Finding Expected Value of Conditional Poisson Distribution

Question

Consider a random variable X ~ Poisson (1). Namely, $P(x=k) = \frac{e^{-1}}{k!}$ , k=0,1,2,...

I'm trying to solve for $\mathbb{E}\{X|X\geq 1\}$.

My approach:

Given that $X\geq1$ then we know that at least one person has arrived (using arrival / no arrival terminology). Using the fact that Poisson is memoryless then $\mathbb{E}\{X\}=\mathbb{E}\{Y+1\}$ where $Y$ has the same distribution as above Poisson distribution. The solution afterwards is really simple.

Is that approach correct?

Answer 1

You can use the equality: $$\mathbb EX=\mathbb E[X\mid X\geq1]P(X\geq1)+\mathbb E[X\mid X=0]P(X=0)$$leading to:$$1=\mathbb E[X\mid X\geq1](1-P(X=0))=\mathbb E[X\mid X\geq1](1-e^{-1})$$

hence to:$$\mathbb E[X\mid X\geq1]=\frac{1}{1-e^{-1}}=\frac{e}{e-1}$$

Answer 2

It is $\frac {\sum\limits_{k=1}^{\infty} ke^{-1} \frac1 {k!}} {1-P(X=0)}$ and $P(X=0)=e^{-1}$. Can you compute the sum?

The answer is $\frac e {e-1}$.