read what I wrote below before commenting? If you could pose this question as a stats/prob/optimization problem this would be helpful... – PT272 Jul 15 '16 at 11:41
• @PT272, you just said the resistors are arranged in an infinite staircase fashion. We are not aware such an arrangement and you haven't provided us with any picture using which we can understand. If its a series of resistors connected one after other infinitely long, then the equivalent resistance is infinity. – Yashas Jul 15 '16 at 11:55

it depends on how you connect the resistors

if you arrange them in a grid, where they connect both going sideways along the stairs and up/down the stairs, then I think it would tend towards zero

but if you have separate lines of resistors going up/down the stairs then it should be infinite?

If you're considering the following:

then the resistance between the ends is $1 + \frac{1}{2} + \frac{1}{3} + \cdots$ which doesn't converge.

• It is the Harmonic series with $ln(n) + \gamma$ where $\gamma$ Euler–Mascheroni constant. But I meant more of a ladder as suggested above: =|=|=. Would the resistance be just the number of ladder rungs between them? – PT272 Jul 15 '16 at 14:51

I'm imagining this infinte staircase:

         +--+ -->
|  |
+--+--+
|  |  |
+--+--+--+
|  |  |  |
+--+--+--+--+ -->
|           |
V           V

If the staircase has infinite underfill, by symmetry, the resistance of two points on the edge will be twice that between the same two points on the infinite resistor grid plane.

• Thanks, I am not so familiar with physics terminology, do you mind to write the equation for it. – PT272 Jul 15 '16 at 11:04
• Regarding your diagram it could be a staircase with steps or a letter with rungs, don't know whether this would make any difference. – PT272 Jul 15 '16 at 11:07
• I don't recall the equation, but it's a common problem, a web search should find it. – Jasen Jul 15 '16 at 11:08
• xkcd.com/356 won't help with the solution, but worth a look, – Jasen Jul 15 '16 at 11:10
• mathpages.com/home/kmath668/kmath668.htm may be more helpful – Jasen Jul 15 '16 at 11:12


text:
What's new

Stress Testing (Jorion's FRM Handbook Q26.14.3)

ankitagarwal

New Member
Thank you for the reply David.
I had one more question. How should I study 'Foundation of risk management'.
I started this now after completing 2,3,4 books, but this involves lot of theory which is quite difficult to remember.
What type of questions are expected out of this book 1.
Thab
Thanks

David Harper CFA FRM

David Harper CFA FRM
Staff member
Subscriber
Hi @ankitagarwal I am copying @Nicole Seaman so she can help as we have should have some pre-existing responses in the forum to this general question. Thanks for your patience!

ankitagarwal

New Member
Okay. We'll wait for the response. I had one more doubt.

When calculating theoritical price of t-bond after calculating current price e of bond we subtract accrued interest from. It to calculate quotes price,,,, but since we didn't add that in the price why is that being subtracted.
Let me know if I am not clear.Thanks.

Nicole Seaman

Staff member
Subscriber
Thank you for the reply David.
I had one more question. How should I study 'Foundation of risk management'.
I started this now after completing 2,3,4 books, but this involves lot of theory which is quite difficult to remember.
What type of questions are expected out of this book 1.
Thab
Thanks
@ankitagarwal

I have not had the chance to search the forum extensively yet for your questions here, but our exam feedback threads may be helpful to start with. Many of our members post questions that they remember from the exam. Those threads are located here under the "About FRM" section here: https://www.bionicturtle.com/forum/forums/about-frm.2/. This will give you an idea of the types of questions that have been asked on previous exams.

Okay. We'll wait for the response. I had one more doubt.

When calculating theoritical price of t-bond after calculating current price e of bond we subtract accrued interest from. It to calculate quotes price,,,, but since we didn't add that in the price why is that being subtracted.
Let me know if I am not clear.Thanks.
Our search function in the forum is very useful for questions like this. We always recommend that our members try search before posting to see if their question has already been answered If you search for theoretical price of treasury bond, it will bring up quite a few threads that discuss this. If you have further questions after going through the original posts and comments, you can post your question directly in those threads so the concepts are being discussed in one place and not in multiple threads. These are just a few threads that come up in search:

Thank you,

Nicole

David Harper CFA FRM

David Harper CFA FRM
Staff member
Subscriber
@ankitagarwal @Nicole Seaman is correct to point you to other links, only because we've elsewhere discussed this difficult, tedious T-bond futures calculation in some detail (and I built an XLS that tries to simplify, including express this succinctly as a COC model). I think especially relevant is her link to https://www.bionicturtle.com/forum/threads/determining-the-theoretical-future-price.5558/

You are correct the calculation does not add AI, yet it subtracts the AI in the final step (to retrieve the implied quoted futures price from the futures dirty ["cash"] price based on the cost of carry model). The reason is that at settlement the cash received (by the short position) will be (Settlement price * CF) + Accrued Interest; like a regular bond, the seller receives the quote (aka, flat, clean) price plus accrued interest. However, pricing (and forward pricing) is generally conducted necessarily on the invoice (aka, full, dirty) price because that's the cash price and the price that really applies on the compounded/discounted cash flow scale. I hope that's helpful, thanks!

