P.d.f. of $XY$, where $X, Y$ are independent uniformly distributed over $[0,1]$

Question

I tried to change the variables:

Let $U=XY$ and $V=Y$; so then the Jacobian is $1/v$.

So joint pdf $g(u,v) = f(x,y)\cdot (1/v) = 1/v$

Would you then integrate over $v$ from $0$ to $1$ to get the p.d.f., of just $U$?

However this gives infinity.

I accidentally moused over Leave Open and did not intend to make that vote. This is, in fact, a duplicate question. — ncmathsadist, May 15 '14 at 15:05

score 0 · Accepted Answer · answered May 15 '14 at 14:36

0

Your joint pdf is wrong. You should get

$$g(u,v) = I(0<v<1) I(0<u<v)/v,$$

where $I$ is the indicator function. Then just do

$$\int_u^1 \frac{1}{v} dv = [\log v]_u^1= -\log u,$$

which for $0<u<1$ is the density of $XY$.

answered May 15 '14 at 14:36

JPi

Why do you integrate from u to 1? Is it the same as from 0 to u? – timni May 15 '14 at 14:40
Because for $v<u$ the indicator function and hence the density function equals zero. Note that $V$ cannot be less than $U$ by construction. – JPi May 15 '14 at 14:41
1

I understand; thank you! – timni May 15 '14 at 14:52

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