Problem source

The staff of Catastrophe College are paid a salary of $A$ pounds per year.
With a Teaching Assessment Exercise impending it is decided to try to lower the student failure rate  by offering each lecturer an alternative salary of $B/(1+X)$ pounds, where $X$ is the number of his or her students who fail the end of year examination. Dr Doom has $N$ students, each with independent probability $p$ of failure. Show that she should accept the new salary scheme if
$$A(N+1)p < B(1-(1-p)^{N+1}).$$
Under what circumstances could $X$, for Dr Doom, be modelled by a Poisson random variable?
What would Dr Doom's expected salary be under this model?

Solution source

\begin{align*}
&& \E[\text{salary}] &= B\sum_{k=0}^N \frac{1}{1+k}\binom{N}{k}p^k(1-p)^{N-k} \\
\\
&& (q+x)^N &= \sum_{k=0}^N \binom{N}{k}x^kq^{N-k} \\
\Rightarrow && \int_0^p(q+x)^N \d x &= \sum_{k=0}^N \binom{N}{k} \frac{p^{k+1}}{k+1}q^{N-k} \\
&& \frac{(q+p)^{N+1}-q^{N+1}}{N+1} &= \frac{p}{B} \E[\text{salary}] \\
\Rightarrow && \E[\text{salary}] &= B\frac{1-q^{N+1}}{p(N+1)}
\end{align*}

Therefore if $Ap(N+1) < B(1-(1-p)^{N+1})$ the expected value of the new salary is higher. (Whether or not the new salary is worth it in a risk adjusted sense is for the birds).

We could model $X$ by a Poisson random variable if $N$ is large and $Np = \lambda $ is small. Suppose $X \approx Po(\lambda)$ then \begin{align*}
\E \left [\frac{B}{1+X} \right] &= B\sum_{k=0}^\infty \frac{1}{1+k}\frac{e^{-\lambda}\lambda^k}{k!} \\
&= \frac{B}{\lambda} \sum_{k=0}^\infty e^{-\lambda} \frac{\lambda^{k+1}}{(k+1)!} \\
&= \frac{B}{\lambda}e^{-\lambda}(e^{\lambda}-1) \\
&= \frac{B(1-e^{-\lambda})}{\lambda} = B \frac{1-e^{-Np}}{Np}

\end{align*}