91Ó°ÊÓ

Let $X$be a binomial random variable with parameters $n$ and $p$.

Using the theorem regarding the expectation of the function of the random variable, we have that

$E\left(\frac{1}{1+X}\right)$

$=\underset{k=0}{\overset{n}{∑}}\frac{1}{1+k}\left(\begin{array}{l}n\\ k\end{array}\right)p^k(1-p{)}^{n-k}$

$=\underset{k=0}{\overset{n}{∑}}\frac{1}{1+k}\frac{n!}{k!(n-k)!}{p}^k(1-p{)}^{n-k}$

$\underset{k=0}{\overset{n}{∑}}\frac{n!}{(k+1)!(n-k)!}{p}^k(1-p{)}^{n-k}$

Multiply these terms in the sum with $(n+1)$and $p$and get that the expression from the above is equal to

$\frac{1}{(n+1)p}\underset{k=0}{\overset{n}{∑}}\frac{(n+1)!}{(k+1)!(n-k)!}{p}^{k+1}(1-p{)}^{n-k}$

Note this expression a bit more different to get

$\frac{1}{(n+1)p}\underset{k=0}{\overset{n}{∑}}\frac{(n+1)!}{(k+1)!\left(\right(n+1)-(k+1\left)\right)!}{p}^{k+1}(1-p{)}^{(n+1)-(k+1)}$

Change index in the summary that it can go from $1$to $n+1$and we have that

$\frac{1}{(n+1)p}\underset{k=1}{\overset{n+1}{∑}}\frac{(n+1)!}{k!\left(\right(n+1)-k)!}{p}^k(1-p{)}^{(n+1)-k}$

Now, evaluate the sum. It is exactly the probability that a Binomial with parameters $p$and $n+1$considers every other value instead of zero. So, that probability is $1-(1-p{)}^{n+1}$. Finally, we get that

$E\left(\frac{1}{1+X}\right)=\frac{1}{(n+1)p}·\left(1-(1-p{)}^{n+1}\right)$

Assume the Binomial with parameters $n+1$and $p$.

$E\left(\frac{1}{1+X}\right)=\frac{1}{(n+1)p}·\left(1-(1-p{)}^{n+1}\right)$