Question

If x~P(λ) show that var(x)= λ without using moment generating function of X ?

Answer 1

`\mbox{var}(X) = E(X^2) - E^2(X)`. If `X ~ P(\lambda), E(X) = \lambda`. So `E^2(X)=\lambda^2`.

`E(X^2) = \sum_{n=0}^{\infty} k^2 1/(k!) \lambda^k e^(-\lambda)=\lambda^2 + \lambda` following the same steps for the formula of the expected value.

So `\mbox{var}(X) = E(X^2) - E^2(X) = \lambda^2 + \lambda - \lambda^2 = \lambda.`