Expectation of inverse non central chi-squared

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 $$\sigma^2$$?

• Given $\mathbb{E}(\sigma^2/X^{'}X) = \sigma^2 \mathbb{E}(1/X^{'}X)$, so you just need to prove $\mathbb{E}(1/X^{'}X) > 0$. – user158565 Oct 29 '18 at 20:09
• @a_statistician I know.. But how? – Qwerty Oct 29 '18 at 20:17
• $X'X = \sum(X_i^2) \ge 0$ ==> $1/X'X \ge 0$ – user158565 Oct 29 '18 at 20:32
• @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 – Qwerty Oct 29 '18 at 21:07
• I am wrong, because I ignored that both parts are related to $\sigma^2$ – user158565 Oct 29 '18 at 21:52

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 $$z$$.