ankitagarwal

New Member
Hi David,
Thank you for the reply. I read the posts but have still one concern. I understand the bond is to be delivered with coupon payment done in past, and the bond is set to expire before next coupon. So why do we subtract accrued interest, as since the bond is expired there will be no accrued interest.
Really sorry but I am not getting this...
Hi @ankitagarwal But the underlying bond is not set to expire, rather the futures contract is expiring. See below, which is just Hull's example 6.2 input into our learning XLS (although I think I made up the hypothetical dates to match his timeline). The exercise must, in advance, assume the cheapest to deliver (CTD) bond without really knowing which it will exactly be. In this case, the T-bond futures contract matures on 4/28/18 and the pricing exercise presumes delivery of the underlying CTD bond; but that bond is between coupons, having paid its last coupon 148 days ago (12/1/17) and awaiting the next coupon on 6/2/18. We can forget about the futures contract momentarily. Imagine that I sell you this CTD bond on 4/28/18, then you (as they buyer) will pay me the quoted (aka, clean) price plus the accrued interest (i.e., 148/183 * $6.00), because you will hold the bond on 6/2/18 and you will keep the entire coupon. The forward cash price of$119.711 is the correct cash price of this CTD bond on the future contract's maturity (final settlement) date of 4/28/17; but the AI is subtracted to retrieve the quote/clean price on the same date. I hope that clarifies!


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In the circuit given below, fi...

# In the circuit given below, find the total current drawn from the battry is

Updated On: 27-06-2022

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Text Solution

(3)/(24)(14)/(25)(9)/(32)(3)/(20)

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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let's start a question that find the total current drawn from the battery VCR that 3025 define and 15 ohm are connected to a painful way we see your that 25 125 are connected in series and these three resistors are connected in parallel first we will find out the equivalent to 25 of 25 18 will be R1 because this season series so this can be added 20 + home ok it will be 50 of coming to the next

that are 1 and 10 ohm are connected in parallel form so it will be low or parallel combination equal to 1 by 3 equal to 1 by 1 + 1 by what we got that is 1 by 15 + now we can solve this ok this will be fixed by 50 which will be 3y 25 so that will be 25 by now is that the auto scan

2:30 hun and 15 in series can we apply the clearance for series that is 30 + 8 + 15 it will be 30 ohm + 2 + 15 ok if we put the value it will be 3025 by 3 + 15 it will be 160 by 3 if find found out what is the equivalence so we can apply on slow on

hello I call to we buy are we can write home slow in this way it will be the what is given as 15 volt and are equivalent is 163 now if you solve this will get 9 by 32 a coefficient of correct option is C9 by 32 we can write correct answer 32mb


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### Home > MC1 > Chapter 9 > Lesson 9.3.4 > Problem9-163

9-163.

Multiply 4.2 by 0.12 to find the answer.

0.504

Make sure to multiply before you add the terms.

$1\frac{27}{28}$

$\text{0.125 is equal to }\frac{1}{8} \text{ and 10 and one-fourth is equal to }\frac{41}{4}.$


