Independance of RVs with same distribution

[nexp]

[Solved] Independance of RVs with same distribution

Hey all,

Let's say we have two Gaussian random variables X, Y, each with zero mean and unit variance. Is it correct to say that $$P(X|Y) = P(X)$$?

In other words, suppose that we want to compute the expectation of their product $$\operatorname{E}[XY]$$. Is the following correct? I.e. does their joint distribution factorise?

$$E[XY] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x y\: p(x) p(y)\: dx dy$$
$$= \int_{-\infty}^{\infty}x \:p(x)\: \operatorname{d} x \int_{-\infty}^{\infty} y \: p(y) \: \operatorname{d} y$$
$$= \operatorname{E}[X]\operatorname{E}[Y] \nonumber$$

Many Thanks.

Update

I have now figured out the answer to the above questions. I'll post it here for anyone who is interested.

If X and Y have the same distribution, then we can write $$P(X|X) = 1 \neq P(X)$$.

Now looking again at expectations. From the above, we have that

$$E[XY]=E[X^2]$$
$$=\int_{-\infty}^{\infty}x^2 p(x^2) \: dx$$

similarly giving a negative answer for the expectation of the product.

 

[mathman]

It looks like you are confusing two different things. X and Y are assumed to be normal with the same distribution.

However whether of not they are independent is a completely separate question. If they are independent then your analysis before the update is correct. On the other hand if X=Y, then the comment after the update is valid.

These are the two extreme possibilities (partial dependence in between).

 

[nexp]

Thanks very much. I understand what's going on better now.