Negative Moments and Binomial Distribution

Question

Let $X$ be a random variable with a distribution given by $\mathrm{Bin}(n,p)$ for $p$ in the open unit interval. Compute the moment $E(1/(X+1))$, where we define the negative moment of order $n$ by $E(X^{-1})$ for $n>0$ integer.

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How would one do the above computation? I've tried for a while to no avail. I was thinking that perhaps one could manipulate the $1/(X+1)$ term so that the evaluation of the sum becomes easier (kind of how we use $X^{2}=X(X-1)+X$ for some computations).

Thanks for the help!

Answer 1

Hint (to be combined with the answer of @Alex):

$$\frac1{k+1}\binom{n}{k}=\frac1{n+1}\binom{n+1}{k+1}$$

Answer 2

Try calculating $$ \sum_{k=0}^n \frac{1}{k+1} \binom{n}{k} p^k (1-p)^{n-k} $$

Answer 3

I came by this question, saw the accepted answer and the hint, and it still took me some time to solve. In favor of future readers, I'm writing the entire answer here below.

Let $B\sim Bin(n,p)$. As @Alex says, let's look at the expression we want to compute: $$ \mathbb E\left[ \frac{1}{1+B}\right]=\sum_{k=0}^n \frac{1}{k+1} \binom{n}{k} p^k (1-p)^{n-k}. $$

Furher, simple algebra reavels the hint @drhab mentioned: $$\frac1{k+1}\binom{n}{k}=\frac1{n+1}\binom{n+1}{k+1}.$$ Using this, we get $$ \mathbb E\left[ \frac{1}{1+B}\right]=\frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{k+1} \binom{n}{k} p^k (1-p)^{n-k}. $$ Multiplying and dividing by $p$, $$=\frac{1}{p(n+1)}\sum_{k=0}^n \binom{n+1}{k+1} p^{k+1} (1-p)^{n+1-(k+1)}.$$ Setting $j=k+1$, we get $$=\frac{1}{p(n+1)}\sum_{j=1}^{n+1} \binom{n+1}{j} p^{j} (1-p)^{n+1-j}.$$ Recall that $$ \sum_{j=0}^{n+1}\Pr(Bin(n+1,p)=j)=\sum_{j=0}^{n+1} \binom{n+1}{j} p^{j} (1-p)^{n+1-j}=1 $$ However, the expression above misses the case for $j=0$. Adding and substracting it, $$=\frac{1}{p(n+1)}\left(-(1-p)^{n+1}+\underbrace{\sum_{j=0}^{n+1} \binom{n+1}{j} p^{j} (1-p)^{n+1-j}}_{1}\right)=\frac{1}{p(n+1)}\left(1-(1-p)^{n+1}\right).$$