6
$\begingroup$

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?

$\endgroup$
0

3 Answers 3

12
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ The expectation of the inverse chi-square is $\frac{1}{\nu-2}$. How can I reach it result from your derivation? $\endgroup$ Commented Sep 25, 2017 at 14:02
5
$\begingroup$

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!

$\endgroup$
1
  • 2
    $\begingroup$ Excellent solution! $\endgroup$
    – reyna
    Commented Oct 3, 2020 at 10:56
2
$\begingroup$

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 $E[1/V^2]=\frac{1}{(n-2)(n-4)}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .