Problem source

The maximum height $X$ of flood water each year on a certain river is a random variable with probability density function $\f$ given by
\[
\f(x) = \begin{cases}
\lambda \e^{-\lambda x} & \text{for $x\ge0$}\,, \\

0 & \text{otherwise,}
\end{cases}
\]
where $\lambda$ is a positive constant.

It costs $ky$ pounds each year to prepare for flood water of height $y$ or less, where $k$ is a positive constant and $y\ge0$.  If $X \le y$ no further costs are incurred but if $X> y$ the additional cost of flood damage is  $a(X - y )$ pounds where $a$ is a positive constant. 
\begin{questionparts}
\item Let $C$ be the total cost of dealing with the floods in the year. Show that the expectation of $C$ is given by 
\[\mathrm{E}(C)=ky+\frac{a}{\lambda}\mathrm{e}^{-\lambda y} \, .
\]
How should $y$ be chosen in order to minimise $\mathrm{E}(C)$, in the different cases that arise according to the value of $a/k$?

\item Find  the variance of $C$,  and show that the more that is spent on preparing for flood water in advance the smaller this variance.

\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ \begin{align*}
&& \mathbb{E}(C) &= \int_0^\infty \text{cost}(x) f(x) \d x \\
&&&= ky + \int_y^{\infty} a(x-y) \lambda e^{-\lambda x} \d x\\
&&&= ky + \int_0^{\infty} a u \lambda e^{-\lambda u -\lambda y} \d x \\
&&&= ky + ae^{-\lambda y} \left( \left [ -ue^{-\lambda u} \right]_0^\infty -\int_0^\infty e^{-\lambda u} \d u\right) \\
&&&= ky + \frac{a}{\lambda}e^{-\lambda y} \\
\\
&& \frac{\d \mathbb{E}(C)}{\d y} &= k - ae^{-\lambda y} \\
\Rightarrow && y &= \frac{1}{\lambda}\ln \left ( \frac{a}{k} \right)
\end{align*}

Since $\mathbb{E}(C)$ is clearly increasing when $y$ is very large, the optimal value will be $\frac{1}{\lambda}\ln \left ( \frac{a}{k} \right)$, if $\frac{a}{k} > 1$, otherwise you should spend nothing on flood defenses.

\item \begin{align*}
&& \mathbb{E}(C^2) &= \int_0^{\infty} \text{cost}(x)^2 f(x) \d x \\
&&&= \int_0^{\infty}(ky + a(x-y)\mathbb{1}_{x > y})^2 f(x) \d x \\
&&&= k^2y^2 + \int_y^{\infty}2kya(x-y)f(x)\d x + \int_y^{\infty}a^2 (x-y)^2 f(x) \d x \\
&&&= k^2y^2 + \frac{2kya}{\lambda}e^{- \lambda y}+a^2e^{-\lambda y}\int_{u=0}^\infty u^2 \lambda e^{-\lambda u} \d u \\
&&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y}+a^2e^{-\lambda y}(\textrm{Var}(Exp(\lambda)) + \mathbb{E}(Exp(\lambda))^2\\
&&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} \\
&& \textrm{Var}(C) &=  k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} - \left ( ky + \frac{a}{\lambda} e^{-\lambda y}\right)^2 \\
&&&= a^2e^{-\lambda y} \frac{2}{\lambda^2} - a^2 e^{-2\lambda y}\frac{1}{\lambda^2} \\
&&&= \frac{a^2}{\lambda^2} e^{-\lambda y}\left (2 - e^{-\lambda y} \right) \\
\\
&& \frac{\d \textrm{Var}(C)}{\d y} &= \frac{a^2}{\lambda^2} \left (-2\lambda e^{-\lambda y} +2\lambda e^{-2\lambda y} \right) \\
&&&= \frac{2a^2}{\lambda} e^{-\lambda y}\left (e^{-\lambda y}-1 \right) \leq 0
\end{align*}

so $\textrm{Var}(C)$ is decreasing in $y$.

\end{questionparts}