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Here's a proof of $\mathrm{Var}(X) = \lambda$ for the Poisson distribution.

Proof: First we work out $E(X(X-1))$

$$E(X(X-1)) = \sum_Xx(x-1)f(x)$$

$$= \sum_{x = 0}^{\infty}x(x-1)e^{-\lambda}\frac{\lambda^x}{x!}$$

Start from $x = 2$

$$= \lambda^2 e^{-\lambda} \sum_{x = 2}^{\infty}\frac{\lambda^{x-2}}{(x-2)!}$$

Letting $y = x - 2$ we have

$$= \lambda^2 e^{-\lambda} \sum_{y = 0}^{\infty}\frac{\lambda^y}{y!}$$

$$= \lambda^2 e^{-\lambda} e^\lambda$$

$$= \lambda^2$$


$$\mathrm{Var}(X) = EX^2 - (EX)^2$$

$$= E(X(X - 1) + EX - (EX)^2$$

$$=\lambda^2 + \lambda - \lambda^2$$



It's the line after "Start from $x = 2$" that I have an issue with. As far as I can see we have to start from $x = 2$ because otherwise we will have a negative factorial in the denominator which is undefined. But shouldn't that mean the equation itself is undefined, rather than just 'dropping' the undefined variables in the summation and continuing on as we do in the above proof? Why are we allowed to drop the undefined variables in the summation?

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You start at $x=2$ because $x(x-1)e^{-\lambda}\lambda^x/x!$ is $0$ for $x=0,1$, so you might as well not include them. Note that $x!$ is in fact well-defined for $x=0,1$ (it is equal to $1$). – Stefan Hansen Nov 27 '12 at 14:00
Without being at all mathematically rigorous (which is wrong for math.stackexchange), you would need at least two elements to have a variance, wouldn't you? – user50600 Nov 27 '12 at 14:06
@nvlass I am not sure your comment applies here. – Did Nov 27 '12 at 22:22

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