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?
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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 $$ \sigma^2=\mu_2-\mu_1^2 = \lambda $$
There are many ways. Since you mention differentiation, let's do it that way. We will calculate the variance, assuming that the mean is $\lambda$. The argument for calculating the mean is similar to the one below, but simpler, and we leave it to you.
Let our random variable $X$ have Poisson distribution with parameter $\lambda$. We start from the familiar Maclaurin series $$e^{\lambda}=1+\frac{\lambda}{1!}+\frac{\lambda^2}{2!}+\frac{\lambda^3}{3!}+\frac{\lambda^4}{4!}+\frac{\lambda^5}{5!}+\cdots.$$ Differentiate twice, and multiply by $\lambda^2e^{-\lambda}$. We get $$\lambda^2=(2)(1)e^{-\lambda}\frac{\lambda^2}{2!}+(3)(2)e^{-\lambda}\frac{\lambda^3}{3!}+(4)(3)e^{-\lambda}\frac{\lambda^4}{4!}+(5)(4)e^{-\lambda}\frac{\lambda^5}{5!}+\cdots.$$ We recognize the right-hand side as $E(X(X-1))$. So $E(X^2)=\lambda^2+\lambda$, and therefore $\text{Var}(X)=E(X^2)-(E(X))^2=\lambda^2+\lambda-\lambda^2=\lambda$.