Expected value of $\exp\big(-\frac{1}{X+1}\big)$ when X have binomial distribution

Question

I have a function $f(X)= e^{-\frac{1}{X+1}}$ where X is $\operatorname{Bin}(n,p)$.

Is there a way to simplify $E[f(X)]=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) p^{k}(1-p)^{n-k} e^{-\frac{1}{k+1}}$.

If not, is there a good approximation when $p<<1, n>100$.

Answer 1

I highly doubt that that can be simplified. An approximation can be obtained by doing a Taylor expansion around the mean.

The second order approximation I get is

$$ E[f(X)] \approx f(\mu) + \frac12f''(\mu) \sigma^2 =\exp\left(- \frac{1}{1+\mu}\right)\left(1- \sigma^2 \frac{\mu + \frac12}{(1+\mu)^4}\right)$$

where $\mu = np$ , $\sigma^2=np(1-p)$

Some values for $n=100$

    p        exact   approx
   0.1      0.90623  0.90721
   0.05     0.82528  0.82942
   0.01     0.55403  0.55024

We could add more terms to the expansion. It's not clear, though, under which conditions this behaves as an asymptotic expansion (so that higher order terms can be neglected). For fixed (and small) $p$, the second term (and, one hopes, the next ones) turns negligible for $n\gg 1/p$.

Furthermore, because $f(X)$ is concave, Jensen gives us the strict bound $E[f(X)] < \exp\left(- \frac{1}{1+\mu}\right)$