Is $E(\frac 1X|X>1)=E(\frac {1}{X+1}) $ where X follows binomial distribution?

Question

Please help with this question.
I tried to directly find $E(\frac 1X), X \sim bin(n,p)$.

Answer 1

No, they are not equal.

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Write

\begin{align} E[1/X\mid X>1]=\sum_{k=2}^n \frac1{k}\cdot P(X=k\mid X>1) =\sum_{k=2}^n \frac1{k}\cdot \frac{P(X=k)}{P(X>1)} \end{align} The $P(X>1)$ is just a constant which we can factor out. When you do, what is left is $$ E[1/X\mid X>1]=\frac1{P(X>1)}\sum_{k=2}^n \frac1k\binom{n}kp^k(1-p)^{n-k} $$ For simplicity, let $x=\frac{p}{1-p}$. Then we can write this as $$ E[1/X\mid X>1]=\frac{(1-p)^n}{P(X>1)}\sum_{k=2}^n \frac1k\binom{n}kx^k\tag1 $$ That last summation kind of looks like the binomial theorem, $(1+x)^n=\sum_{k=0}^n \binom{n}k x^k$. We can use this similarity to find a closed form for that sum. Rearrange the binomial theorem like this (using a different variable, $t$): $$ \frac{(1+t)^n-1-nt}{t}=\sum_{k=2}^n\binom{n}k t^{k-1} $$ Then, integrate both sides from $0$ to $x$ in $t$. $$ \int_0^x \frac{(1+t)^n-1-nt}t\,dt = \sum_{k=2}^n\binom{n}k \frac1k x^{k}\tag2 $$ Notice how integration causes the $1/k$ to appear. As long as you can carry out the integration in $(2)$, then you can plug that formula in to $(1)$ to get the answer.