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?
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?
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.
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!