# Expected value of g(x)=1/(x+c) under gamma distribution

1. Nov 28, 2012

### PMRDK

1. The problem statement, all variables and given/known data

Let $g(x) = \frac{1}{x+c}$, where $c$ is a positive constant, and $x$ is a random variable distributed according to the Gamma distribution
$x\sim f(x)=\frac{1}{\Gamma(\alpha) \beta^\alpha} x^{\alpha \,-\, 1} e^{-\frac{x}{\beta}}$.

I wish to calculate the expected value of $g(x)$ with respect to the probability density function $f(x)$.

2. Relevant equations

The expected value can be calculated as
$$E(g(x))=\int_0^∞g(x)f(x)dx = \int_0^∞ \frac{1}{x+c} \frac{1}{\Gamma(\alpha) \beta^\alpha} x^{\alpha \,-\, 1} e^{-\frac{x}{\beta}}dx$$

3. The attempt at a solution
I have problems with calculating the integral. If $g(x)=\frac{1}{x}$, then the integral would not be too difficult. But the constant in the denominator gives me problems. I have attempted with variable substitutions and integration by parts. However, I have not been able to come up with a solution.
Actually, this is not a homework problem. I posted it here since it is `homework style', so I do not know if it is event possible to calculate the expectation.

Any help is much appreciated.

2. Nov 28, 2012

### Mute

The integral may not have a closed form. Do you know for sure if it does?

I'm not sure if this will help much, but you could try splitting up the integral into two, one from $[0,c)$ and the other on $[c,\infty)$, and then expand the 1/(x+c) as a power series and integrate term by term to get a series expression. In the first integral, since x is smaller than c, you would use

$$\frac{1}{x+c} = \frac{1}{c} \frac{1}{1+x/c} = \frac{1}{c} \sum_{n=0}^\infty \left(\frac{x}{c}\right)^n,$$

while in the second term x is greater than c, so you would write

$$\frac{1}{x+c} = \frac{1}{x} \frac{1}{1+c/x} = \frac{1}{x} \sum_{n=0}^\infty \left(\frac{c}{x}\right)^n.$$

In principle one needs to be careful about switching the integral and the sum since the geometric series is only uniformly convergent on a domain $|z-c| \leq R$ for R < 1, so if you want to be rigorous I guess you could split up the integral into three regions, $[0,c-\epsilon), [c-\epsilon,c+\epsilon], (c+\epsilon,\infty)$ and do the series expansion for the first and the last, then estimate the error from neglecting the $[c-\epsilon,c+\epsilon]$ integral when taking $\epsilon \rightarrow 0$, but my guess is that it won't cause problems.

3. Nov 28, 2012

### Ray Vickson

Using b instead of 1/β, Maple evaluates this integral in terms of the incomplete Gamma function:

f:=b^a*x^(a-1)/GAMMA(a)*exp(-b*x);
f = b^a*x^(a-1)/GAMMA(a)*exp(-b*x)

J1:=int(f/(x+c),x=0..infinity) assuming a>0,b>0,c>0;
J1 = b^a*c^(a-1)*exp(c*b)*GAMMA(1-a,c*b)

That is,
$$\int_0^{\infty} \frac{b^a}{\Gamma(a)} \frac{x^{a-1}}{x+c} e^{-bx} \, dx = b^a c^{a-1} e^{bc} \Gamma(1-a,bc),$$
where
$$\Gamma(u,z) = \int_z^{\infty} e^{-t} t^{u-1} \, dt \text{ if } z > 0$$
is the incomplete Gamma function. I believe it has been proven that the incomplete Gamma function is non-elementary if u is not a positive integer.

Last edited: Nov 28, 2012
4. Dec 4, 2012

### PMRDK

Thank you for your help. So Maple was able to provide an answer in terms of a closed form solution (I tried in Maxima but did not succeed).

5. Aug 19, 2014

### anuszka

A more general problem has been solved here, for $1/(1+cx^n)$ and $1/(1+cx)^n$, for any $n$:
http://www.pnas.org/content/107/51/22096.full.pdf+html?with-ds=yes

You can find there the explicit formula for the probability density function, the monotonicity properties of the pdf, the first and second moments for $n=1$ and $n=2$.