1
$\begingroup$

If $X\sim \mathcal{N}_p(\theta,\sigma^2)$ then $X^{'}X/\sigma^2\sim$ non-central $\chi^2_p$ with non-centrality parameter $\theta^{'}\theta/\sigma^2$.

How to find out the expectation of $\sigma^2/X^{'}X$?

More specifically, how to show that $\mathbb{E}(\sigma^2/X^{'}X)$ is an increasing function of $\sigma^2$?

$\endgroup$
  • $\begingroup$ Given $\mathbb{E}(\sigma^2/X^{'}X) = \sigma^2 \mathbb{E}(1/X^{'}X)$, so you just need to prove $\mathbb{E}(1/X^{'}X) > 0 $. $\endgroup$ – user158565 Oct 29 '18 at 20:09
  • $\begingroup$ @a_statistician I know.. But how? $\endgroup$ – Qwerty Oct 29 '18 at 20:17
  • $\begingroup$ $X'X = \sum(X_i^2) \ge 0$ ==> $1/X'X \ge 0$ $\endgroup$ – user158565 Oct 29 '18 at 20:32
  • 2
    $\begingroup$ @a_statistician I think you are missing a point. Even if (say) $\mathbb{E} (1/X^{'}X)=1/\sigma^4$ , then $\mathbb{E} (1/X^{'}X)>0$ but overall its a decreasing function of $\sigma^2$. We need to find theexact representation of $\sigma$ in $\mathbb{E} (1/X^{'}X)$ to be able to proceed $\endgroup$ – Qwerty Oct 29 '18 at 21:07
  • $\begingroup$ I am wrong, because I ignored that both parts are related to $\sigma^2$ $\endgroup$ – user158565 Oct 29 '18 at 21:52
3
$\begingroup$

I can give the expectation but I don't know for the monotonicity.

Let $C \sim \chi^2(d,\theta)$ (where $d$ degrees of freedom, and $\theta$ non-centrality parameter).

If $d > 2$, then $$ \mathbb{E}(1/C) = \frac{1}{2}\exp(-\theta/2)\frac{\Gamma(d/2-1)}{\Gamma(d/2)}{}_1\!F_1(d/2-1,d/2,\theta/2). $$ (I don't know for $d \leq 2$, perhaps the expectation is infinite in this case).

Check with R:

> df <- 10
> ncp <- 2
> x <- rchisq(1e7, df, ncp)
> mean(1/x)
[1] 0.1036415
> 
> 1/2 * gamma(df/2-1)/gamma(df/2) * exp(-ncp/2) * gsl::hyperg_1F1(df/2-1, df/2, ncp/2)
[1] 0.1036383

I got this formula from Gupta & Nagar's book Matrix variate distributions. This book provides a formula for the expectation of $\det(W)^h$ where $W$ follows a noncentral Wishart distribution, and the formula for the expectation of $1/C$ is a particular case.

Note that ${}_1\!F_1(d/2-1,d/2,\theta/2)$ has form ${}_1\!F_1(a,a+1,z)$. Wolfram provides some simplification formulas for ${}_1\!F_1(a,a+1,z)$, but they make sense for $z<0$ only, as far as I can see.

EDIT

Wait... The family of the noncentral $\chi^2$ distributions is stochastically increasing in the noncentrality parameter. Therefore $X'X$ is stochastically decreasing in $\sigma$, and $1/X'X$ is stochastically increasing in $\sigma$. Consequently, $\mathbb{E}(1/X^{'}X)$ is an increasing function of $\sigma$, as well as $\mathbb{E}(\sigma^2/X^{'}X)$.

EDIT 2

Thanks to a relation given in Wikipedia, the formula for the expectation can be simplified to: $$ \mathbb{E}(1/C) = \frac{1}{2}\frac{\Gamma(d/2-1)}{\Gamma(d/2)}{}_1\!F_1(1,d/2,-\theta/2). $$ Now, in view of the integral representation of ${}_1\!F_1$, it is clear that ${}_1\!F_1(a,b,z)$ is increasing in $z$.

$\endgroup$

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.