Expectation of iid Bernoulli Function

Question

For $X_1$, $X_2$,..., $X_n$ iid Bernoulli$(p)$, I am looking to calculate $E\left[\dfrac{1}{n - \sum_{i=1}^{n} X_i}\right]$, but am having trouble doing so. I know the sum of iid Bernoulli's is Binomial, but I am not sure how this applies to this expectation, besides the fact that E[$\sum_{i=1}^{n} X_i$] = np. Any ideas?

Answer 1

The below shows the steps for $n+1-\sum X_{i}$ in the numerator - as the comment by Misha Lavrov says $n$ does not work.

{Step 1 - rewrite the expectation for a general X:}

Let $X$ be a random variable on $[0,1]$ with $X<1$ (we don't want to deal with a $0\cdot\infty$ type product in an expectation in full generality - the underlying integrand's behaviour around the singularity of the function should be observed more carefully). To calculate the expectation we can use Taylor expansion (everything's non-negative):

$$ \mathsf{E}\left(\frac{X}{1-X}\right)=\mathsf{E}\left(\sum_{i=1}^{\infty}X^{i}\right)=\sum_{i=1}^{\infty}\mathsf{E}X^{i} $$

{Step 2 - Calculate $\mathsf{E}X^{i}$ for the specific case:}

MGF calculation with Bernulli, we can plug in the following to the above:

$$ X=\frac{1}{n+1}\left(X_{1}+\ldots+X_{n}\right) $$

$$ G_{p}(t)=\left(1-p\right)+\exp\left(\frac{t}{n+1}\right)p $$

Hence we have to calculate the derivatives of the $n$th power of this - by independence, so use the binomial expansion:

$$ G_{p,n}\left(t\right)=\left(\left(1-p\right)+\exp\left(\frac{t}{n+1}\right)p\right)^{n}=\sum_{j=0}^{n}\binom{n}{j}\left(1-p\right)^{n-j}p^{j}\exp\left(\frac{tj}{n+1}\right) $$

Which is simple enough for terms of the form $d\exp\left(ct\right)$:

$$ \frac{\partial}{\partial t^{i}}\left(\sum_{j=0}^{n}\binom{n}{j}\left(1-p\right)^{n-j}p^{j}\exp\left(\frac{tj}{n+1}\right)\right)=\left(\sum_{j=0}^{n}\binom{n}{j}\left(1-p\right)^{n-j}p^{j}\frac{j^{i}}{\left(n+1\right)^{i}}\exp\left(\frac{tj}{n+1}\right)\right) $$

For the expectation, we need to use $t=0$.

Hence:

$$ \mathsf{E}X^{\ell}=\left(\sum_{j=0}^{n}\binom{n}{j}\left(1-p\right)^{n-j}p^{j}\frac{j^{\ell}}{\left(n+1\right)^{\ell}}\right) $$

Plug this in, switch the sums - everything's non-negative:

$$ \sum_{\ell=1}^{\infty}\mathsf{E}X^{\ell}=\sum_{\ell=1}^{\infty}\left(\sum_{j=0}^{n}\binom{n}{j}\left(1-p\right)^{n-j}p^{j}\frac{j^{\ell}}{\left(n+1\right)^{\ell}}\right)= $$ Geometric series: $$ =\sum_{j=0}^{n}\binom{n}{j}\left(1-p\right)^{n-j}p^{j}\sum_{\ell=1}^{\infty}\frac{j^{\ell}}{\left(n+1\right)^{\ell}}=\sum_{j=0}^{n}\binom{n}{j}\frac{\left(1-p\right)^{n-j}p^{j}}{1-\cfrac{j}{n+1}}\cfrac{j}{n+1}= $$ And finally take what the binomial coefficient is to arrive at: $$ =\frac{p \left(1-p^n\right)}{1-p} $$