# Variance of "modified" inverse chi-square distribution

Let $\chi^2(n)$ be a chi-squared random variable with $n$ degrees of freedom. It is known that $\frac{1}{\chi^2(n)}$ has an "inverse chi-square distribution" with mean $\frac{1}{n-2}$ (if $n$ is big enough).

However, what if we rather are interested in the distribution of $$\frac{1}{\chi^2(n)+k},$$ where $k>0$ is a constant? How do the mean and variance change in this case?

I'm kind of thinking it should be close to $\frac{1}{n-2+k}$.

• Can we safely assume $k>0$ or must we deal with the possibility that the denominator can have nonzero density at $0$? With $k>0$, I believe that the expectation will be substantially lower than you suggest. Oct 20, 2015 at 1:33
• Yes, $k>0$, thanks for the observation. Why do you think it's lower? Oct 20, 2015 at 2:02
• Because I used Jensen's inequality to derive the bounds Xi'an presents below (though not so neatly in the case of the upper bound). Oct 20, 2015 at 9:21

By Jensen's inequality, since the function $f(x)=1/\{x+k\}$ is convex, $$\mathbb{E}\left[\frac{1}{\chi^2(n)+k}\right]\ge\frac{1}{\mathbb{E}[‌​\chi^2(n)]+k}=\frac{1}{n+k}$$ and since the function $f(x)=1/\{x^{-1}+k\}$ is concave, $$\mathbb{E}\left[\frac{1}{\{1/\chi^{-2}(n)\}+k}\right]\le\frac{1}{\{1/\mathbb{E}[‌​\chi^{-2}(n)]\}+k}=\frac{1}{n-2+k}$$Hence, $$\frac{1}{n+k}\le\mathbb{E}\left[\frac{1}{\chi^2(n)+k}\right]\le\frac{1}{n-2+k}$$ The following graph shows the position of the expectation wrt both bounds for four values of $n$ and a range of values of $k$: 