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 1
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:
 * $$E \left({X^2}\right) = \sum_{x \in \Omega_X} x^2 \Pr \left({X = x}\right)$$

So:

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

Then:

$$ $$ $$

Proof 2
From Variance of Discrete Random Variable from P.G.F., we have:
 * $$\operatorname{var} \left({X}\right) = \Pi''_X \left({1}\right) + \mu - \mu^2$$

where $$\mu = E \left({x}\right)$$ is the expectation of $$X$$.

From the Probability Generating Function of Poisson Distribution, we have:
 * $$\Pi_X \left({s}\right) = e^{-\lambda \left({1-s}\right)}$$

From Expectation of Poisson Distribution, we have:
 * $$\mu = \lambda$$

From Derivatives of PGF of Poisson Distribution, we have:
 * $$\Pi''_X \left({s}\right) = \lambda^2 e^{- \lambda \left({1-s}\right)}$$

Putting $$s = 1$$ using the formula $$\Pi''_X \left({1}\right) + \mu - \mu^2$$:
 * $$\operatorname{var} \left({X}\right) = \lambda^2 e^{- \lambda \left({1-1}\right)} + \lambda - \lambda^2$$

and hence the result.

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