Expected value with negative exponent I am trying to solve identify the expected value of a statistic that involves a fraction. I have simplified the expression to:
$E[\frac{1}{1+ \sum_i x_i}] = E[\frac{1}{1+ T}]$
However, I am not sure how to proceed. Is there anyway to simplify this through algebra, i.e. simplify the expression to the point that I have $E[x_i]$ and then substitute in a known expression for the expected value of $x_i$? Or, should I attempt to find the expected value of the expression by working with the pmf as
$\sum \frac{1}{1+ T} * f_T(t)$ ?
 A: This is really a long, possibly useful, comment. Not a straightforward answer.
If the distribution of $X_i$ allows you to find $f_T$ easily, and if
$T$ has a continuous distribution with support $S$, then
it will likely be easier to integrate to find $E[(1-T)^{-1}]$ by evaluating $\int_S \frac{1}{1+t}f_T(t)\,dt.$ (Not to quibble, but
note the slight changes in notation from what you wrote.) Perhaps
a change of variable will enable you to get the integrand into
a recognizable form (except for a constant that can be 'factored out'), or to break the integral into two terms.
It would have been easier to give a straightforward answer if you
had told us the distribution of the $X_i.$
A: For T distributed Negative Binomial (r,p) where $r>1$:
\begin{align*}
E \left( \frac{1}{1+T} \right) &= \sum \limits_{t=0}^{\infty} \frac{1}{1+t} \binom{r+t-1}{t} p^r (1-p)^t \\
&= \sum \limits_{t=0}^{\infty} \frac{1}{t+1} \cdot \frac{(r+t-1)!}{(t)!(r-1)!} p^r (1-p)^t \\
&= \sum \limits_{t=0}^{\infty} \frac{(r+t-1)!}{(t+1)!(r-1)!} p^r (1-p)^t \\
&= \sum \limits_{t=0}^{\infty} \frac{1}{r-1} \cdot \frac{(r+t-1)!}{(t+1)!(r-2)!} p^r (1-p)^t \\
&= \frac{1}{r-1} \sum \limits_{t=0}^{\infty} \binom{r+t-1}{t+1} p^r (1-p)^t \\
&= \frac{1}{r-1} \sum \limits_{t=0}^{\infty} \binom{r+(t+1)-2}{t+1} p^r (1-p)^t \\
&= \frac{1}{r-1} \sum \limits_{y=1}^{\infty} \binom{r+y-2}{y} p^r (1-p)^{y-1}, \,\,\, \text{where  } y = t+1 \\
&= \frac{p}{(r-1)(1-p)} \sum \limits_{y=1}^{\infty} \binom{(r-1)+y-1}{y} p^{r-1} (1-p)^y \\
&= \frac{p}{(r-1)(1-p)} \left[ \sum \limits_{y=0}^{\infty} \binom{(r-1)+y-1}{y} p^{r-1} (1-p)^y \, - \, p^{r-1} \right] \\
&= \boxed{\frac{p ( 1 - p^{r-1} )}{(r-1)(1-p)}} \\
\end{align*}
Alternatively, if someone knows of a simple form for the Negative Binomial $n$th moment, then another solution might be found by the following Taylor expansion about 0:
$E \left( \frac{1}{1+T} \right) = E \left( 1 - T + T^2 - T^3 + \cdots \right) = \sum \limits_{n=0}^{\infty} (-1)^n E(T^n)$