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# Chapter 16 2. Calculate the WACC of a firm with EBIT = $50 million, Tax rate... ###### Question: Chapter 16 2. Calculate the WACC of a firm with EBIT =$50 million, Tax rate = 40%, Debt = $100 million, cost of debt = 9%, and unlevered cost of capital - 12%? ## Answers #### Similar Solved Questions 1 answer ##### A. What is the relationship between real interest rate, nominal interest rate and inflation rate? b.... a. What is the relationship between real interest rate, nominal interest rate and inflation rate? b. What are the reasons for very high nominal interest rates in the 1980s? c. Explain ex-ante real rate and ex-post real rate.... 5 answers ##### SerPur?-13 Poines ]Detng Jhead taveling with velocity 5.00 MV/s Sue one-lare tunnel , She then oosenves slow-moving van 170 5 Specdy Sue, driving at 34.0 m/s, enters rad wet: Will there be a-ollision? but can ercelerate cnly a = 00 m/s? beczuse apphes her brekesdetcrmine the distance closcst approsch bctxeen Sue'$ cardctenlte hon far Into thc tunnel &nd nhet time te collision Occurs . Kiyes, the van and enter zero for the Ume_ Oistancd
SerPur? -13 Poines ] Detng Jhead taveling with velocity 5.00 MV/s Sue one-lare tunnel , She then oosenves slow-moving van 170 5 Specdy Sue, driving at 34.0 m/s, enters rad wet: Will there be a-ollision? but can ercelerate cnly a = 00 m/s? beczuse apphes her brekes detcrmine the distance closcst appr...
##### Homework: S 7.1 HWHms0k, ot i0860e 0 0]dO_ EnTne7,1.21-7mhrclarEh Garda Eint LElili tolaritna neatonronttot59
Homework: S 7.1 HW Hms0k, ot i0 860e 0 0]d O_ EnTne 7,1.21-7 mhrclar Eh Garda Eint LElili tolaritna neaton ronttot 59...
##### Solve each triangle. $a=4, c=8, \beta=60^{\circ}$
Solve each triangle. $a=4, c=8, \beta=60^{\circ}$...
##### PrileOly HtlIencn Gurcr-n 056u7d ta culuala Heillete uuuMule enmnee< nnd &nicr Hluticr rulowis nntnt dient > ficncicn unvgy Indalte Dlet styICE Icn Uxtsulteet Arias 10 ,ers Of UXTICTe_ whercr; conallut Dhx [ m =oteno Upmtl IndcreroaLcic ANDc Tnnt tornn hnnIn corrl Itnta nr fumted lert 4c8 ud {5Ito Mcct cunsutant nith molc exnoreha; the hahct popor N1e4% Blc Tring RrNevn ` 0f Irerutn [0 Qltiolis wrrolu37Rl d74in 74Fucum aul ILQNe lurnntd phatect Wdoltu' Lnlaenmtat antennetCnttn (
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##### Find the limit, if it exists. (If an answer does not exist, enter DNE:) lim (e-X + 9 cos(6x))
Find the limit, if it exists. (If an answer does not exist, enter DNE:) lim (e-X + 9 cos(6x))...
##### Problem 3 (10 points) In the figure, a solid sphere of radius a = 1 . Ocm is concentric with a spherical conducting shell of inner radius b = 2 . Oa and outer radius â‚¬ = 2 . Sa. The sphere has a net uniform charge Q1 = + 2.OfC; the shell has a net charge Q2 Q1: What is the magnitude of the electric field at radial distances (a) r = 0 (b) r =a/2, (c)r=a (d) r = 1.5a, and (e) r = 3.5a?
Problem 3 (10 points) In the figure, a solid sphere of radius a = 1 . Ocm is concentric with a spherical conducting shell of inner radius b = 2 . Oa and outer radius â‚¬ = 2 . Sa. The sphere has a net uniform charge Q1 = + 2.OfC; the shell has a net charge Q2 Q1: What is the magnitude of the ele...
##### QUESTION 3 (15marks) There are three possible states of an economy a. Describe the relationship between...
QUESTION 3 (15marks) There are three possible states of an economy a. Describe the relationship between Real GDP and Natural Real GDP in the three states of an economy by using graph. (8 marks b. Describe the relationship between the unemployment rate and the natural unemployment rate in the three s...
##### Solution. HCI M 0.003 of a [OH:] continued and E the Ii (Ipt) Pre-Laboratory
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Question 19 4 pts How should this transformation be done? HO OH A_ 1. Na/NHz; 2. H3O+ B. 1. Hz/Pd/CaCO3/Pb; 2. mCPBA; 3. HaOt C. 1. mCPBA; 2. Hz/Pd/CaCO3/Pb D 1. Hz/Pd/CaCO3/Pb; 2. OsO4 E 1 Na/NHz; 2. OsO4...
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text:
# Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of benzene. The freezing point of benzene is 5.5oC, and the freezing point of the mixture is 2.8oC. What is the molal freezing point constant for benzene?

## Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of benzene. The freezing point of benzene is 5.5oC, and the freezing point of the mixture is 2.8oC. What is the molal freezing point constant for benzene?

May 11, 2016

Your experimental molal freezing point depression constant for benzene is $\text{4.3 °C·kg·mol"^"-1}$.

#### Explanation:

The formula for calculating freezing point depression is

color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = K_fm color(white)(a/a)|)))" "

where

ΔT_"f" is the decrease in freezing point

${K}_{\text{f}}$ is the molal freezing point depression constant

$\text{m}$ is the molality of the solution.

We can rearrange the above formula to get

K_"f" = (ΔT_"f")/m

ΔT_"f" = T_"f"^° -T_"f" = "5.5 °C – 2.8 °C" = "2.7 °C"

${\text{Moles of C"_10"H"_8 = 1.60 color(red)(cancel(color(black)("g C"_10"H"_8))) × ("1 mol C"_10"H"_8)/(128.17 color(red)(cancel(color(black)("g C"_10"H"_8)))) = "0.012 48 mol C"_10"H}}_{8}$

$\text{Molality" = "moles of solute"/"kilograms of solvent" = "0.012 48 mol"/"0.0200 kg" = "0.6242 mol/kg}$

${K}_{\text{f" = (ΔT_"f")/m = "2.7 °C"/"0.6242 mol/kg" = "4.3 °C·kg·mol"^"-1}}$

The actual value of ${K}_{\text{f}}$ for benzene is $\text{5.1 °C·kg·mol"^"-1}$.

Are you sure that the observed freezing point wasn't 2.3 °C?


text:
# Let $M_{\alpha}, \alpha \in \mathbb{C}$ be the subgroup of the Möbius transformation mapping $\alpha$ to itself. Calculate $M_i$

Let $M_{\alpha}, \alpha \in \mathbb{C}$, be the subgroup of $M$ mapping $\alpha$ to itself, that is, the stabilizer of $\alpha$. Given that

$$M_0 = \left \{w = \frac{z}{cz + d}, d \neq 0 \right \}$$

compute the subgroup $M_i, i = \sqrt{-1}$.

$M$ is the Möbius transformation on the extended complex plane.

What I have said is that in order to compute this subgroup, I want some transformation, $L$, which will send map $z \mapsto z + i$. This subgroup can be worked out using the composition

$$M_i = L^{-1} \circ M_0 \circ L$$

So I know that $L = \frac{(z + i)}{c(z + i) + d}$, but I'm not sure how I get $L^{-1}$. I'm guessing from here, it'll be fairly simple to work out the composition.

