Finding the mean of the Poisson distribution using the moment generating function

Problem:
Let $$X$$ have a Poisson distribution with parameters $$\lambda$$. Use moment generating function to find the mean and variance of $$X$$.
For the Poisson distribution we have $$M_x(t) = e^{\lambda(t-1)}$$. Now, to find the mean, $$u$$, we compute $$M_x'(0)$$. \begin{align*} M_x'(t) &= \lambda e^{\lambda(t-1)} \\ M_x'(0) &= \lambda e^{\lambda(0-1)} = \lambda e^{-\lambda} \\ u &= \lambda e^{-\lambda} \end{align*} Where did I go wrong? The answer should be $$\lambda$$.

Here is a revised solution based upon the comments of Ekesh Kumar. I now correctly find the mean, but I get the wrong value for the variance. For the Poisson distribution we have $$M_x(t) = e^{\lambda(e^t-1)}$$. Now, to find the mean, $$u$$, we compute $$M_x'(0)$$. \begin{align*} M_x'(t) &= \lambda e^t e^{\lambda(e^t-1)} \\ M_x'(0) &= \lambda e^0 e^{\lambda(e^0-1)} = \lambda e^{\lambda(1-1)} \\ M_x'(0) &= \lambda \\ u &= \lambda \end{align*} Now to find $$E(x^2)$$, I find $$M''_x(0)$$. \begin{align*} M_x''(t) &= \lambda e^t ( \lambda e^t e^{\lambda(e^t-1) } ) + \lambda e^t \lambda e^t e^{\lambda(e^t-1)} \\ M_x''(0) &= \lambda e^0 ( \lambda e^0 e^{\lambda(e^0-1) } ) + \lambda e^0 \lambda e^0 e^{\lambda(e^0-1)} \\ M_x''(0) &= \lambda ( \lambda e^{\lambda(1-1) } ) + \lambda \lambda e^0 e^{\lambda(1-1)} \\ M_x''(0) &= \lambda ( \lambda e^{\lambda(1-1) } ) + \lambda \lambda e^{\lambda(0)} \\ M_x''(0) &= \lambda \lambda + \lambda \lambda = 2{\lambda}^2 \\ \sigma_2 &= M_x''(0) - u^2 = 2{\lambda}^2 - {\lambda}^2 \\ \sigma_2 &= {\lambda}^2 \end{align*}

• That's the PGF. It only tells you anything useful when you input $t=1$. – J.G. Jan 26 at 22:31
• @J.G. Are you saying to find the mean using a moment generating function, I should find $M'_x(1)$ not $M'_x(0)$? – Bob Jan 26 at 23:11
• MGF at $0$, PGF at $1$, but what you calculated is actually the PGF. – J.G. Jan 26 at 23:34

Let's clarify the difference between two approaches. The probability-generating function$$G_X(t):=\Bbb Et^X=\sum_{k\ge0}t^ke^{-\lambda}\frac{\lambda^k}{k!}=e^{-\lambda}e^{\lambda t}$$satisfies$$G_X^{(n)}(t)=\Bbb Et^X(X)_n=\lambda^ne^{-\lambda}e^{\lambda t}$$in terms of falling Pochhammer symbols, so$$\Bbb E(X)_n=G_X^{(n)}(1)=\lambda^n.$$In particular\begin{align}\Bbb EX&=\lambda,\\\Bbb E(X^2-X)&=\lambda^2,\\\Bbb EX^2&=\lambda^2+\lambda,\\\operatorname{Var}X&=\Bbb EX^2-(\Bbb EX)^2=\lambda^2+\lambda-\lambda^2=\lambda.\end{align}The moment-generating function$$M_X(t):=\Bbb Ee^{tX}=G_X(e^t)=e^{-\lambda}e^{\lambda e^t}$$satisfies$$M_X^{(n)}(t)=\Bbb EX^ne^{tX}=e^{-\lambda}\frac{d^n}{dt^n}e^{\lambda e^t}$$so$$\Bbb EXe^{tX}=M_X^\prime(t)=\lambda e^{-\lambda}e^te^{\lambda e^t}$$and$$\Bbb EX^2e^{tX}=M_X^{\prime\prime}(t)=\lambda e^{-\lambda}e^te^{\lambda e^t}\left(1+\lambda e^t\right).$$Hence$$\Bbb EX=M_X^{\prime}(0)=\lambda e^{-\lambda}e^0e^\lambda=\lambda$$and$$\Bbb EX^2=M_X^{\prime\prime}(0)=\lambda e^{-\lambda}e^0e^{\lambda}\left(1+\lambda\right)=\lambda^2+\lambda,$$so $$\operatorname{Var}X=\lambda$$ by the same logic as before.
• If $EX^2 = \lambda^2 + \lambda$ then I do not understand how you find that $Var X = \lambda$. – Bob Jan 27 at 17:40
• @Bob Your calculation of $M_X^{\prime\prime}(t)$ is wrong due to bungling the product rule. Let $u:=\lambda e^t,\,v:=e^{\lambda(e^t-1)}$; then $M_X^{\prime\prime}=uv^\prime+u^\prime v$, but you calculated $2uv^\prime$ (or perhaps you were thinking of $uv^\prime+u^\prime v^\prime$). – J.G. Jan 28 at 6:49
Your approach is right, but your moment generating function is wrong. It should be $$M_X(t) = e^{\lambda(e^t-1)}$$ as derived here.