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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 $\sigma^2$?

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  • $\begingroup$ 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, 2018 at 20:09
  • $\begingroup$ @a_statistician I know.. But how? $\endgroup$
    – Qwerty
    Oct 29, 2018 at 20:17
  • $\begingroup$ $X'X = \sum(X_i^2) \ge 0$ ==> $1/X'X \ge 0$ $\endgroup$
    – user158565
    Oct 29, 2018 at 20:32
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    $\begingroup$ @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, 2018 at 21:07
  • $\begingroup$ I am wrong, because I ignored that both parts are related to $\sigma^2$ $\endgroup$
    – user158565
    Oct 29, 2018 at 21:52

1 Answer 1

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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 $z$.

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  • $\begingroup$ Your formula is supported by Equation (10.9) in Paolella's Intermediate Probability, which gives arbitrary moments of chi-square variates. $\endgroup$
    – steveo'america
    Jan 21, 2020 at 18:08

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