EDIT: Also, does order matter? I.e is

$$M_i = L^{-1} \circ M_0 \circ L = L \circ M_0 \circ L^{-1}$$

-
If $w = z+i$, then $z=w-i$. This should help you to compute $L^{-1}$. –  mrf Jan 8 '13 at 11:44
@mrf So would I say my $L^{-1}$ maps $z \mapsto z - i$? –  Kaish Jan 8 '13 at 11:46
Yes.${}{}{}{}{}$ –  mrf Jan 8 '13 at 11:49

Yes: you want to move $i$ to $0$, then apply a map that stabilizes $0$, then move $0$ back to $i$ -- in this order. The simplest maps to use are translations by $\pm i$ (suggested by mrf). Thus, $$M_i = \left \{w = i+\frac{z-i}{c(z-i) + d}, d \neq 0 \right \}$$ which you can simplify if desired.


text:
Keywords: energy |  hamiltonian monte carlo |  nuts |  leapfrog |  canonical distribution |  microcanonical distribution |  transition distribution |  marginal energy distribution |  data augmentation |  classical mechanics |  detailed balance |  statistical mechanics |  divergences |  step-size |  non-centered hierarchical model |  hierarchical |  Download Notebook

## Contents

%matplotlib inline
import numpy as np
import scipy as sp
import matplotlib as mpl
import matplotlib.cm as cm
import matplotlib.pyplot as plt
import pandas as pd
pd.set_option('display.width', 500)
pd.set_option('display.max_columns', 100)
pd.set_option('display.notebook_repr_html', True)
import seaborn as sns
sns.set_style("whitegrid")
sns.set_context("poster")
import pymc3 as pm

## Centered parametrization

We’ll set up the modelled in what is called a “Centered” parametrization which tells us how $\theta_i$ is modelled: it is written to be directly dependent as a normal distribution from the hyper-parameters.

J = 8
y = np.array([28,  8, -3,  7, -1,  1, 18, 12])
sigma = np.array([15, 10, 16, 11,  9, 11, 10, 18])

We set up our priors in a Hierarchical model to use this centered parametrization. We can say: the $\theta_j$ is drawn from a Normal hyper-prior distribution with parameters $\mu$ and $\tau$. Once we get a $\theta_j$ then can draw the means from it given the data $\sigma_j$ and one such draw corresponds to our data.

where $j \in {1, \ldots, 8 }$ and the ${ y_{j}, \sigma_{j} }$ are given as data

with pm.Model() as schools1:

mu = pm.Normal('mu', 0, sd=5)
tau = pm.HalfCauchy('tau', beta=5)
theta = pm.Normal('theta', mu=mu, sd=tau, shape=J)
obs = pm.Normal('obs', mu=theta, sd=sigma, observed=y)

with schools1:
trace1 = pm.sample(5000, init=None, njobs=2, tune=1000)

Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 6000/6000 [00:11<00:00, 509.61it/s]
There were 914 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.445177634049, but should be close to 0.8. Try to increase the number of tuning steps.
There were 88 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.71464739587, but should be close to 0.8. Try to increase the number of tuning steps.
The gelman-rubin statistic is larger than 1.05 for some parameters. This indicates slight problems during sampling.
The estimated number of effective samples is smaller than 200 for some parameters.

pm.diagnostics.gelman_rubin(trace1), pm.diagnostics.effective_n(trace1)

({'mu': 1.0214138101971797,
'tau': 1.0890758192035539,
'theta': array([ 1.00109614,  1.00574629,  1.01928611,  1.01160475,  1.03728323,
1.01563653,  1.00039723,  1.00409768])},
{'mu': 30.0,
'tau': 9.0,
'theta': array([ 1826.,  1694.,    37.,   220.,    25.,    36.,  1430.,  2591.])})

The Gelman-Rubin statistic seems fine, but notice how small the effective-n’s are? Something is not quite right. Lets see traceplots.

pm.traceplot(trace1);

Its hard to pick the thetas out but $\tau$ looks not so white-noisy. Lets zoom in:

trace1.varnames

['mu', 'tau_log__', 'theta', 'tau']

pm.traceplot(trace1, varnames=['tau_log__'])

array([[<matplotlib.axes._subplots.AxesSubplot object at 0x1192ec438>,
<matplotlib.axes._subplots.AxesSubplot object at 0x1185e6470>]], dtype=object)

There seems to be some stickiness at lower values in the trace. Zooming in even more helps us see this better:

plt.plot(trace1['tau_log__'], alpha=0.6)
plt.axvline(5000, color="r")
#plt.plot(short_trace['tau_log_'][5000:], alpha=0.6);

<matplotlib.lines.Line2D at 0x1184761d0>

### Tracking divergences

divergent = trace1['diverging']
divergent

array([False, False, False, ..., False, False, False], dtype=bool)

print('Number of Divergent %d' % divergent.nonzero()[0].size)
divperc = divergent.nonzero()[0].size/len(trace1)
print('Percentage of Divergent %.5f' % divperc)

