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.
Best Answer
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 $k$, $\Gamma(k) = (k-1)!$.