Finding the PDF of a product of two r.v.s

Question

Let $X,Y \sim \operatorname{Unif}(0,1)$ be independent of each other. Find the PDF of $V:= XY$.

This is was I did $$\mathbb{P}(V \leq v) = \mathbb{P}(YX \leq v) = \mathbb{P}\left(Y\leq \frac v X \right) \\ = \int_0^{v/x} 1 \, dy = \frac v x.$$ I'm not sure if this is right since I can't think of the restrictions and also there involves an $x$ term in the answer...

Siong Thye Goh · Accepted Answer · 2018-03-23T04:20:30.207

1

Let $v \in (0,1)$,

\begin{align} P( Y \le \frac{v}{X})&= \int_0^1 P(Y \le \frac{v}{x})\, dx \\ &= \int_0^1 \min\left( 1, \frac{v}{x}\right)\, dx \\ &= \int_0^v \, dx + \int_v^1 \frac{v}{x}\, dx \\ &= v-v\log v \\ \end{align}

Differentiate it to obtain the pdf.

Edit:

$$f_V(v) = \begin{cases} -\log v &, v \in (0,1) \\ 0 & \text{, otherwise.} \end{cases}$$

edited Mar 23 '18 at 04:20

answered Mar 23 '18 at 03:57

Siong Thye Goh

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How did you get the first equality (the upper bound being $1$)? – Mr. Bromwich I Mar 23 '18 at 03:58
I consider every single value that $x$ can possibly take. there is actually an invisble $f_X(x)$ inside the integral. – Siong Thye Goh Mar 23 '18 at 04:00
Oh is it actually $\int_0^1 \mathbb{P}(Y\leq v/X | X=x) f_X(x) dx$? – Mr. Bromwich I Mar 23 '18 at 04:01
yes, you get it. – Siong Thye Goh Mar 23 '18 at 04:03
Thank you, and one last Q: why does $P(Y\leq v/x) = min(1,v/x)$? I thought you'd need to do another integral using the joint distribution – Mr. Bromwich I Mar 23 '18 at 04:03
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$Y$ is a uniform distribution, we know the CDF of the uniform distribution if $y$ is positive is $F_Y(y) = \min(y,1)$. – Siong Thye Goh Mar 23 '18 at 04:04
In the first step actually, we looked at $P(Y\leq v/x) = F_Y(v/x)$.
This looks like I'm finding the probability cumulation of $Y$ aren't I? – Mr. Bromwich I Mar 23 '18 at 04:12
Yup, but $X$ is a random variable there. – Siong Thye Goh Mar 23 '18 at 04:14
Hmm, so in the end the answer is $F_Y(v/x) = v-v log(v)$ and diff. wrt. $v$, i get $f_Y(v/x) \frac{1}{x} = -log(v)$ which seems really confusing – Mr. Bromwich I Mar 23 '18 at 04:17
$-\log v$ is the correct solution. – Siong Thye Goh Mar 23 '18 at 04:19
What should I be differentiating with respect by? Since when I differentiate wrt $v$, I get the extra factor of $x$ giving $-xlog(v)$ – Mr. Bromwich I Mar 23 '18 at 04:21
Let us continue this discussion in chat. – Siong Thye Goh Mar 23 '18 at 04:22

score 0 · Answer 2 · answered Mar 23 '18 at 04:28

\begin{align} & \Pr(XY\le v) = \operatorname E(\Pr(XY\le v \mid X)) = \operatorname E\left( \Pr \left( Y \le \frac v X \,\middle|\, X \right) \right) \\[10pt] = {} & \operatorname E \left.\begin{cases} \dfrac v X & \text{if } \dfrac v X <1 \\[12pt] 1 & \text{otherwise} \end{cases} \right\} = \int_v^1 \frac v x \cdot 1 \, dx + \int_0^v 1 \,dx = v(1-\log_e v). \end{align}

Finding the PDF of a product of two r.v.s

2 Answers2