# Calculating variance of poisson distributed random variable

I am calculating variance of a Poisson distributed random variable with mean $$\lambda$$. I am doing it in the following way:

$$\mathbb{V}(X) = \mathbb{E}(X^2) - \lambda^2 \\ = \sum_{x\geq 0} \quad x^2\frac{e^{-\lambda} \quad\lambda^x}{x!} - \lambda^2\\ = 0 + \lambda e^{-\lambda} + \sum_{x\geq 2} \quad x^2\frac{e^{-\lambda} \quad\lambda^x}{x!}-\lambda^2\\ = \lambda e^{-\lambda} + e^{-\lambda}\lambda^2\sum_{x\geq 2} \quad \frac{\lambda^{x-2}}{(x-2)!}\\ =\lambda e^{-\lambda} + \lambda^2 -\lambda^2\\ =\lambda e^{-\lambda}$$

The answer is wrong. But I cannot understand what am I doing wrong in breaking the summation on line 3.

Edit: shown step 4 to answer a comment

• How do you get that sum over $x \geq 2$ to be equal to $\lambda^2$? – jbowman Feb 10 at 21:29
• I have added a step. – anotherone Feb 10 at 21:37
• Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x \geq 2$, which is not true. Also, your $-\lambda^2$ term disappears on line 4 and then reappears on line 5. – Alex Feb 10 at 21:52
• Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further. – StatsStudent Feb 10 at 21:55
• @Alex I totally missed that! The $\lambda^2$ is a typo though. – anotherone Feb 10 at 21:59

$$\mathbb{E}((X)_r) = \mathbb{E} \bigg( X(X-1) \cdots (X-r+1) \bigg) = \sum_{x=0}^\infty x(x-1) \cdots (x-r+1) \cdot \frac{\lambda^x}{x!} e^{-\lambda}.$$
Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $$X$$. These are all polynomials in $$X$$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $$r=2$$.