# What is the relationship between Forward measures and LMM?

I am reading the book "Term-structure Models: A Graduate Course" by D. Filipovic. In chapter 7 they define the $T$-Forward measure through the density process $$Z_t:=E_Q[\frac{P(T,T)}{P(0,T)B(T)}|\mathcal{F}_t]=\frac{P(t,T)}{P(0,T)B(t)}$$ or simply on $\mathcal{F}_T$ $$\frac{dQ^T}{dQ}=\frac{1}{P(0,T)B(T)}\tag{1}$$ This is called the $T$-forward measure. However, introducing in chapter 11 the LIBOR Market Model (LMM), we define equivalent measures through $$\frac{dQ^{T_{M-1}}}{dQ^{T_M}}=\mathcal{E}_{T_{M-1}}(\sigma_{T_{M-1},T_M}\bullet W^{T_M}) \tag{2}$$ where $\sigma_{T_{M-1},T_M}(t)=\frac{\delta L(t,T_{M-1})}{\delta L(t,T_{M-1)}+1}\lambda(t,T_{M-1})$ which is a bounded and progressive process (hence $(2)$ really defines a equivalent measure). Continue by backward induction to obtain a whole finite family of such measures. They call any such $Q^{T_m}$ a forward measure. I do not see why the call them forward measures, since the radon nikodym derivative in $(2)$ is obviously different from the one in $(1)$. So why do the call this measure forward measure?

-

## 3 Answers

Another definition of the $T_{i+1}$-forward measure: Under a $T_{i+1}$-forward measure the forward (LIBOR) rate with (natural) payment in $T_{i+1}$ is a martingale. Since there is a famility of forward (LIBOR) rates, there is a familiy of such measures.

I do not have the book at hand, but as far as I understand the situation is as follows.

${Q}_{T_M}$ is the $T$ forward measure for $T=T_{M}$. This is the induction start. Assumed that ${Q}_{T_M}$ is the $T=T_{M}$-forward measure, then equation (2) will indeed give that ${Q}_{T_{M-1}}$ is the $T$ forward measure for $T=T_{M-1}$. Without this initial assumption (which is given at the start of the Section, afaik) it would indeed not make sense to call that family $T$-forward measures.

What is the relation between T-forward measure and LMM

With respect to the question in the title I would take a didactically different approach than the author (and actually, the equation (2) is used in the reversed direction $M-1 \rightarrow M$ in the following): If you look at a single forward rate $L_{T_{i}}$ in the LMM, then its SDE is a Black model. Looking at this is enough if you just consider simple options in that rate, like a caplet. This SDE becomes very simple under ${Q}_{T_{i+1}}$ - the corresponding forward measure, since the drift is zero under ${Q}_{T_{i+1}}$. Hence we know the drift of each single rate of the LMM, when we consider each forward rate under its respective forward measure. However, to have a consistent model, we need to write down the SDE-system under one measure. For example, the terminal measure ${Q}_{T_{M}}$, that is the forward measure of the last rate. Now equation (2) provides the step vise change of measure moving from ${Q}_{T_{i+1}}$ towards ${Q}_{T_{M}}$ getting all SDEs under one single measure.

-

Thanks for your answer. Just to be sure I understand your argument: If we assume that $Q^{T_M}$ is the $T_M$ forward measure, and the radon nikodym derivative of $dQ^{T_{M-1}}/Q^{T_M}$ is given as a stochastic exponential, hence we call this again a forward measure? I think this is indeed the motivation of the author. However, for me it seems from a mathematical viewpoint a little bit ambiguous. – hulik Feb 5 at 9:42
Yes. If if you have that assumption as an induction start, then (2) will give you that you have a family of forward measures. I update my answer a bit.... – Christian Fries Feb 5 at 13:50

If you look in Chapter 7.1 where you find equation (1), you will see just below that:

$$\frac{d\mathbb{Q}^T}{d\mathbb{Q}} \mid_\mathcal{F_t}=\mathcal{E}_t(v(\cdot,T) \bullet W^*)$$

where $W^*$ is a $\mathbb{Q}$-Brownian motion.

Besides, you'll notice on the book that your equation (2) is described as a way to induce the probability measure $\mathbb{Q}_{T_M} \sim \mathbb{Q}_{T_{M-1}}$ on $\mathcal{F}_{T_{M-1}}$. I think the "$\mid_{\mathcal{F}_{T_{M-1}}}$" is implied here.

So there you can see your relationship and why it is called a forward measure.

-

1
Thank you for you answer. I know this particular form of the density process as a stochastic exponential. However this is more an assumption on HJM framework. But I do not see how this answer the question. Maybe I do not understand you conclusion. If you say that a Forward measure has a density process of a particular form, i.e. a stochastic exponential, then I do not agree. They define the forward measure as $\frac{dQ^T}{dQ}=\frac{1}{P(0,T)B(T)}$. Otherwise assuming the filtration is generated by Brownian Motion, every martingale would introduce a forward measure – hulik Feb 3 at 12:20
@hulik I thought you were saying that the two formulas didn't look the same at all in your question, so I tried to show that actually they look very much the same. – SRKX♦ Feb 3 at 23:16
thanks a lot for your comment. So you would define a forward measure as a measure, which has a density process of a particular form? Sorry for bothering you and thanks for your patience! – hulik Feb 4 at 8:58

A Forward measure is simply a measure such that the $T$-discounted forward bond starting at time $t$ is a martingale. This allows you to work with reference underlying $S(t) = P(t,T_1)/P(t,T_2)$ with $T_1$ and $T_2$ fixed.

Flipovic says that $\frac{dQ^T}{dQ} = \frac{1}{P(0,T)B(T)}$ is "called" a forward measure and then proceeds to show that the discounted $t$-forward bond is a martingale under that measure. In my opinion it would have been better, in order to avoid confusion, to define the forward measure as the one that makes the discounted forward bond to be a martingale.

-

Thanks for your comment! It seems that it is didactically probably not the best way how the author introduced forward measure. – hulik Feb 6 at 8:01
By $T$-discounted you mean just dividing by $B(T)$. However, this is not enough. You also have to divide by $P(0,T)$ to obtain martingales. Or what exactly do you mean by "A Forward measure is simply a measure such that the T-discounted forward bond starting at time t is a martingale."? Is it: $P(t,T)/B(T)$ should be a martingale under this measure? – hulik Feb 6 at 8:41
I would stick to the other definition: The forward measure is the measure under which the forward is a martingle. The advantage is that this definition is also valid in a more general context (like multi-curve / OIS-discounting). – Christian Fries Feb 6 at 10:27
@hulk: I believe the "$T$-discounted" in this answer is misleading. The forward bond (without disocunting!) is a martingale. This is the case if you choose $P(t,T_{2})$ as numeraire (not B(t)!). – Christian Fries Feb 6 at 10:29
@ChristianFries You are right about the numeraire. That was a mistake in my previous comment. By forward rate, do you mean instantaneous forward rate, i.e. $f(t,T)$? If so, why should this be a martingale under $Q^T$? For that you need further assumption on $\sigma(\cdot,T)$ – hulik Feb 6 at 12:22
show 4 more comments
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 52, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9609942436218262, "perplexity_flag": "head"}