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I found this problem interesting but I don't see clearly how to solve it. Any hint or ideas on how to get this one started?

BlueRedem1
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  • See https://math.stackexchange.com/q/659254/321264. If you use conditional probability, you will also get conditional expectation. Not sure what you are looking for exactly. – StubbornAtom May 25 '20 at 21:25

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For example like that:

Take any $t \in (0,1)$. Then: $$\mathbb P(XY \le t) = \mathbb E [ \mathbb P(XY \le t|Y)] = \mathbb E[\mathbb P(Xy \le t)|_{y=Y}] = \mathbb E((\frac{t}{y}1_{\{y \ge t\}}+1_{\{y<t\}})|_{y=Y}) = $$ $$= \mathbb E[\frac{t}{Y}1_{\{Y \ge t\}} + 1_{\{Y < t\}}] = t \int_{t}^1\frac{dy}{y}+\int_0^tdy=t-t\ln(t) $$

Which gives us $F_Z(t) = \begin{cases} 0 & t \le 0 \\ t-t\ln(t) & t \in(0,1) \\ 1 & t \ge 1\end{cases}$

Where $1_A$ means the indicator function and I used the fact that $\mathbb E[f(X,Y)|Y] = \mathbb E[f(X,y)]|_{y = Y}$ when $X,Y$ are independent and $f$ is such that $\mathbb E[|f(X,Y)|] < \infty$.

Presage
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  • Why we can obtain the first equality? Is the only thing I Don't see very clear – BlueRedem1 May 25 '20 at 23:53
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    $\mathbb P(A) = \mathbb E[1_A] = \mathbb E[\mathbb E[1_A | \mathcal G]] = \mathbb E[\mathbb P(A|\mathcal G)]$. It is just by tower property of conditional expectation, with $A={XY \le t}$ and $\mathcal G = \sigma(Y)$ – Presage May 25 '20 at 23:55