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!   
additional information:
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
 A: 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?
A: 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 $\frac{4}{2\pi}\int_{-\pi/2}^{\pi/2} \sin^2 u\; du=1$.  
