Problem:
Let $X$ have a Poisson distribution with parameters $\lambda$. Use moment generating function to find the mean and variance
of $X$.
Answer:
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*}
Best Answer
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.