Showing 1-1 of 1 problems
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.
Show Solution
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 \(\displaystyle \int_{-\frac12}^{\frac12} \frac{1}{\sqrt{2\pi \lambda}} e^{-\frac{x^2}{2\lambda}} \d x \approx \frac{1}{\sqrt{200\pi}} = \frac{1}{\sqrt{630.\ldots}} \approx \frac{1}{25} = 0.04 \)