How to find $P(W \le w)$ for $W=XY$ knowing the following

Question

Given the joint PDF for $(X,Y)$

$$ p(x, y)=\left\{\begin{array}{cl} \frac{6}{5}\left(x+y^{2}\right) & \text { for } 0<x<1,0<y<1 \\ 0 & \text { ellers } \end{array}\right. $$

I found marginal distribution of $X$ and of $Y$. I found out that $X \not \perp Y$.

But I do not know how use this information to find CDF for $W=XY$.

Kavi Rama Murthy · Answer 1 · 2020-03-12T23:21:21.763

1

$P(W \leq w)=E(P(X\leq \frac w Y|Y))=\int_0^{1} \int_0^{\min ({\frac w y ,1)\}}} p(x,y) dx dy$. This can be written as $\int_w^{1} \int_0^{\frac w y } p(x,y) dx dy+\int_0^{w} \int_0^{1} p(x,y) dx dy$. Now you can carry out the integration.

edited Mar 12 '20 at 23:21

answered Mar 12 '20 at 23:17

Kavi Rama Murthy

311,013

Integration is the easy part. What you did is the hard part. – Sorry Mar 12 '20 at 23:20
And what is $EP$ ? a mistake ? – Sorry Mar 12 '20 at 23:20
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@Xenusi I took the conditional probability of $W \leq w$ given $Y$. That is a random variable and I took expectation of that variable. – Kavi Rama Murthy Mar 12 '20 at 23:22
I have never seen this before. – Sorry Mar 12 '20 at 23:23
@Xenusi I don't think the is any other procedure for finding the distribution of $XY$. – Kavi Rama Murthy Mar 12 '20 at 23:24
Which theorems and definitions have you been using? – Sorry Mar 12 '20 at 23:25
@Xenusi For any event $A$ we always have $P(A)=E(P(A|Y))$. A more general version of this says $E(E(Z|\mathcal G))=EZ$ where $E(E(Z|\mathcal G))$ denotes the conditional expectation of $X$ given the sigma field $\mathcal G$. If you are not familar with conditional expectations you can take this intuitively. – Kavi Rama Murthy Mar 12 '20 at 23:36

How to find $P(W \le w)$ for $W=XY$ knowing the following

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