3
$\begingroup$

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 $\frac{1}{n-2+k}$.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Glen_b Oct 20 '15 at 1:33
  • $\begingroup$ Yes, $k>0$, thanks for the observation. Why do you think it's lower? $\endgroup$ – Almost Shirley Oct 20 '15 at 2:02
  • $\begingroup$ 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). $\endgroup$ – Glen_b Oct 20 '15 at 9:21
3
$\begingroup$

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 $k$:

Monte Carlo approximations to $\mathbb{E}\left[\frac{1}{\chi^2(n)+k}\right]$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.