Calculation of inverse of chi-square's expectation I don't know where to begin to calculate the expectation value of the random variable $1/V$, where $V$ is a random variable with chi-square distribution $\chi^2(\nu)$.
Could somebody help me?
 A: I try to help this question. 
A random variable $X$ with inverse chi-square distribution has p.d.f
$$\frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} x^{-\frac{v}{2}-1} exp\Big(-\frac{1}{2x}\Big), x>0$$
Since it is a proper distribution, we have 
$$\int_0^{\infty}  x^{-\frac{v}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx=1 \rightarrow 2^{\frac{v}{2}} \Gamma(\frac{v}{2})$$
Therefore, the expectation for inverse chi-square distribution is:
$$E(X) = \int_0^{\infty}  x \cdot \frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})}x^{-\frac{v}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx = \frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} \int_0^{\infty} x^{-\frac{v-2}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx$$
$$\frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} \cdot 2^{\frac{v-2}{2}}\cdot \Gamma\Big(\frac{v-2}{2}\Big) = \frac{1}{2 \cdot \frac{v-2}{2}} = \frac{1}{v-2}$$
$$E(X^2) = \int_0^{\infty}  x^2 \cdot \frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})}x^{-\frac{v}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx = \frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} \int_0^{\infty} x^{-\frac{v-4}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx$$
$$\frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} \cdot 2^{\frac{v-4}{2}}\cdot \Gamma\Big(\frac{v-4}{2}\Big) = \frac{1}{2^2 \cdot \frac{v-2}{2} \cdot \frac{v-4}{2}} = \frac{1}{(v-2)\cdot (v-4)}$$
Therefore, the variance of inverse chi-square distribution is:
$$V(X) = E(X^2) - [E(X)]^2 = \frac{1}{(v-2)\cdot (v-4)} - \Big(\frac{1}{v-2}\Big)^2 = \frac{}{}\frac{2}{(v-2)^2 (v-4)}$$
Hope this helps!
A: Let $V=z_1^2+...+z_\nu^2$ with the $z_i\sim N(0,1)$. By integration by parts writing the density of $z_i$, for a differentiable function
$$E[z_i f(z_1,...,z_\nu)] = E[\frac{\partial}{\partial z_i} f(z_1,...,z_\nu)].$$
Applying this to $f=1/V$ gives for each $i$
$$
E[\frac{z_i^2}{V}] = E[\frac{1}{V}] - E[\frac{2z_i^2}{V^2}].
$$
Summing over $i=1,...,\nu$ gives $1 = \nu E[1/V] - 2E[1/V]$ so that $E[1/V] =1/(\nu-2)$.

For the second moment of $1/V$ (or variance of $1/V$), similarly
$$
E[z_i^2/V^2]
= E[1/V^2] - 4 E[z_i^2/V^3]
$$
and summing over $i=1,...,\nu$ gives
$E[1/V] = \nu E[1/V^2] - 4[V^2]$ hence
$E[1/V^2]=\frac{1}{(n-2)(n-4)}$.
A: The pdf of a chi-square distribution is $$\frac{1}{2^{\nu/2} \Gamma(\nu/2)} x^{\nu/2-1} e^{-x/2}.$$
So you want to calculate 
$$\int_0^{\infty} \frac{1}{x} \frac{1}{2^{\nu/2} \Gamma(\nu/2)} x^{\nu/2-1} e^{-x/2} dx = \int_0^{\infty} \frac{1}{2^{\nu/2} \Gamma(\nu/2)} x^{\nu/2-2} e^{-x/2} dx.$$
Rewrite the integrand so that it is the pdf of a $\chi^2(\nu-2)$ random variable, which will then integrate to 1.  The leftover constant factor will be the expected value you're looking for.
If you want a more detailed hint, just ask.  
