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{\bf Question}

The following is a simple model for the spread of an epidemic
through an infinite population.  In a small time interval $(t, t +
\delta t]$ each infected individual has, independently of other
individuals, a chance $\lambda \, \delta t$ of infecting one
healthy individual and a chance $1 - \lambda \, \delta t$ of
infecting no-one. He has a chance of recover which increases
linearly with calendar time and is given by $\mu t \delta t$.
$\lambda$ and $\mu$ are both positive constants.

Show that the probability $p_n(t)$, that there are $n$ infected
individuals at time $t$, satisfies the differential-difference
equation $$p_n'(t) = (n-1)\lambda p_{n-1}(t) - n(\lambda +\mu
t)p_n(t) + \mu t(n+1)p_{n+1}(t) \, \, n = 1, 2,\ldots$$

By constructing a differential equation, or otherwise, show that
the expected number of infected individuals at time $t,\ \  M(t),$
is $$M(0) e^{(\lambda t - \frac{1}{2} \mu t^2)}.$$ Describe the
behaviour of $M(t)$ as $t$ varies.



\vspace{.25in}

{\bf Answer}

Let $p_n(t) = P(N(t)=n)$ where $N(t)$ = no. of infected patients.

\begin{eqnarray*} p_n(t+\delta t) & = &P(N(t+\delta t) = n | N(t)
= n-1) P(N(t)=n-1)\\ & + & P(N(t+\delta t)=n|N(t) = n) P(N(t) =
n)\\& + & P(N(t+\delta t)=n|N(t) = n+1) P(N(t) = n+1)\\
\\ & = & (n-1)\lambda \delta t (1-\mu t(n-1)\delta t)p_{n-1}(t) \\
& & + (n \lambda \delta t n \mu t \delta t + (1 - n\lambda \delta
t)(1 - n \mu t \delta t))p_n(t) \\ & & + ((1 - (n+1)\lambda\delta
t)\mu t(n+1)\delta t) p_{n+1}(t) + o(\delta t) \end{eqnarray*}

\begin{eqnarray*} \frac{p_n(t+\delta t) - p_n(t)}{\delta t} & = &
(n-1) \lambda p_{n-1}(t) - p_n(t)n(\lambda+\mu t) \\ & &
+p_{n+1}(t) \mu t(n+1) + \frac{o(\delta t)}{\delta t}
\end{eqnarray*}

Thus $\ds p_n'(t) = (n-1)\lambda p_{n-1}(t) - n(\lambda + \mu
t)p_n(t) + \mu t(n+1)p_{n+1}(t)$

\begin{eqnarray*} M(t) & = & \sum_{n=1}^\infty np_n(t)
\hspace{.2in} {\rm so} \hspace{.2in} M'(t) = \sum_{n=1}^\infty
np'_n(t)
\\ & = & \lambda \sum_{n=1}^\infty n(n-1)p_{n-1}(t) - (\lambda +
\mu t) \sum_{n=1}^\infty n^2 p_n(t) \\ & & + \mu t
\sum_{n=1}^\infty n(n+1)p_{n+1}(t) \\ & = & \lambda
\sum_{n=1}^\infty (n+1)np_n(t) - (\lambda + \mu
t)\sum_{n=1}^\infty n^2p_n(t)\\ & & +\mu t
\sum_{n=1}^\infty(n-1)np_n(t) \\
 & = & \sum_{n=1}^\infty
p_n(t) n(\lambda - \mu t) \\ & = & (\lambda - \mu  t) M(t)
\end{eqnarray*}


i.e. $$\ds M'(t) = (\lambda - \mu t) M(t);\ \ \  M(t) =
M(0)\exp\left(\lambda t - \frac{\mu t^2}{2}\right)$$

$M(t)$ has a maximum of $\ds
M(0)\exp\left(\frac{\lambda^2}{2\mu}\right)$ when $\ds
t=\frac{\lambda}{\mu}.$



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