Given a Poisson distributed random variable with parameter $\lambda$ that take the values $0,1,\ldots$ Show that mean and variance both equal to $\lambda$.
I differentiated the Taylor series and then tried to proved but I am not able to figure it out. I am stuck what to do after differentiation. Please help me. How to solve this question?
Best Answer
since, for $n \in \mathbb{N}$ $$ p_n = e^{-\lambda}\frac{\lambda^n}{n!} $$ form the probability generating function: $$ f(x) = \sum_{n=0}^{\infty}p_nx^n= \sum_{n=0}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!}x^n = e^{\lambda(x-1)} $$ now we have for the first and second moments: $$ \mu_1 = \sum_{n=0}^{\infty}np_n=f'(1) = \lambda \\ \mu_2 = \sum_{n=0}^{\infty}n^2p_n=\frac{d}{dx}(xf'(x))|_{x=1}=f'(1)+f''(1) =\lambda+\lambda^2 $$ so finally, for the standard deviation $$ \sigma^2=\mu_2-\mu_1^2 = \lambda $$