1

Let $\text X$ be a binomial random variable with paramets $n$ and $p$. Show that $$E\left(\dfrac{1}{1+\text X}\right)=\dfrac{1-(1-p)^{n+1}}{(n+1)p}.$$

Would anyone mind telling me how to solve this question?

user110858
  • 259
  • 1
    This looks a lot like a homework problem (for example, it is Theoretical Exercise 4.10 in Sheldon Ross's A First Course in Probability (8th ed.)) and so I will give only a hint. What Ross used to call the Law of the Unconscious Statistician in earlier editions allows you to write $$E\left[\frac{1}{X+1}\right]=\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}p^k(1-p)^{n-k}=\frac{1}{n+1}\sum_{k=0}^n\frac{n+1}{k+1}\binom{n}{k}p^k(1-p)^{n-k}.$$ Now try to write $\frac{n+1}{k+1}\binom{n}{k}$ as a binomial coefficient. – Dilip Sarwate Nov 23 '13 at 14:35

2 Answers2

2

Hint: Do you know that $E(\varphi(X))=\sum \varphi(x_i) P\{X=x_i\}$?

user110822
  • 311
1

Hint: $$\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\binom{n+1}{k+1}.$$

The rest will use the Binomial Theorem.

André Nicolas
  • 507,029