1. ## Expectation of functions

Suppose X has binomial distribution with probability of success p over n trials,

How do I show that:

E[1/(1+X)] = [1-(1-p)^(n+1)]/p(n+1)?

Suppose X has binomial distribution with probability of success p over n trials,

How do I show that:

E[1/(1+X)] = [1-(1-p)^(n+1)]/p(n+1)?
Use the definition, like...

$\displaystyle E[\frac{1}{1+X}] = \sum_x \frac{1}{1+x} p_X = \sum_x \frac{1}{1+x} {n\choose x} p^x (1-p)^{n-x}$

Now you want to manipulate this so that you can use the fact that $\displaystyle \sum p_X=1.$

Some tips:
$\displaystyle a)$ Play with the $\displaystyle {n\choose x}$ to cancel with the $\displaystyle \frac{1}{1+x}.$
$\displaystyle b)$ Set $\displaystyle j=x+1.$

Show me where you are stuck and I will help you proceed. To get started:

$\displaystyle \frac{1}{1+x} {n\choose x} = \frac{1}{1+x} \cdot \frac{n!}{x!(n-x)!} =\frac{n!}{(1+x)!(n-x)!} =$ $\displaystyle \frac{n!}{(1+x)!(n-x)!}= \frac{1}{n+1}\frac{(n+1)!}{(1+x)!((n+1)-(x+1))!} = \frac{1}{1+n} {n+1 \choose x+1}$

So we have that,

$\displaystyle \sum_x \frac{1}{1+x} {n\choose x} p^x (1-p)^{n-x} =$ $\displaystyle \frac{1}{1+n}\sum_x {n+1 \choose x+1} p^x (1-p)^{n-x} = \frac{1}{p(1+n)}\sum_x {n+1 \choose x+1} p^{x+1} (1-p)^{(n+1)-(x+1)}$

Now set j=x+1, then

$\displaystyle \frac{1}{p(1+n)}\sum_{j=2}^n {n+1 \choose j} p^{j} (1-p)^{(n+1)-j}$