# Cauchy and $\chi^{2}$ dist

I have been working with this question and can't really figure it out!

Let $Y$ be Cauchy distributed and let $X$ be $\chi^{2}$ distrubted with 1 degree of freedom

Show that $E(XY^{2}) = 1$

I have tried to show that X and $Y^{2}$ is independent in order to find the density function for the vector $(X,Y^{2})$ but with no luck!

The function $p(x,y) = \dfrac{e^{-\frac{x}{2}(1+y^{2})}}{2\pi}$ for $x>0, y\in \mathbb{R}$

is a densitiy function on $(0,\infty )$x$\mathbb{R}$

Let $(X,Y)$ be a continuous random variable with densitiy p

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Unless you know the joint distribution of $X$ and $Y$, you cannot compute $E[XY^2]$. You say that you have been trying to show that $X$ and $Y^2$ are independent. There is no way of doing this; unless you have been given some information from which it is possible to deduce that $X$ and $Y$ are independent and you have not shared that with us. If $X$ and $Y$ are independent, $E[XY^2] = E[X]E[Y^2]$ and $E[Y^2]$ does not exist, so the result you are trying to prove is false. What additional information do you have that you are not telling us? – Dilip Sarwate Jan 20 '12 at 17:11
Given the joint density function, you just need to integrate. $E(XY^2)=\int xy^2p(x,y)dxdy$. – Aaron Jan 20 '12 at 18:52
@Aaron Is it possible for you to do the exact calculations? – user11775 Jan 20 '12 at 20:53
>Is it possible for you to do the exact calculations? How about you try and see if you can work out the value of $$\int_0^\infty \frac{1}{2\pi} x e^{-ax}\mathrm dx$$ and save Aaron the trouble of doing the calculation of the inner integral in $$\int \left[\int xy^2p(x,y)\mathrm dx \right] \mathrm dy$$ – Dilip Sarwate Jan 20 '12 at 21:45
Yes, I have tried. But I ended up with $\infty$ which is far from 1. – user11775 Jan 20 '12 at 22:11

Here is one case where $E[XY^2] = 1$.

Suppose that $W$ and $Z$ are independent standard normal random variables. Then $W^2 = X$ is a $\chi^2$ random variable with one degree of freedom while $\frac{Z}{W} = Y$ is a Cauchy random variable. $X$ and $Y$ are not independent random variables, but $$E[XY^2] = E\left[W^2\frac{Z^2}{W^2}\right] = E[Z^2] = 1.$$

As I asked in my comment, what information have you been given that you are not sharing with us?

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(+1) Your example is prescient. An elementary calculation shows that the density now provided in the OP's question matches exactly with the example you've provided. – cardinal Jan 20 '12 at 18:22
It is worth noting that the technique of describing your random variables as coming from other random variables which are easier to understand is a form of coupling, which is definitely a technique worth knowing. – Aaron Jan 20 '12 at 22:00

Given the joint distribution of a number of variables $X_1, \ldots, X_n$ as a function $p(x_1,\ldots, x_n)$, and given some function $g$ of $n$-variables, we have that

$$E[g(X_1,\ldots, X_n)]=\int_{\mathbb R^n} g(x_1,\ldots, x_n)p(x_1,\ldots, x_n)dx_1\cdots dx_n.$$

In our particular case, we have

$$E[XY^2]=\frac{1}{2\pi}\int_{\mathbb R} \int_0^{\infty} xy^2e^{-x(1+y^2)/2)}dx dy$$

Integration by parts shows that $\int_0^{\infty} xe^{-\alpha x}dx=\frac{1}{\alpha^2}$, and so we can simplify the inner integral to get

$$E[XY^2]=\frac{1}{2\pi}\int_{\mathbb R} \frac{4y^2}{(1+y^2)^2}dy$$

One way to do this integral is by using the trig substitution $y=\tan u$, which simplifies the integral to $\frac{4}{2\pi}\int_{-\pi/2}^{\pi/2} \sin^2 u\; du=1$.

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Thank you! That helped me alot. – user11775 Jan 20 '12 at 22:15