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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

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

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

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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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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    Rasmus: The OP wants to show that the variance exists, not to compute its value (as the OP said twice, once in the question, the other in the comments).2012-01-23
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    @DidierPiau: I was hoping that my second paragraph takes care of this.2012-01-23
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    Despite your *In particular* (which directs the reader to the independence property, as if it were crucial), independence is irrelevant here. That is, unless one wants to compute the value of the variance, which is not what the OP asks for...2012-01-23
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    Unless $X$ and $Y$ are bounded it is not true, in general, that $XY$ has second moment if $X$ and $Y$ have. This follows from independence though. You choose to use boundedness instead of independence in your comments above, which is as irrelevant as independence (one needs one of them to get to the conclusion).2012-01-23
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    True, but how do you know that X and Y have finite second moments? Another argument is needed for this step, which might be the boundedness you could have used in the first place. But hey, no big deal, to have different solutions is fine...2012-01-23
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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