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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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    @user: I expanded my answer to account for your request.2012-01-22

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Since $0\lt X\lt2\lt Y\lt4$ almost surely and $V=XY$, $0\lt V\lt 8$ almost surely. Being almost surely bounded, $V$ has moments of all orders, in particular $V$ is square integrable. This means that the variance of $V$ (exists and) is finite.

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Observe that $X$ and $Y$ are independent since their joint density has product form.

In particular, given that $X$ and $Y$ have second moments, the second moment of $XY$ exists and is equal to $E(X^2)E(Y^2)$.

Using independence, we get $ \mathrm{Var}(XY)=E((XY)^2)-(E(XY))^2=E(X^2)E(Y^2)-(E(X))^2 (E(Y))^2. $ Now you only have to compute those four expectation values and plug them in.

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    @Didier Piau: I agree that another argument is needed to show that $X$ and $Y$ have second moments. For some reason I thought that that was clear to the OP.2012-01-23