Number of Divergent 1002
Percentage of Divergent 0.20040

def biasplot(trace):
logtau = trace['tau_log__']
mlogtau = [np.mean(logtau[:i]) for i in np.arange(1, len(logtau))]
plt.figure(figsize=(8, 2))
plt.axhline(0.7657852, lw=2.5, color='gray')
plt.plot(mlogtau, lw=2.5)
plt.ylim(0, 2)
plt.xlabel('Iteration')
plt.ylabel('MCMC mean of log(tau)')
plt.title('MCMC estimation of cumsum log(tau)')

biasplot(trace1)

def funnelplot(trace):
logtau = trace['tau_log__']
divergent = trace['diverging']
theta_trace = trace['theta']
theta0 = theta_trace[:, 0]
plt.figure(figsize=(5, 3))
plt.scatter(theta0[divergent == 0], logtau[divergent == 0], s=10, color='r', alpha=0.1)
plt.scatter(theta0[divergent == 1], logtau[divergent == 1], s=10, color='g')
plt.axis([-20, 50, -6, 4])
plt.ylabel('log(tau)')
plt.xlabel('theta[0]')
plt.title('scatter plot between log(tau) and theta[0]')
plt.show()

funnelplot(trace1)

You can also get an idea of your acceptance rate. 65% is decent for NUTS.

np.mean(trace1['mean_tree_accept'])

0.57991251495943541

### Where are the divergences coming from?

Divergences can be a sign of the symplectic integration going off to infinity, or a false positive. False positives occur because instead of waiting for infinity, some heuristics are used. This is typically true of divergences not deep in the funnel, where the curvature of the target distribution is high.

## The effect of step-size

Looking at the docs for the NUTS sampler at https://pymc-devs.github.io/pymc3/api/inference.html#module-pymc3.step_methods.hmc.nuts , we see that we can co-erce a smaller step-size $\epsilon$, and thus an ok symplectic integration from our sampler by increasing the target acceptance rate.

If we do this, then we have geometric ergodicity (we go everywhere!) between the Hamiltonian transitions (ie in the leapfrogs) and the target distribution. This should result in the divergence rate decreasing.

But if for some reason we do not have geometric ergodicity, then divergences will persist. This can happen deep in the funnel, where even drastic decreases in the step size are not able to explore the highly curved geometry.

with schools1:
step = pm.NUTS(target_accept=.85)
trace1_85 = pm.sample(5000, step=step, init=None, njobs=2, tune=1000)
with schools1:
step = pm.NUTS(target_accept=.90)
trace1_90 = pm.sample(5000, step=step, init=None, njobs=2, tune=1000)
with schools1:
step = pm.NUTS(target_accept=.95)
trace1_95 = pm.sample(5000, step=step, init=None, njobs=2, tune=1000)
with schools1:
step = pm.NUTS(target_accept=.99)
trace1_99 = pm.sample(5000, step=step, init=None, njobs=2, tune=1000)

Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 6000/6000 [00:16<00:00, 363.88it/s]
There were 158 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.718139302733, but should be close to 0.85. Try to increase the number of tuning steps.
There were 245 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.7030685137, but should be close to 0.85. Try to increase the number of tuning steps.
The number of effective samples is smaller than 10% for some parameters.
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 6000/6000 [00:21<00:00, 284.50it/s]
There were 76 divergences after tuning. Increase target_accept or reparameterize.
There were 156 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.717631407103, but should be close to 0.9. Try to increase the number of tuning steps.
The number of effective samples is smaller than 10% for some parameters.
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 6000/6000 [00:32<00:00, 186.34it/s]
There were 87 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.906593617309, but should be close to 0.95. Try to increase the number of tuning steps.
There were 167 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.834705258178, but should be close to 0.95. Try to increase the number of tuning steps.
The number of effective samples is smaller than 10% for some parameters.
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 6000/6000 [00:52<00:00, 114.73it/s]
There were 22 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.95631951392, but should be close to 0.99. Try to increase the number of tuning steps.
There were 23 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.959504633927, but should be close to 0.99. Try to increase the number of tuning steps.
The number of effective samples is smaller than 10% for some parameters.

for t in [trace1_85, trace1_90, trace1_95, trace1_99]:
print("Acceptance", np.mean(t['mean_tree_accept']), "Step Size", np.mean(t['step_size']), "Divergence", np.sum(t['diverging']))

Acceptance 0.710603908216 Step Size 0.320560550265 Divergence 403
Acceptance 0.780963159621 Step Size 0.239954261044 Divergence 232
Acceptance 0.870649437743 Step Size 0.174223895957 Divergence 254
Acceptance 0.957912073924 Step Size 0.0766605241909 Divergence 45

for t in [trace1_85, trace1_90, trace1_95, trace1_99]:
biasplot(t)
funnelplot(t)

plt.plot(trace1_99['tau_log__'])

[<matplotlib.lines.Line2D at 0x11bb497f0>]

df99 = pm.trace_to_dataframe(trace1_99)

mu theta__0 theta__1 theta__2 theta__3 theta__4 theta__5 theta__6 theta__7 tau
0 0.476766 5.244940 11.947961 -2.145423 -4.379043 -2.582646 -6.725088 0.876145 1.316700 5.273161
1 4.609032 12.076996 -7.616970 6.384172 5.940715 3.010732 13.464991 3.902260 -2.996996 6.836384
2 3.430340 10.423970 -0.889389 7.728908 8.366453 5.533233 15.195454 5.423708 5.681029 6.246578
3 3.950691 3.769489 -2.915153 4.353163 -3.110655 2.558563 -3.668601 7.396706 3.781000 3.469858
4 1.124394 8.179717 11.401375 13.333014 4.985015 -2.360172 3.140696 7.872551 1.341216 6.758684
df99.corr()

mu theta__0 theta__1 theta__2 theta__3 theta__4 theta__5 theta__6 theta__7 tau
mu 1.000000 0.496149 0.575298 0.605714 0.605968 0.617708 0.600656 0.507750 0.559291 -0.137465
theta__0 0.496149 1.000000 0.354088 0.268464 0.340208 0.229898 0.277801 0.459301 0.356955 0.363872
theta__1 0.575298 0.354088 1.000000 0.329506 0.361995 0.330835 0.359453 0.350181 0.313994 0.061695
theta__2 0.605714 0.268464 0.329506 1.000000 0.376094 0.400555 0.378128 0.265483 0.329640 -0.178865
theta__3 0.605968 0.340208 0.361995 0.376094 1.000000 0.377131 0.393790 0.323031 0.366408 -0.004266
theta__4 0.617708 0.229898 0.330835 0.400555 0.377131 1.000000 0.400101 0.228025 0.324586 -0.277632
theta__5 0.600656 0.277801 0.359453 0.378128 0.393790 0.400101 1.000000 0.251256 0.332723 -0.169251
theta__6 0.507750 0.459301 0.350181 0.265483 0.323031 0.228025 0.251256 1.000000 0.380259 0.379399
theta__7 0.559291 0.356955 0.313994 0.329640 0.366408 0.324586 0.332723 0.380259 1.000000 0.080647
tau -0.137465 0.363872 0.061695 -0.178865 -0.004266 -0.277632 -0.169251 0.379399 0.080647 1.000000
trace1_99['diverging'].shape, df99.shape

((10000,), (10000, 10))

df99si = df99.reset_index()

index mu theta__0 theta__1 theta__2 theta__3 theta__4 theta__5 theta__6 theta__7 tau
0 0 0.476766 5.244940 11.947961 -2.145423 -4.379043 -2.582646 -6.725088 0.876145 1.316700 5.273161
1 1 4.609032 12.076996 -7.616970 6.384172 5.940715 3.010732 13.464991 3.902260 -2.996996 6.836384
2 2 3.430340 10.423970 -0.889389 7.728908 8.366453 5.533233 15.195454 5.423708 5.681029 6.246578
3 3 3.950691 3.769489 -2.915153 4.353163 -3.110655 2.558563 -3.668601 7.396706 3.781000 3.469858
4 4 1.124394 8.179717 11.401375 13.333014 4.985015 -2.360172 3.140696 7.872551 1.341216 6.758684
dvs = sum(trace1_99['diverging']==True)

from pandas.plotting import parallel_coordinates as pc
plt.figure(figsize=(16, 6))
orderedticks = ['index', 'mu', 'tau']+["theta__{}".format(i) for i in range(8)]
div_colors = list(islice(cycle(['g', 'g', 'g', 'g', 'g']), None, int(dvs)))
pc(df99si[trace1_99['diverging']==True][orderedticks], 'index', color=div_colors, alpha=0.1);
from itertools import cycle, islice
undiv_colors = list(islice(cycle(['k', 'k', 'k', 'k', 'k']), None, 200))
pc(df99si[trace1_99['diverging']==False].sample(200)[orderedticks], 'index', color=undiv_colors, alpha=0.05);
plt.gca().legend_.remove()

The divergences decrease, but dont totally go away, showing that we have lost some geometric ergodicity. And as we get to a very small step size we explore the funnel much better, but we are now taking our sampler more into a MH like random walk regime, and our sampler looks very strongly autocorrelated.

We know the fix, it is to move to a

## Non-centered paramerization

We change our model to:

Notice how we have factored the dependency of $\theta$ on $\phi = \mu, \tau$ into a deterministic transformation between the layers, leaving the actively sampled variables uncorrelated.

This does two things for us: it reduces steepness and curvature, making for better stepping. It also reduces the strong change in densities, and makes sampling from the transition distribution easier.

with pm.Model() as schools2:
mu = pm.Normal('mu', mu=0, sd=5)
tau = pm.HalfCauchy('tau', beta=5)
nu = pm.Normal('nu', mu=0, sd=1, shape=J)
theta = pm.Deterministic('theta', mu + tau * nu)
obs = pm.Normal('obs', mu=theta, sd=sigma, observed=y)

with schools2:
trace2 = pm.sample(5000, init=None, njobs=2, tune=1000)

Multiprocess sampling (2 chains in 2 jobs)
NUTS: [nu, tau_log__, mu]
100%|██████████| 6000/6000 [00:08<00:00, 693.86it/s]
There were 8 divergences after tuning. Increase target_accept or reparameterize.

And we reach the true value better as the number of samples increases, decreasing our bias

biasplot(trace2)

How about our divergences? They have decreased too.

divergent = trace2['diverging']
print('Number of Divergent %d' % divergent.nonzero()[0].size)
divperc = divergent.nonzero()[0].size/len(trace2)
print('Percentage of Divergent %.5f' % divperc)

Number of Divergent 8
Percentage of Divergent 0.00160

funnelplot(trace2)

The divergences are infrequent and do not seem to concentrate anywhere, indicative of false positives. Lowering the step size should make them go away.

