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My lecturer in his notes uses this definition of Poisson distribution:

$ P_{X}(t)= \exp( \lambda (t-1)) $

You differentiate once and set equal $t=1$ to get $E[X]=\lambda$, but in the notes to get $E[X^2]$, he doesn't differentiate $P_{X}(t) = \exp( \lambda (t-1))$ twice and set equal $t=1$.

Instead the formulae used is this E[X^2-X]=P_{X}''(1).

Can someone explain that? As it confusing the hell out of me. Like I don't know when to just differentiate twice and make it equal to $E[X^2]$ or I have to do that trick.

Or maybe you can't actually differentiate twice to get $E[X^2]$.

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    @Mike Oh thanks that makes perfect sense now.2011-11-04

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The formula $P_X(t)$ is your notes is the factorial moment generating function, defined as $ P_X(t) = \mathbb{E}( t^X ) $ Therefore: $ P_X^\prime(t) = \mathbb{E}(X t^{X-1} ) \qquad P_X^{\prime\prime}(t) = \mathbb{E}(X (X-1) t^{X-2} ) \qquad P_X^{(k)}(t) = \mathbb{E}( X(X-1) \cdots (X-k+1) t^{X-k} ) $ Setting $t=1$ afterwords makes $t^{X-k}$ factor equal 1 and you are left with expectations of linear combination of moments. Thus $ m_1 = P^\prime(1) \qquad m_2 = P_X^{\prime\prime}(1) + P_X^\prime(1) \qquad m_k = \sum_{m=0}^k \mathsf{S}^{(2)}_{k,m} P_X^{(m)}(1) $ where $\mathsf{S}^{(2)}_{k,m}$ is the Stirling number of the second kind.

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    @FanZhang Yes, this is exactly that. Notice that $P_X(\exp(u)) = M_X(u)$, where $M_X$ is the moment generating function.2011-11-04
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If you were thinking about the moment generating function rather than the factorial moment generating function it would look like this:

$M_{X}(t)=\exp( \lambda (e^t-1))$ is the moment generating function of a Poisson distribution, not $\exp( \lambda (t-1))$.

Differentiate it once to M'_{X}(t)=\lambda \exp(\lambda (e^t-1)+t) and set $t=0$ (not $1$) to get E[X]=M'_{X}(0)=\lambda.

Differentiate it again to M''_{X}(t)=\lambda^2 \exp(\lambda (e^t-1)+2t)+\lambda \exp(\lambda (e^t-1)+t) and set $t=0$ to get E[X^2]=M''_{X}(0)=\lambda^2+\lambda.

So $\operatorname{Var}(X)=E[X^2]-(E[X])^2=\lambda^2+\lambda-\lambda^2=\lambda$ as expected.

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    As Sasha points out, you are not using the moment generating function, but I will leave this here anyway as it may be of interest to somebody. $M_X(t)$ is the usual notation for the moment generating function2011-11-04