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