So I'm trying to derive an expected value (related to Bayesian risk/loss function) and I've derived all except one final part. To finish the final part I need to derive one of the following expected values (either will work)
Define the probability mass function
$$p_N (n) = {n-1 \choose x-1} \frac{\Gamma(a+b)}{\Gamma(a) + \Gamma(b)}\frac{\Gamma(a+x)+\Gamma(b+n-x)}{\Gamma(a+b+n)}$$
for $n =x,x+1,x+2,\dots$ and also define the conditional pmf
$$p_N(n|p) = {n-1 \choose x-1} p^x (1-p)^{n-x}$$
To complete the final step I need either one of:
$$E_N \left[ \left(\frac{x-1}{N-1} \right)^2\right]$$
or
$$E_{N} = \left[ \left(\frac{x-1}{N-1} \right)^2 \bigg| p \right]$$
In previous questions, I've derived the necessary expected values by absorbing the terms into the probability functions in order to construct a new distribution function, and then obtaining the expected value by normalizing it so that it sums/integrates to $1$. But for these ones I'm stuck due to the fact it's squared and you're left with a single fraciton that can't be absorbed into the combination.
Does anyone see a way forward?
Hint using generating functions :
\begin{align*} \sum_{n=x}^\infty {n-1\choose x-1} \frac{1}{(n-1)^2} q^n&=q\sum_{n=x}^\infty {n-1\choose x-1} \frac{1}{(n-1)^2} q^{n-1}\\ &=q\sum_{n=x}^\infty {n-1\choose x-1} \int \frac{1}{(n-1)} q^{n-2} dq\\ &=q\sum_{n=x}^\infty {n-1\choose x-1} \int \frac{1}{q} \frac{1}{(n-1)} q^{n-1} dq\\ &=q\sum_{n=x}^\infty {n-1\choose x-1} \int \frac{1}{q} \int q^{n-2} dq dq\\ &=q\int \frac{1}{q} \int \left[\sum_{n=x}^\infty {n-1\choose x-1} q^{n-2}\right] dq dq\\ &=q\int \frac{1}{q} \int (1-q)^{-x} q^{x-2} dq dq\\ \end{align*}
The combinatorics identity I use can be proved using the following argument : \begin{align*} S(x,q)&=\sum_{n=x}^\infty {n-1\choose x-1} q^n\\ &= \sum_{n=x}^\infty \left [ {n\choose x} - {n-1 \choose x}\right] q^n\\ &=\sum_{n=x}^\infty {n\choose x}q^n - \sum_{n=x}^\infty{n-1 \choose x} q^n\\ &= \frac{1}{q}\sum_{n=x+1}^\infty {n-1\choose x}q^n - \sum_{n=x+1}^\infty {n-1 \choose x} q^n\\ &= \left( \frac{1}{q} - 1 \right) S(x+1, q) \end{align*}
Which means that $S(x,q) = \left( \frac{q}{1-q} \right)^{x-1} S(1,q)=\left( \frac{q}{1-q} \right)^{x-1} \sum_{n=1}^\infty {n-1\choose 0} q^n=\left( \frac{q}{1-q} \right)^x$