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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?

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  • $\begingroup$ Poisson distribution is not memoryless. $\endgroup$ Commented Aug 6, 2020 at 12:17
  • $\begingroup$ Sorry I meant the process $\endgroup$ Commented Aug 6, 2020 at 12:25

2 Answers 2

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

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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 $\frac e {e-1}$.

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  • $\begingroup$ You are using the formula of conditional probabilities right ? Denominator is clear its P(X>=1) which is 1-P(X=0) which is the evaluation of our distribution at X=0. For the numerator P(X=x and X>=1) how do you get that though? $\endgroup$ Commented Aug 6, 2020 at 12:00
  • $\begingroup$ @HasanHammoud $E(X|X\geq1 )=\frac {E(X 1_{X\geq 1})} {P(X \geq 1)}$. For the numerator multiply the values $X=k$ ($k \geq 1$ )by their probabilities and sum over $k$. $\endgroup$ Commented Aug 6, 2020 at 12:02
  • $\begingroup$ But when you are working in the conditional space we have a new probability mass function don't we ? You are using the same pmf in the expectation sum as that of X but in the conditional space its different isnt it ? $\endgroup$ Commented Aug 6, 2020 at 12:09
  • $\begingroup$ @HasanHammoud The two pmfs are equal up to the multiplicative constant $P(X\geq 1)$, which is exactly the denominator. $\endgroup$ Commented Aug 6, 2020 at 12:10
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    $\begingroup$ @HasanHammoud You seem to think that under condition $X\geq1$ random variable $X-1$ has Poisson distribution. This however is not the case. Just explore $$P(X-1=k\mid X\geq1)=\frac{P(X=k+1)}{P(X\geq1)}=\frac{P(X=k+1)}{1-P(X=0)}$$ to see that. $\endgroup$ Commented Aug 6, 2020 at 12:44

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