Problem source

The random variable $X$ takes the values $k=1$, $2$, $3$, $\dotsc$,
and has probability distribution 
$$
\P(X=k)= A{{{\lambda}^k\e^{-{\lambda}}} \over {k!}}\,,
$$
where $\lambda $ is a positive constant.
Show that $A = (1-\e^{-\lambda})^{-1}\,$. Find
the mean ${\mu}$ in terms of  ${\lambda}$  and show that
$$
\var(X) = {\mu}(1-{\mu}+{\lambda})\;.
$$
Deduce that ${\lambda} < {\mu} < 1+{\lambda}\,$.
Use a normal approximation to find
the value of $P(X={\lambda})$
in the case where ${\lambda}=100\,$, giving your answer to 2 decimal places.

Solution source

Let $Y \sim Po(\lambda)$
\begin{align*}
&& 1 &= \sum_{k=1}^\infty \mathbb{P}(X = k ) \\
&&&=  \sum_{k=1}^\infty A \frac{\lambda^k e^{-\lambda}}{k!}\\
&&&= Ae^{-\lambda} \sum_{k=1}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} \\
&&&= Ae^{-\lambda} \left (e^{\lambda}-1 \right) \\
\Rightarrow && A &= (1-e^{-\lambda})^{-1} \\
\\
&& \E[X] &=  \sum_{k=1}^{\infty} k \cdot \mathbb{P}(X=k) \\
&&&= A\sum_{k=1}^{\infty} k  \frac{\lambda^k e^{-\lambda}}{k!} \\
&&&= A\E[Y] = A\lambda = \lambda(1-e^{-\lambda})^{-1} \\
\\
&& \var[X] &= \E[X^2] - (\E[X])^2 \\
&&&= A\sum_{k=1}^{\infty} k^2  \frac{\lambda^k e^{-\lambda}}{k!} - \mu^2 \\
&&&= A\E[Y^2] - \mu^2 \\
&&&= A(\var[Y]+\lambda^2) - \mu^2 \\
&&&= A(\lambda + \lambda^2) - \mu^2 \\
&&&= A\lambda(1+\lambda) - \mu^2 \\
&&&= \mu(1+\lambda - \mu)
\end{align*}

Since $A > 1$ we must have $\mu > \lambda$ and since $\var[X] > 0$ we must have $1 + \lambda > \mu$ as required.

If $\lambda = 100$, then $A \approx 1$ and $P(X=\lambda) \approx P(Y = \lambda)$ and $Y \approx N(\lambda, \lambda)$ so the value is approximately $\displaystyle \int_{-\frac12}^{\frac12} \frac{1}{\sqrt{2\pi \lambda}} e^{-\frac{x^2}{2\lambda}} \d x \approx \frac{1}{\sqrt{200\pi}} = \frac{1}{\sqrt{630.\ldots}} \approx \frac{1}{25} = 0.04 $