# Probability, variance and integrals

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

$p(x,y)= \frac{3}{2} x^{2} y^{-2}$ when $0<x<2$ and $2<y<4$ or else $p(x,y) = 0$

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

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

$q(z) = \frac{3}{2} z^{2}\cdot$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. – Did Jan 22 '12 at 13:20
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. – Dilip Sarwate Jan 22 '12 at 13:21
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? – user Jan 22 '12 at 13:37
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 (0<V<8 with full probability) hence V has moments of all orders, in particular V is square integrable hence the variance of V exists and is finite. – Did Jan 22 '12 at 14:07
@user: I expanded my answer to account for your request. – Rasmus Jan 22 '12 at 15:10

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). – Did Jan 23 '12 at 2:05
@DidierPiau: I was hoping that my second paragraph takes care of this. – Rasmus Jan 23 '12 at 6:38
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... – Did Jan 23 '12 at 6:52
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). – Rasmus Jan 23 '12 at 11:15
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... – Did Jan 23 '12 at 11:39