Find the expected value of $\frac{1}{X+1}$ where $X$ is binomial

Question

The problem:

X is a binomial random variable, find $E[\frac{1}{X+1}]$ n and p are not given

PDF for a binomial distribution is $\binom{n}{k}p^k(1-p)^{n-k}$

Expected value is $\sum{x_ip(x_i)}$

But this is where I get stuck, I'm really rusty on my statistics and I'm not sure exactly how to structure it in the next step? I think I want to get the form of the following out of the summation

$\sum _{k=0}^{n} \binom{n}{k}p^k(1-p)^{n-k} = (p + 1 - p)^n = 1$

But I'm not sure if it should look like

$\sum \frac{1}{xp(x)+1} $

and if it should where to go from here?

What certain reasons are there, except perhaps ignoring the fact that people put effort into posting answers? — Asaf Karagila, Feb 18 '13 at 08:59
Please do not delete your question. The answers given here can be beneficial for future visitors. — Stefan Hansen, Feb 18 '13 at 08:59

score 16 · Accepted Answer · answered Feb 18 '13 at 07:59

16

By definition of Expectation, $\mathbb{E}(\frac{1}{1+X})$ should look like $\sum \frac{1}{1+x}\cdot p(x)$. In fact, $$\mathbb{E}(\frac{1}{1+X})=\sum_{k=0}^{n}\frac{1}{1+k}\cdot {n \choose k}p^k(1-p)^{n-k}=\frac{1}{(n+1)p}\cdot \sum_{k=0}^{n}{n+1 \choose k+1}\cdot p^{k+1}(1-p)^{n-k}=\frac{1}{(n+1)p}\cdot(1-(1-p)^{n+1}) $$

answered Feb 18 '13 at 07:59

Coiacy

1,640

Hello~ May I know how you get from the secondline to the last result？ I‘m a bit unclear of how you did it. I understand that term looks like bin(n+1, p), which usually sums up to one. But here we missed one portion, right? How did you find that portion? Thanks!!! – JungleKing May 31 '21 at 03:48
@JungleKing introduce $j = k+1$, and you can see that it is binomial(n+1,p), but with the first term missing, since $j$ goes from 1 to $n+1$. – Attila1177298 Sep 08 '21 at 08:54

score 9 · Answer 2 · answered Feb 19 '13 at 16:36

9

Another way to solve:

Use the fact $E\left[ \dfrac{1}{X+a}\right]=\int_{0}^{1}t^{a-1}. P_X(t)\, dt$ where $P_X(t)$ is PGF(Probability Generating Function) .

So $$\begin{align} E\left[ \dfrac{1}{X+1}\right]&=\int_{0}^{1}t^0. P_X(t)\, dt \\ &=\int_{0}^{1} (q+pt)^n\, dt \\ &=\dfrac{1-q^{n+1}}{(n+1)p}\end{align}$$ where $q=1-p$.

answered Feb 19 '13 at 16:36

Argha

4,671

Where can I see the proof of this formula? – Dominic Joseph Apr 27 '22 at 04:53

score 4 · Answer 3 · answered Feb 18 '13 at 07:57

4

Using the law of the unconscious statistician we get that $$ E\left[\frac{1}{X+1}\right]=\sum_{k=0}^n\frac{1}{1+k}\binom{n}{k}p^k(1-p)^{n-k} $$ which should be computed. To do that, try and write $\frac{1}{k+1}$ as part of the term $\binom{n}{k}$ by noting that $$ \frac{1}{k+1}\binom{n}{k}=\binom{n+1}{k+1}\frac{1}{n+1}. $$

answered Feb 18 '13 at 07:57

Stefan Hansen

25,582
7
59
91

2

Is this a law applied to statisticians who got smacked on the head with a cast iron frying pan? :-) – Asaf Karagila Feb 18 '13 at 21:26
@AsafKaragila: Haha, indeed it is :) – Stefan Hansen Feb 19 '13 at 06:09

score 2 · Answer 4 · answered Dec 31 '20 at 10:15

we know :$\frac{1}{X+1}=\int_{0}^{1}s^{X}ds$

$E(\frac{1}{X+1})=\int_{0}^{1}E(s^{X})ds$

$MGF=E(e^{tx})=(q+pe^t)^n;(p+q=1)$

Also

$E(s^{X})=(q+ps)^n$

Thus

$E(\frac{1}{X+1})=\int_{0}^{1}(q+ps)^nds=\dfrac{(q+ps)^{n+1}}{p(n+1)}\bigg]_{0}^{1}=\dfrac{(q+p)^{n+1}}{p(n+1)}-\dfrac{q^{n+1}}{p(n+1)}$ $=\dfrac{1}{p(n+1)}-\dfrac{q^{n+1}}{p(n+1)}$

score 1 · Answer 5 · answered Feb 18 '13 at 07:57

1

Law of the unconscious statistician: if $X$ has probability mass function $p(x)$, then $E[f(X)] = \sum_x p(x) f(x)$.

answered Feb 18 '13 at 07:57

Robert Israel

448,999

Find the expected value of $\frac{1}{X+1}$ where $X$ is binomial

5 Answers5

Linked

Related