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{\bf Question}

Suppose that $X_1,\ldots,X_n$ is a random sample from a Poisson
distribution with parameter $\lambda$.  Using the mgf approach
show that the sample sum $Y=\ds\sum_{i=1}^n X_i$ has a Poisson
distribution with parameter $n\lambda$.


\vspace{.25in}

{\bf Answer}

Let $X \sim {\rm Poisson}(\lambda)$ then mgf of $X$ is
$M_X(t)=e^{\lambda(e^t-1)},\ \ -\infty<t<\infty$

Here $Y=\ds\sum_{i=1}^{n} X_i$

\begin{eqnarray*} M_Y(t) & = & E(e^{tY})\\ & = &
E\{e^{t(X_1+...X_n)}\}\\ & = & E\left\{e^{tX_1}. e^{tX_2}\ldots
e^{tX_n}\right\}\\ & = & E\left(e^{tX_1}\right).
E\left(e^{tX_2}\right)\ldots E\left(e^{tX_n}\right)\ \ \ {\rm
(independence)}\\ & = & \left\{E(e^{tX_1})\right\}^n\\ & = &
\left\{e^{\lambda(e^t-1)}\right\}^n\\ & = & e^{n\lambda(e^t-1)}
\end{eqnarray*}

But this is the mgf of Poisson with parameter $n\lambda$.

Therefore $Y \sim {\rm Poisson}(n\lambda)$ by using the uniqueness
theorem of mgf.


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