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 $\sigma^2$?
 A: 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 $z$. 
