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Conditioning on other random variables is something I don't have a great grasp of. I came across example 3.2 in 'Random Processes for Engineers' by professor hajek and this concept seems to be what prevents me from finishing off the example. In the problem we have Y=XU where X and U are independent random variables, U~Unif[0,1] and X has the Raleigh distribution $fx(x) = \frac{x}{\sigma^2}e^\frac{-x^2}{{2}{\sigma^2}}$ for x>=0 else 0. Part of this problem involves finding E[Y|X] where first step is finding the joint distribution $fxy = f_X(x)*f_{Y|X}(y|x)$. In this type of problem I don't understand how we find the conditional density. I believe we say that $P(Y\le y)$ becomes $P(XU\le y)$ which is $P(U\le \frac{y}{X}|X=x)$ but how do we get from this step to the final step of resolving $f_{Y|X}(y|x)$?

tommik
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Bolpincal
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  • It would help to know whether or not $X$ and $U$ are independent. After that, you might do a 2-d PDF transformation $(x,u)\rightarrow(x,y)$, or you could compute $P[Y\leq y|X=x] = P[XU\leq y|X=x]=...$ – Michael Dec 06 '21 at 16:39
  • Sorry, I forgot to mention that X and U are independent. I've updated the original post accordingly. Yea, I understand that we can compute P[Y≤y|X=x]=P[XU≤y|X=x] via law of total probability but that is the calculation that I struggle to understand. I believe that P[XU≤y|X=x] = P(U≤yX|X=x) but I'm not sure how to resolve from there. – Bolpincal Dec 06 '21 at 18:06
  • The conditional CDF is $F_{Y|X=x}(y|x)$, you can differentiate to get conditional PDF. This is not a law of total probababilty problem. Fix $x>0, y>0$. Conditional CDF is calculated via substitution $$P[Y\leq y|X=x] = P[XU\leq y|X=x] = P[xU\leq y|X=x] = ...$$ where I have simply substituted $x$ for $X$ since we are given $X=x$. Notice that the event ${xU\leq y}$ has nothing to do with $X$ and so by independence... – Michael Dec 06 '21 at 18:31

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The main step is to find the joint density $f_{XY}(x,y)$. This can be done using the jacobian method

To do that, set

$$\begin{cases} y=xu\\ z=x \end{cases}\rightarrow\begin{cases} x=z\\ u=y/z \end{cases} $$

the jacobian is $|J|=1/z$ thus

$$f_{YZ}(y,z)=\frac{1}{\sigma^2}e^{-z^2/(2\sigma^2)}\cdot\mathbb{1}_{0<y<z<\infty}$$

Thus by definition

$$f_{Y|X}(y|x)=\frac{1}{x} \cdot\mathbb{1}_{0<y<x}$$

That is a uniform density with expectation

$$\mathbb{E}[Y|X=x]=x/2$$

tommik
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