I am trying to find following expectation

\begin{equation} \mathbb{E} [\frac{1}{1+X}] \end{equation}

where $X$ is a non central Chi Squared distribution with $K$ degree of freedom and $K$ non-centrality parameter. I know it must be related to Marqum Q function however I can't find something like this in integral tables and I don't have Mathematica.

  • $\begingroup$ @wolfies how did you find that? actually I needed a closed form solution to address my problem. originally I needed the solution for $\mathbb{E}[\frac{1}{1+a X}]$ , where $a$ is a positive real number $\endgroup$ – Alireza Aug 14 '17 at 16:34
  • $\begingroup$ @wolfies hello? $\endgroup$ – Alireza Aug 18 '17 at 21:56
  • $\begingroup$ I've posted some thoughts below $\endgroup$ – wolfies Aug 19 '17 at 17:39

To define notation, let $X \sim \text{NoncentralChisquared}(k,k)$ with pdf $f(x)$:

enter image description here

Symbolic solution to $\mathbb{E} [\frac{1}{1+X}]$

A closed form solution to $\mathbb{E} [\frac{1}{1+X}]$ does not seem readily obtainable, whether by transforming the pdf to $Y = 1/(1+X)$ or otherwise.

The neatest form I have obtained is first by deriving the mgf $M_X(t) = \mathbb{E}[e^{t X}]$ as:

enter image description here

where I am using the Expect function from mathStatica/Mathematica to automate the nitty gritties (I should note I am one of the authors of the function).

The mgf can be used to derive this negative moment (for non-negative random variables) via:

$$\mathbb{E} [\frac{1}{1+ aX}] \quad = \quad \int_0^{\infty } e^{-t} M(-a t) \;dt$$

When $a = 1$, this is: $$\mathbb{E}[\frac{1}{1+X}] \quad = \quad \int_0^{\infty } \exp\big(-\frac{k t}{1+2t}-t\big) (1+2t)^{-k/2} \;dt$$

which can be solved for odd-valued integers.

For example, when $k = 1$, the exact solution is: $$\frac{e^{-i} \sqrt{\pi } \left(\text{erf}\left((-1)^{3/4}\right)+e^{2 i} \text{erfc}\left(\sqrt[4]{-1}\right)+1\right)}{2 \sqrt{2}} \quad \approx \quad 0.520862$$


The above seems far too messy. Fortunately, an exceptional and simple approximation is: $$\mathbb{E} [\frac{1}{1+X}] \; \; \approx \; \frac{1}{2(k-1)} \quad \quad \text{for} \quad k> 7$$

The following figure compares:

  • $\mathbb{E} [\frac{1}{1+X}]$ (calculated by numerical integration) - BLUE curve
  • ... to the simple $\frac{1}{2(k-1)}$ approximation - ORANGE dashed curve

enter image description here

For $k> 10$, the approximation appears to be accurate to at least 3 decimal places.


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