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{\bf Question}

A population consists of $b$ individuals.  During a time interval
$(t, t + \delta  t]$ each individual in the population has,
independently of all the other individuals and of what happened on
$[0,t],$ a probability $\mu \delta t + o (\delta t)$ of dying.
Thus the population decreases from $b$ to 0.  Let $X(t)$ be the
size of the population after time $t$ and $p_n(t)$ denote the
probability that $X(t)= n \ \ (n = 0,1,\ldots,b).$  Show that:
\begin{description}
\item[(i)] $P\{X(t + \delta t) = n - 1| X(t) = n\} = n \mu \delta t
+ o(\delta t),$ and

$P\{X (t + \delta t) = n | X(t) = n \} = 1 - n \mu \delta t+
o(\delta t)$ as $\delta t \rightarrow 0$
\item[(ii)] $p_b'(t) = -b\mu p_b(t).$
\item[(iii)] $p_n'(t) = -p_n(t)n \mu + p_{n+1}(t)(n+1) \mu, $ for $n =
0,1,\ldots, b-1.$
\item[(iv)] the generating function, $G(z,t),$ of $X(t)$ satisfies the
differential equation $$ \frac{\pl G}{\pl t} = \mu (1-z)\frac{\pl
G}{\pl z}$$
\item[(v)] $\ds G(z,t) = e^{-\mu tb} (z + e^{\mu t} - 1)^b.$
\end{description}
Hence find the distribution of $X(t)$ and comment on whether or
not this result is surprising.  The process $\{X(t) ; t \geq 0\}$
is called a \emph{linear death process.}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(i)]
$\ds P(X(t + \delta t) = n-1 | X(t) = n) = P$(only 1 out of $n$
individuals dies) \begin{eqnarray*} &amp; = &amp; n[\mu \delta t +
o(\delta t)][1- \mu \delta t + o(\delta t)]^{n-1} \\ &amp; = &amp; n \mu
\delta t + o(\delta t) \hspace{.2in} {\rm as\ } \delta t \to 0 \\
&amp; &amp; \left[P(X(t + \delta t) = n | X(t) = n) = P({\rm no\
individual\ dies})\right] \\ &amp; = &amp; [1 - \mu \delta t + o(\delta
t)]^n = 1 - n \mu \delta t + o(\delta t) \hspace{.2in} {\rm as\ }
\delta t \to 0
\end{eqnarray*}

\item[(ii)]
$\ds P_b(t+ \delta t) = P(X(t) = b$ and no individual dies in $(t,
t + \delta t])$

$= p_b(t)[1-b \mu \delta t + o(\delta t)]$ by independence (Markov
property)

Thus $\ds P_b'(t) = -b \mu P_b(t)$



\item[(iii)]
for $n = 0, 1, \ldots, b-1$

\begin{eqnarray*} p_n(t + \delta t) &amp; = &amp; P(X(t) = n {\rm \ and\ no\
deaths\ in\ }(t, t + \delta t]) \\ &amp; &amp; + P(X(t) = n+1 {\rm \ and\
1\ death\ in\ }(t, t + \delta t]) \\ &amp; &amp; + P(X(t) &gt; n+1 {\rm \
and\ &gt; 1\ death\ in\ }(t, t + \delta t]) \\ &amp; = &amp; p_n(y)[1 - n \mu
\delta t + o(\delta t)] \\ &amp; &amp; +p_{n+1}(t)[(n+1)\mu \delta t +
o(\delta t)] + o(\delta t) \end{eqnarray*}

Thus $\ds p_n'(t) = -n \mu p_n(t) + (n+1) \mu p_{n+1}(t)$


\item[(iv)] $\ds G(z,t) = \sum_{n-0}^\infty p_n(t)z^n$

\begin{eqnarray*}\frac{\pl G}{\pl t} &amp; = &amp; \sum_{n=0}^\infty
p_n'(t)z^n \\ &amp; = &amp; \sum_{n=0}^{b-1}-n\mu p_n(t)z^n +
\sum_{n=0}^{b-1}(n+1)\mu p_{n+1}(t) z^n - b \mu p_b(t)z^b \\ &amp; = &amp;
-\sum_{n=1}^b n \mu p_n(t)z^n + \sum_{n=1}^bn \mu p_n(t)z^{n-1} \\
&amp; = &amp; \mu(1 - z) \frac{\pl G}{\pl z} \end{eqnarray*}




\item[(v)]Let $\ds G(z, t) = e^{-\mu t b}(z + e^{\mu t} - 1)^b$

\begin{eqnarray*}\frac{\pl G}{\pl t} &amp; = &amp; -\mu b e^{-\mu t b}(z +
e^{\mu t} - 1)^b + e^{-\mu t b}b(z + e^{\mu t} - 1)^{b-1}\mu
e^{\mu t} \\ &amp; =&amp; \mu be^{-\mu t b}(z+e^{\mu t} - 1)^{b-1}(e^{\mu
t} - (z+e^{\mu t}-1)) \\ &amp; = &amp; \mu (1-z) be^{-\mu t b}(z + e^{\mu
t} - 1)^{b-1} \\ &amp; = &amp; \mu(1-z) \frac{\pl G}{\pl z}
\end{eqnarray*}

Now $p_b(0) = 1$ and $p_n(0)= 0$ for $n \not= b$ so $G(z, 0)$
should be $z^b$, which is the case.

To find $p_n(t)$ we need to expand $G(z,t)$ as a polynomial in
$z.$ The coefficient of $z^n$ is \begin{eqnarray*} p_n(t) &amp; = &amp;
\left(
\begin{array}{c} b \\ n \end{array} \right)e^{-\mu t b}(e^{\mu t}
-1)^{b-n} \\ &amp; = &amp; \left( \begin{array}{c} b \\ n \end{array}
\right)e^{-\mu t n}(1 - e^{\mu t})^{b-n} \end{eqnarray*}

So $X(t) \sim B(b,e^{-\mu t})$ - binomial

$G(z, t) = (e^{-\mu t}z + (1 - e^{-\mu t}))^b$ - binomial p.g.f.

Each of the $b$ members of the population has a probability
$e^{-\mu t}$ of surviving longer than time $t$, independently  of
the others.

So we have a binomial situation, with survival being a Bernouilli
trial.


\end{description}


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