Variance of Poisson Distribution

Theorem
Let $$X$$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.

Then the variance of $$X$$ is given by:
 * $$\operatorname{var} \left({X}\right) = \lambda$$

Proof
From the definition of Variance as Expectation of Square minus Square of Expectation:
 * $$\operatorname{var} \left({X}\right) = E \left({X^2}\right) - \left({E \left({X}\right)}\right)^2$$

From Expectation of Function of Discrete Random Variable:
 * $$E \left({X^2}\right) = \sum_{x \in \operatorname{Im} \left({X}\right)} x^2 \Pr \left({X = x}\right)$$

So:

$$ $$ $$ $$ $$ $$ $$ $$

Then:

$$ $$ $$

Comment
The interesting thing about the Poisson distribution is that its expectation and its variance are both equal to its parameter $$\lambda$$.