Finding Expected Value of Conditional Poisson Distribution 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?

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