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?
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3 Answers
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1 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.
- 1$\begingroup$ The expectation of the inverse chi-square is $\frac{1}{\nu-2}$. How can I reach it result from your derivation? $\endgroup$garciparedes– garciparedes2017-09-25 14:02:16 +00:00Commented Sep 25, 2017 at 14:02
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1 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!
- 2$\begingroup$ Excellent solution! $\endgroup$zaira– zaira2020-10-03 10:56:41 +00:00Commented Oct 3, 2020 at 10:56
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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 $E[1/V^2]=\frac{1}{(n-2)(n-4)}$.
