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 {\bf Question}

 Suppose that a random variable $X$ has the pmf

$$f(x)=C\frac{e^{-\lambda}\lambda^x}{x!},\ \
x=1,2,....,\lambda>0.$$

 Note that X takes values 1,2,3, .... Find $C$ and
$E(X)$.


\medskip

 {\bf Answer}

Since $X$ is discrete and $f(x)$ is a pmf we must have

\begin{eqnarray*} & & \ds \sum_{x=1}^{\infty} f(x)=1\\ &
\Rightarrow & \ds \sum_{x=1}^{\infty}
c\frac{e^{-\lambda}\lambda^x}{x!} = 1\\ & \Rightarrow &
ce^{-\lambda}\left\{\ds \sum_{x=1}^{\infty}
\frac{\lambda^x}{x!}+\frac{\lambda^0}{0!}-\frac{\lambda^0}{0!}
\right\} = 1\\
 & \Rightarrow & ce^{-\lambda}\left\{e^{-\lambda}-1\right\} = 1\\
 & \Rightarrow & c\left\{1-e^{-\lambda}\right\} = 1\\
 & \Rightarrow & c = \frac{1}{1-e^{-\lambda}} \end{eqnarray*}

\begin{eqnarray*}
E(X) & = & \frac{e^{-\lambda}}{1-e^{-\lambda}}\ds
\sum_{x=1}^{\infty} \frac{x\lambda^x}{x!}\\ & = &
\frac{e^{-\lambda}}{1-e^{-\lambda}}\ds \sum_{x=1}^{\infty} \frac
{\lambda^x}{(x-1)!}\\ & = & \frac{e^{-\lambda}}{1-e^{-\lambda}}
\lambda \ds \sum_{x=1}^{\infty} \frac{\lambda^{x-1}}{(x-1)!}\\ & =
& \frac{e^{-\lambda}}{1-e^{-\lambda}} \lambda \ds
\sum_{t=0}^{\infty} \frac{\lambda^t}{t!}\\ & = &
\frac{e^{-\lambda}}{1-e^{-\lambda}} \lambda e^{\lambda}\\ & = &
\frac{\lambda}{1-e^{-\lambda}}.
\end{eqnarray*}


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