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The following summation arises is a Bayesian model where Beta distribution is used as the prior for Negative Binomial distribution. The summation is used to assess the risk of the resulting estimator. However, to this stage it is only a pure mathematical problem. My question is how to find the following sum, please? Thank you!

\begin{align*} \sum_{n=2}^\infty \frac{(n+1)(n-1)}{(n+7)^2(n+8)} (1-\theta)^{n-2}. \end{align*}

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  • $\begingroup$ $(x-1)/(x+7)=1-(8/(x+7))$. $(1-\theta)^{x-2}=(1-\theta)^{-9}(1-\theta)^{x+7}$. $\sum u^r/r$ is a logarithm. $\endgroup$ May 6, 2014 at 7:47
  • $\begingroup$ I don't quite understand why some are voting to close... $\endgroup$ May 6, 2014 at 8:37
  • $\begingroup$ @Jean-ClaudeArbaut Have you ever read some kind of how-to-ask page on the site? $\endgroup$
    – Did
    May 6, 2014 at 8:53
  • $\begingroup$ @Jean-ClaudeArbaut This summation is from a negative binomial(r=2, p=1-$\theta$) with beta(4,3) as prior. I found the posterior distribution of $\theta$ as beta(6, $x$+1) and the posterior mean is $6/(x+7)$. But I could not calculate the Bayes Risk for this estimator which involves the above sum. Do you know any way to find the Bayes Risk without using this summation, please? Thank you! $\endgroup$
    – LaTeXFan
    May 6, 2014 at 9:05
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    $\begingroup$ Yes, you get a dilogarithm. $\endgroup$ May 6, 2014 at 12:19

1 Answer 1

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First, $${{\left(n-1\right)\,\left(n+1\right)}\over{\left(n+7\right)^2\, \left(n+8\right)}}={{63}\over{n+8}}-{{62}\over{n+7}}+{{48}\over{\left(n+7\right)^2}}$$

Thus, with $\xi=1-\theta$

$$f(\theta)=\sum_{n=2}^\infty \frac{(n+1)(n-1)}{(n+7)^2(n+8)} (1-\theta)^{n-2}\\= 63 \sum_{n=2}^\infty \frac{(1-\theta)^{n-2}}{n+8} -62\sum_{n=2}^\infty \frac{(1-\theta)^{n-2}}{n+7} +48 \sum_{n=2}^\infty \frac{(1-\theta)^{n-2}}{(n+7)^2} \\= 63 \frac{1}{\xi^{10}} \sum_{n=10}^\infty \frac{\xi^{n}}{n} -62 \frac{1}{\xi^{9}} \sum_{n=9}^\infty \frac{\xi^{n}}{n} +48 \frac{1}{\xi^{9}} \sum_{n=9}^\infty \frac{\xi^{n}}{n^2} $$

And

$$\sum_{n=1}^\infty \frac{\xi^n}{n}=-\log (1-\xi)=-\log \theta$$ $$\sum_{n=1}^\infty \frac{\xi^n}{n^2}=\mathrm{Li_2}(\xi)$$

Where $\mathrm{Li_2}$ is the dilogarithm.

Therefore,

$$f(\theta)=-63\frac{\log \theta}{(1-\theta)^{10}}+62\frac{\log \theta}{(1-\theta)^{9}}+48\frac{\mathrm{Li_2}(\theta)}{(1-\theta)^{9}}-g(\theta)$$

With

$$g(\theta)=63 \sum_{n=1}^{9} \frac{(1-\theta)^{n-10}}{n} -62 \sum_{n=1}^{8} \frac{(1-\theta)^{n-9}}{n} +48 \sum_{n=1}^{8} \frac{(1-\theta)^{n-9}}{n^2}$$

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