### A smaller step size

with schools2:
step = pm.NUTS(target_accept=.95)
trace2_95 = pm.sample(5000, step=step, init=None, njobs=2, tune=1000)

Multiprocess sampling (2 chains in 2 jobs)
NUTS: [nu, tau_log__, mu]
100%|██████████| 6000/6000 [00:13<00:00, 434.59it/s]

biasplot(trace2_95)

funnelplot(trace2_95)

Indeed at a smaller step-size our false-positive divergences go away, and the lower curvature in our parametrization ensures geometric ergodicity deep in our funnel

## Path length L

If we choose too small a $L$ we are returning our HMC sampler to a random walk. How long must a leapfrog run explore a level set of the Hamiltonian (ie of the canonical distribution $p(p,q)$ beofre we force an accept-reject step and a momentum resample?

Clearly if we go too long we’ll be coming back to the neighborhood of areas we might have reached in smaller trajectories. NUTS is one approach to adaptively fix this by not letting trajectories turn on themselves.

In the regular HMC sampler, for slightly complex problems, $L=100$ maybe a good place to start. For a fixed step-size $\epsilon$, we can now check the level of autocorrelation. If it is too much, we want a larger $L$.

Now, the problem with a fixed $L$ is that one $L$ does not work everywhere in a distribution. To see this, note that tails correspond to much higher energies. Thus the level-set surfaces are larger, and a fixed length $L$ trajectory only explores a small portion of this set before a momentum resampling takes us off. This is why a dynamic method like NUTS is a better choice.

## Tuning HMC(NUTS)

This requires preliminary runs. In pymc3 some samples are dedicated to this, and an adaptive tuning is carried out according to algorithm 6 in the original NUTS paper: http://www.stat.columbia.edu/~gelman/research/published/nuts.pdf .

But we have seem how to play with step-size within the context of the NUTS sampler, something which we might need to do for tetchy models. Clearly too large an $\epsilon$ can lead to loss of symplecticity. And too small a step-size will get us to a random walk for finite sized $L$. Clearly the adaptive approach in NUTS is a good idea.

In pymc3’s regular HMC sampler a knob step_rand is available which allows us a distribution to sample a step-size from. In the case that we are not adaptively tuning as in NUTS, allowing for this randomness allows for occasionally small values of $\epsilon$ even for large meaned distributions. Note that this should be done at the star of leapfrog, not in-between.

Another place where this is useful is where the exact solution to the Hamiltonian equations (gaussian distribution, harmonic oscillator hamiltonian resulting) has periodicity. If $L\epsilon$ is chosen to be $2\pi$ then our sampler will lack ergodicity. In such a case choose $\epsilon$ from a distribution (or for that matter, $L$).

Finally there are multiple ways to tweak the mass matrix. One might use a variational posterior to obtain a approximate covariance matrix for the target distribution $p(q)$. Or one could use the tuning samples for this purpose. But choosing the mass matrix as the inverse of the covariance matrix of the target is highly recommended, as it will maximally decorrelate parameters of the target distribution.

The covariance matrix also establishes a scale for each parameter. This scale can be used to tweak step size as well. Intuitively the variances are measures of curvature along a particular dimension, and choosing a stepsize in each parameter which accomodates this difference is likely to help symplecticity. I do not believe this optimization is available within pymc3. This does not mean you are out of luck: you could simply redefine the parameters in a scaled form.

If you are combining HMC with other samplers, such as MH for discrete parameters in a gibbs based conditional scheme, then you might prefer smaller $L$ parameters to allow for the other parameters to be updated faster.

## Efficiency of momentum resampling

When we talked about the most Hamiltonian-trajectory momentum resampling, we talked about its efficiency. The point there was that you want the marginal energy distribution to match the transition distribution induced by momentum resampling.

pymc3 gives us some handy stats to calculate this:

def resample_plot(t):
sns.distplot(t['energy']-t['energy'].mean(), label="P(E)")
sns.distplot(np.diff(t['energy']), label = "p(E | q)")
plt.legend();
plt.xlabel("E - <E>")

So let us see this for our original trace

resample_plot(trace1);

Awful. The momentum resamples here will do a very slow job of traversing this distribution. This is indicative of the second issue we were having with this centered model (the first was a large step size for the curvature causing loss of symplectivity): the momentum resampling simply cannot provide enough energy to traverse the large energy changes that occur in this hierarchical model.

Note the caveat with such a plot obtained from our chains: it only tells us about the energies it explored: not the energies it ought to be exploring, as can be seen in the plot with trace1_99 below. Still, a great diagnostic.

resample_plot(trace1_99)

The match is much better for the non-centered version of our model.

resample_plot(trace2)

resample_plot(trace2_95)


text:
I'm trying to write the following block matrix but i have some points i need a help to fix it. first, in the block matrix where i write some headings (k_1, k_2 n-k_1-k_2) there is between them vertical line i don't want to keep it. i need it only in the block matrix inside and for the diagonal matrix \lambda_1 \ddots \lambda_1 and the others i don't want the space before the first row i.e. before the \lambda_1 and \lambda_2.

