$$X \sim \text{Gamma}(\alpha, \beta)$$ $$Y = \frac{1}{X+c}, \ c > 0$$

I am interested in $E(Y)$, which I'm pretty sure is intractable...

$$E(Y) = \frac{\beta^\alpha}{\Gamma(\alpha)}\int_0^\infty \frac{x^{\alpha-1}e^{-\beta x}}{x+c} dx $$

If an exact analytic answer is unavailable, I would like to approximate this in terms of $\alpha$, $\beta$ and $c$.

By Jensen's Inequality, we have:

$$E\left[\frac{1}{X+c}\right] \geq \frac{1}{E(X) + c}$$

Or equivalently, for some $\epsilon(\alpha, \beta, c) > 0$:

$$E(Y) = \frac{1}{E(X) + c} + \epsilon(\alpha, \beta, c)$$

Can we say anything about the magnitude of $\epsilon(\alpha, \beta, c)$. If not, is there a better way to approach this problem?

Thanks in advance!

  • 1
    $\begingroup$ It all depends on what you mean by "intractable": with a simple change of variable, this is easily expressible in terms of the exponential integral function. A lot is known about it and values are readily computable in various ways. $\endgroup$
    – whuber
    May 5, 2017 at 23:08
  • $\begingroup$ @whuber I would be fine with a solution in terms of the exponential integral. But I'm not seeing the change of variables. $t = x + c$ doesn't seem to get us anywhere. $\endgroup$
    – knrumsey
    May 5, 2017 at 23:55
  • 3
    $\begingroup$ Using a computer algebra system returns: $\frac{1}{b}e^{c/b} E_a\left(\frac{c}{b}\right)$ where $E_a()$ is the ExpIntegralE function $E_a(z)=\int _1^{\infty }\frac{e^{-z t}}{t^a} dt$ $\endgroup$
    – wolfies
    May 6, 2017 at 5:15

1 Answer 1


According to Rockower (Integral Identities for Random Variables, $\it{The \ American \ Statistician}$, February 1988, Vol. 42, No. 1), the expectation is given by $$E \left[ \frac{1}{X+c} \right] =\int_0^{\infty} e^{-cs} \cal{L}(s)ds,$$ where $\cal{L} (s)$ is the Laplace transform of the pdf of the random variable $X$.

From a table of Laplace transforms we find $$\cal{L} \left( x^{\alpha-1} e^{-x/{\beta}} \right)=\frac{\Gamma \left( \alpha \right)}{\left(s+\frac{1}{\beta} \right)^{\alpha} } $$

So for the Laplace transform of the pdf we get $$\cal{L} \left( \frac{x^{\alpha-1} e^{-x/{\beta}}}{\beta^{\alpha} \ \Gamma \left( \alpha \right) }\right)=\frac{1}{\beta^{\alpha} \left(s+\frac{1}{\beta} \right)^{\alpha} } $$

Then $$E \left[ \frac{1}{X+c} \right] =\int_0^{\infty} \frac{e^{-cs}}{\beta^{\alpha} \left( s + \frac{1}{\beta} \right)^{\alpha}} ds=\int_0^{\infty} \frac{e^{-cs}}{ \left( \beta s + 1 \right)^{\alpha}} ds$$

Now make the substitution $q=\beta s + 1.$ The result is

$$E \left[ \frac{1}{X+c} \right] =\int_1^{\infty} \frac{e^{-c \left( \frac{q-1}{\beta} \right)}}{\beta \ q ^{\alpha}} dq=\frac{1}{\beta}e^{c/{\beta}}\int_1^{\infty} \frac{e^ { - \left( {\frac{c}{\beta}} \right) q } }{ q^{\alpha}} dq$$

From the definition for the generalized exponential integral function given in the comment by wolfies, we have finally $$ E \left[ \frac{1}{X+c} \right]= \frac{e^{c/{\beta}}}{\beta} E_{\alpha} \left( \frac{c}{\beta} \right), $$

which agrees with his answer using the computer algebra system.

  • 2
    $\begingroup$ Thanks, I had already verified this numerically. But I appreciate the answer. It's nice to see the derivation as well. $\endgroup$
    – knrumsey
    May 12, 2017 at 0:37

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