# Thread:

1. ## Independent Random Variables

Question 1)

Note: iid=idependent and identically dsitributed

I don't get the part underlined in red. Why are they independent? How can we prove that, or at least intuitively make sense of it?

Question 2) If X1,X2,...,X6 are independent random varaibles, then my textbook has a theorem saying that g1(X1), g2(X2), ..., g6(X6) are also independent, where the gi's are any function of a single random variable.
But how about f1(X1,X2,...,X5), f2(X6)? If X1,X2,...,X6 are independent, are any function of X1,...,X5 and any function of X6 independent?

Any help is greatly appreciated!

2. Originally Posted by kingwinner
Question 1)

Note: iid=idependent and identically dsitributed

I don't get the part underlined in red. Why are they independent? How can we prove that, or at least intuitively make sense of it?
This is because for iid $\mathcal{N}(0,1)$ random variables $Y_1,\ldots,Y_n$, the random variables $\overline{Y}=\frac{1}{n}\sum_{i=1}^n Y_i$ and $U=\sum_{i=1}^n (Y_i- \overline{Y})^2$ are independent ("the empirical mean and variance of independent gaussian r.v. are independent").
I only rewrote what you said, but the point is that this is very specific to Gaussian random variables, and you probably have it in your course notes.
I don't have a simple intuitive way to think about it (this can be understood geometrically thanks to Cochran's theorem using orthogonal projections of Gaussian random vectors, but this doesn't make it much more intuitive; anyway I can develop if you wish). As for a proof, it requires knowledge about Gaussian random vectors; the idea is that the covariance of $\overline{Y}$ and $U$ is found to be zero, and that for Gaussian vectors this implies the independence; again I can develop if you're really interested, but you may have seen it in your course.

Question 2) If X1,X2,...,X6 are independent random varaibles, then my textbook has a theorem saying that g1(X1), g2(X2), ..., g6(X6) are also independent, where the gi's are any function of a single random variable.
But how about f1(X1,X2,...,X5), f2(X6)? If X1,X2,...,X6 are independent, are any function of X1,...,X5 and any function of X6 independent?
The answer is yes. It is probably in your textbook as well, but I don't know the English name, it is "indépendance par paquets" in French, like "independence by packets" or "packets independence" perhaps?

3. 1) This is my first statistics course in university and I haven't done anything on random vectors yet, so I won't be likely to understand the proof.
But there is a theorem in my textbook saying that for iid normal random variables, the "sample mean" and "sample variance" are independent random variables.
So assuming this fact, how can we prove the independence in the red underlined part? There we have (Ybar)^2 + (Y6)^2

2) Let X1,X2,...,X6 be independent random varaibles.
Then for example X1+(X2)^2 and X6 would be independent, and something like X1+X2+(X3)^4 and (X4)^3 + X5 would also be independent, am I right?

Thanks for helping!

4. Originally Posted by kingwinner
1) This is my first statistics course in university and I haven't done anything on random vectors yet, so I won't be likely to understand the proof.
But there is a theorem in my textbook saying that for iid normal random variables, the "sample mean" and "sample variance" are independent random variables.
So assuming this fact, how can we prove the independence in the red underlined part? There we have (Ybar)^2 + (Y6)^2
This is the right theorem. You get that $U$ and $\overline{Y}$ are independent. On the other hand, both $U$ and $\overline{Y}$ depend only on $Y_1,\ldots,Y_5$, so that they are independent of $Y_6$ (using your second question!).
More precisely, by this argument, the couple $(\overline{Y},U)$ is independent of $Y_6$.
This proves that the three r.v. $\overline{Y}$, $U$ and $Y_6$ are independent.
In particular, $\overline{Y}^2+(Y_6)^2$, as a function of two of them, is independent of the third one, namely $U$. This is again related to your second question.

Note: there's a slight subtlety in the argument. I said the couple $(\overline{Y},U)$ is independent of $Y_6$, because it is not sufficient to say that $X,Y,Z$ are pairwise independent (i.e. $X$ independent of $Y$ and $Z$, and $Y$ independent of $Z$) in order to conclude that $X,Y,Z$ are independent. But if $(X,Y)$ is independent of $Z$ and $X$ is independent of $Y$, then $X,Y,Z$ are independent... If this is your first course in probability, don't worry too much about this anyway.

2) Let X1,X2,...,X6 be independent random varaibles.
Then for example X1+(X2)^2 and X6 would be independent, and something like X1+X2+(X3)^4 and (X4)^3 + X5 would also be independent, am I right?
You are! More generally, if you gather independent random variables in several groups, then the functions of random variables in different groups are independent.

5. So we have:
$\overline{Y}$ and U independent (by theorem)
$\overline{Y}$ and Y6 independent (f(Y1,Y2,...Y5) and g(Y6) are independent)
U and Y6 independent (h(Y1,Y2,...Y5) and k(Y6) are independent)

Does this imply $\overline{Y}$, U, and Y6 are independent?

6. Originally Posted by kingwinner
So we have:
$\overline{Y}$ and U independent (by theorem)
$\overline{Y}$ and Y6 independent (f(Y1,Y2,...Y5) and g(Y6) are independent)
U and Y6 independent (h(Y1,Y2,...Y5) and k(Y6) are independent)

Does this imply $\overline{Y}$, U, and Y6 are independent?
No, confer to my note. This proves that $\overline{Y},U,Y_6$ are pairwise independent, but not that they are independent.

However, the couple $(\overline{Y},U)$ is a function of $Y_1,\ldots,Y_5$, so that it is independent of $Y_6$. In addition, $\overline{Y}$ and $U$ are independent. From there, it follows rigorously that the three r.v. are independent. I show you why: Suppose $X,Y,Z$ are real r.v. such that $(X,Y)$ is independent of $Z$ and $X$ is independent of $Y$. Then for any (measurable) subsets $A,B,C$ of $\mathbb{R}$, $P(X\in A,Y\in B,Z\in C)=P((X,Y)\in A\times B, Z\in C)$ $=P((X,Y)\in A\times B)P(Z\in C)=P(X\in A)P(Y\in B)P(Z\in C)$.

To see why "pairwise independence" and "independence" are not equivalent: consider two independent dices, $X$ is the parity of the first dice (0 if even, 1 if odd), [tex]Y[/Math] is the parity of the second dice, and $Z$ is the parity of the sum of the results. After a thought, you'll understand why $X$ and $Z$ are independent. In the same way, $Y$ and $Z$ are independent. At last, it is straightforward that $X$ and $Y$ are independent. But $X,Y,Z$ can't be independent: $Z$ is a function of $X$ and $Y$.
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