Hope someone can help me. Here is the code that i write it:

$\begin{blockarray}{cc|c|c} & k_{1}&k_2&n-k_{1}-k_{2}\\ \begin{block}{c(c|c|c)} &\lambda_1 & & \\ k_1 & \hspace{0.75cm} \ddots \hspace{0.75cm} & & \\ & \hspace{2cm} \lambda_1 & & \\\cline{2-4} & &\lambda_2 & \\ k_2 & \Phi &\hspace{0.75cm} \ddots \hspace{0.75cm} & \\ & & \hspace{1.5cm} \lambda_2 & \\ \cline{2-4} & & &\\ n-k_1-k_2 & \Phi & \Phi & \bar{\bar{M}}_3\\ & & &\\ \end{block} \end{blockarray}$

Here is a solution. We have to use 10 columns and booktabs to have horizontal rules that do not get across the parentheses:

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{fourier}
\usepackage{mathtools, blkarray, booktabs}

\begin{document}

$\aboverulesep = 0pt \belowrulesep = 0pt \begin{blockarray}{r*{3}{>{\scriptstyle}c>{\scriptstyle}c@{\enspace}>{\scriptstyle\quad}c}} & & {k_{1}} & & & k_2& & & \mathclap{n-k_{1}-k_{2}} \\% \begin{block}{>{\scriptstyle}r(*{2}{ccc|}cc>{\enspace}c<{\enspace})} &\rule{0pt}{3ex}\:\lambda_1 & & & & & & & & \\ k_1 & & \ddots & & & & & & & \\ & & & \lambda_1 & & & & & & \\ \cmidrule(lr){2-10} & & & &\lambda_2 & & & & & \\ k_2 & & \Phi & & & \ddots \hspace{0.75cm} & & & & \\ & & & & & & \lambda_2 & \\ \cmidrule(lr){2-10} & & & & & & & & & \\ n-k_1-k_2 & & \Phi & & & \Phi & &\enspace &\overline{\overline{M}}_3 & \\[0.5ex] \end{block} \end{blockarray}$

\end{document}

You can use the package nicematrix which has been designed to draw such matrices.

\documentclass{article}
\usepackage{nicematrix}
\NiceMatrixOptions{code-for-first-row = \scriptstyle , code-for-first-col = \scriptstyle}
\begin{document}
$\begin{pNiceArray}{CCC|CCC|C}[first-row,first-col] & & k_1 & & & k_2 & & n-k_1-k_2 \\ & \lambda_1 & & & & & & \hspace*{2cm} \\ k_1 & & \Ddots & & & & \\ & & & \lambda_1 & & & \\ \hline & & & & \lambda_2 & & \\ k_2 & & \Phi & & & \Ddots & &\\ & & & & & & \lambda_2 & \\ \hline & & & & & & & \\ n-k_1-k_2 & & \Phi & & & \Phi & & \overline{\overline{M}}_3 \\ & & & & & & & \\ \end{pNiceArray}$
\end{document}


text:
## week 6 – precalc 11

This week we started a new unit of solving quadratic equations.  last week was mostly a review for the harder lessons we had this week.  in lesson 3.2 we started by solving quadratic equations by using the zero product law.

$x^2-6x+5=0$

$(x-1)(x-5)$

$x= 1, 5$

Verify :

$5^2-6(5)+5=0$

In lesson 3.3 we learned how to solve perfect square trinomials by completing the square.  remembering to put a plus and a minus symbol when adding a radical is an important step when solving and verifying.

$x^2+4x=2$

\$latex x^2+4x-2=0

$(x+2)^2-6=0$

$(x+2)^2=6$

$x=-2 \pm \sqrt{6}$


text:
# Thread: Exemplify why this definition of a limit does not hold

1. ## Exemplify why this definition of a limit does not hold

I've been thinking in circles over this problem so I thought I'd give this forum a try - I hope someone can help.

I've been given a definition:
For all epsilon>0 , there exists a natural number N such that if |Xn-L| < epsilon, then n is greater than or equal to N.

I need to either find an infinite sequence Xn which converges but does not satisfy the conditon above OR
find a sequence Xn and L which do satisfy the condition above but Xn does not converge to L.

I'd be grateful if anybody could talk me through what features such a sequence should have.

Thanks

2. I'll give you the sequence and it's limit from the new definition, you show that it 'converges' to that limit:

$a_n = n \ \forall n \in \mathbb{N}$ and $L = 0$
Hint: Take $N = 1$ for every $\epsilon$

After doing that, see if you can do it for any sequence.

Does this sequence guarantee $n \geq N$ when $|x_n|< \epsilon$ but ${x_n}$ does not converge?

or

Would this show that $|x_n|< \epsilon$ does not guarantee $n \geq 1$ ?

If the latter, does that apply for all possible epsilon greater than zero?

Thanks.

4. Well, is there any time where $n < N$?

5. Ah, okay we have N=1 so n will always be greater than, or equal to it.

I guess my problem is, how does this work for all possible values of epsilon? e.g what if epsilon = 1/2, i.e. for n=1,
$|1|< \frac{1}{2} => n \geq 1$

and how does this converge to 0 when n goes to infinity?

I appreciate your patience but I'm still missing something here.

6. You have (roughly speaking) a proposition of the form $P \Rightarrow Q$ where here, P means $|x_n-L| < \epsilon$ and Q means $n \ge N$
By definition, if P is false then the proposition is true, so you don't care about values of n for which $|x_n-L| \ge \epsilon$.
Now assume P is true. Is Q always true?

7. Ahh I've answered my own question Xn= n does satisfy the condition for L= 0 but does not converge to L which is what I was initially looking for.

8. Correct.