I'm having some confusion over this statement here. Let $T_i \sim Exp(\lambda + \theta)$ and if they are all iid then $\sum_n T_i \sim Gamma(\alpha = n, \beta = 1/(\lambda + \theta))$

I want to find $E(\frac{1}{\sum_n T_i})$. I know I can't just do the reciprocal of $E(\sum_n T_i)$

Do I integrate over $\int_0^\infty \frac{t}{f(t)} dt$?

If I do this I get $\Gamma(\alpha) \beta^\alpha \int_0^\infty \frac{e^{t/\beta}}{t^\alpha} dt$ and I don't know if this is the right way to proceed since I end up with incomplete gamma?

I know that I'm supposed to end up with $\frac{\lambda + \theta}{n-1}$ as the answer but I'm not sure how the gamma is supposed to cancel out.

  • $\begingroup$ the density goes in the numerator...... $\endgroup$ Commented Feb 26, 2015 at 18:19
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    $\begingroup$ and I think you mean the answer should be $(\lambda + \theta)/(n-1)$ $\endgroup$ Commented Feb 26, 2015 at 18:27
  • $\begingroup$ The density should go in the numerator but would it still be the density function I'm integrating over if I want 1/X and not X? And yes you are correct about the answer, I'll fix that typo. Thanks $\endgroup$
    – datatista
    Commented Feb 26, 2015 at 18:28
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    $\begingroup$ If $X$ is a random variable with density $f$, then for any (nice) function $G$, $\mathbb E[ G(X) ] = \int G(x)f(x)dx$ $\endgroup$ Commented Feb 26, 2015 at 18:31
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    $\begingroup$ One easy way to do it is to apply the abovementioned law, and play 'spot the density', pull out the appropriate constants and cross out the integral of the density (since that integrates to 1). $\endgroup$
    – Glen_b
    Commented Feb 26, 2015 at 21:31

1 Answer 1


The calculation of $E\left[X^{-1}\right]$ when $X$ is a Gamma random variable with order parameter $n$ and rate parameter $\lambda$ requires recognition of the density of another Gamma random variable (with order parameter $n-1$ and rate parameter $\lambda$) in the integral given by the law of the unconscious statistician for $E\left[X^{-1}\right]$. We have

$$\begin{align} E\left[X^{-1}\right]&= \int_0^\infty \frac 1x \cdot \underbrace{\lambda \frac{(\lambda x)^{n-1}}{\Gamma(n)}e^{-\lambda x}}_{\Gamma(n,\lambda)~\text{density}}\,\mathrm dx\\ &= \lambda\frac{\Gamma(n-1)}{\Gamma(n)}\int_0^\infty \underbrace{\lambda \frac{(\lambda x)^{n-2}}{\Gamma(n-1)}e^{-\lambda x}}_{\Gamma(n-1,\lambda)~\text{density}}\,\mathrm dx\\ &= \frac{\lambda}{n-1} \end{align}$$ since for positive integer $k$, $\Gamma(k) = (k-1)!$.

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    $\begingroup$ Yea, the gamma's through me off quite a bit since I wasn't quite sure how to go about doing the integral. This explains how the gammas cancelled out so we get just n-1. Thanks! $\endgroup$
    – datatista
    Commented Feb 27, 2015 at 2:51

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