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Let (X,Y) be a continuous random vector with probability density p where

$p(x,y)= \frac{3}{2} x^{2} y^{-2}$ when $0 and $2 or else $p(x,y) = 0$

I know that X has the density $p_{1}(x) = \frac{3}{8} x^{2}$ when $0 (else $p_{1}(x)=0$)

and Y has the density $p_{2}(y) = 4y^{-2}$ when $2 or else $p_{2}(y)=0$

I now define V=XY and my question is how can I show that V has variance?

I tried to find the probability density function for V and ended up with:

$q(z) = 0$ for $z\leq 0$

$q(z) = \frac{3}{2} z^{2}\cdot$log(z) for $0

$q(z) = \frac{3}{2} z^{2}\cdot$log(2) for 2$\leq$z

and then i want to calculate $\int_\infty^\infty z^{2}q(z) dz$

in order to show that $\int_\infty^\infty z^{2}q(z) dz$

But I end up with $\infty$ when I try to calculate $\int_\infty^\infty z^{2}q(z) dz$

What is wrong?

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    The random variables X and Y are bounded, hence every moment of V=XY is finite.2012-01-22
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    You can compute $E[V] = E[XY]$ and $E[V^2] = E[X^2Y^2]$ without computing the probability density function of $Z$. LOTUS gives us the wonderful result $$E[g(X,Y)] = \int\int g(x,y)f_{X,Y}(x,y) dx dy$$ that can be used in this instance.2012-01-22
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    Ok. All of the answers are about calculating the variance. I want to show that V has variance. So the question is now: If I can calculate the variance, can I then deduce that V actually has variance?2012-01-22
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    *All of the answers are about calculating the variance*... No. My comment explains why the variance of V exists. To repeat: V is almost surely bounded (02012-01-22
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    @user: I expanded my answer to account for your request.2012-01-22

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