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This is a part of lecture note. Reviewing it I am not clear with the process. How can I expand $\operatorname{E}\left[\dfrac 1 {\left(\sum X_i\right)}\right]$ to make it $\displaystyle\int {(\lambda^n)(x^{n-1}) \exp(-\lambda x)\,dx \over x(n+1)!}$? It's usung MLE of Gamma distribution for $X_i$'s from exponential distribution.

Michael Hardy
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zac.jung
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Let $X_1,\dots,X_n$ be a random sample from Exponential rate $\lambda$ distribution.

Note that $\sum_{i=1}^n X_i \sim \Gamma(n,\lambda$). You can see this with moment generating functions.

So let $Z \sim \Gamma(n,\lambda)$. \begin{align*} E\left[1/\sum_{i=1}^n X_i \right] = E[1/Z] &= \int_0^\infty \dfrac{1}{z} \dfrac{\lambda^n}{\Gamma(n)} z^{n-1} e^{- \lambda z} \; dz \\ &= \int_0^\infty \dfrac{\lambda^n}{\Gamma(n)} z^{n-2} e^{-\lambda z } \; dz \\ &= \lambda \dfrac{\Gamma(n-1)}{\Gamma(n)} \underbrace{\int_0^\infty \dfrac{\lambda^{n-1}}{\Gamma(n-1)} z^{n-2} e^{-\lambda z } \; dz}_{=1} \\ &= \dfrac{ \lambda \Gamma(n-1)}{\Gamma(n)} = \dfrac{ \lambda}{n-1} \end{align*}

user365239
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  • I appreciate you prompt response! – zac.jung Dec 12 '16 at 03:18
  • Sure - I had a typo I just corrected, and also $\Gamma(n) = (n-1)!$ for positive integers $n$, which looks more like your formula – user365239 Dec 12 '16 at 03:19
  • I just trabaformed to factorial for easier tyoing. Rrgarding Mgf, you mean if I comute integral {Sum(Xi) * exponentil distribution} then the result would be gamma, is this correct? – zac.jung Dec 12 '16 at 03:22
  • Take a look here for the distribution of $\sum_{i=1}^n X_i$. http://math.stackexchange.com/questions/655302/gamma-distribution-out-of-sum-of-exponential-random-variables – user365239 Dec 12 '16 at 03:24
  • I'll look into the post. Thanks! – zac.jung Dec 12 '16 at 03:32