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{\bf Question}

Describe briefly what is meant by a linear birth-death process.

Amoeba,a single cell animal, reproduces itself by dividing into
two. A flask of water contains a number, $b$, of amoeba. The
probability that an amoeba divides into two in a time interval of
length $\delta t$ is $\lambda\delta t+o(\delta t)$, and the
probability that it dies is $\mu\delta t+o(\delta t)$. Let $p_n(t)
(n=0,\ 1,\ 2,\ \cdots)$ denote the probability that the flask
contains $n$ amoebae at times $t$, and $p_n'(t)$ denote its
derivative with respect to time. Show that

$p_n'(t)=\lambda(n-1)p_{n-1}(t)-(\lambda+\mu)np_n(t)+\mu(n+1)p_{n+1}(t),$

$n=1,\ 2,\ 3,\ \cdots .$

Suppose that the mean number of amoebae at time $t$ is

$$M(t)=\ds\sum_{n=0}^\infty np_n(t).$$

Show that $M(t)$ satisfies the differential equation

$$M'(t)=(\lambda-\mu)M(t),$$

and hence find $M(t)$.

If $W(t)$ denotes the mean of the square of the number of amoebae
at time $t$ prove that

$$W'(t)=2(\lambda-\mu)W(t)+(\lambda+\mu)M(t).$$

Explain, without performing any calculations, how the result could
be used to find the variance of the number of amoebae at time $t$.






\vspace{.25in}

{\bf Answer}

A linear birth-death process is a $s.p(X(t): t \geq 0)$ where
$X(t)$ is the number of individuals in the population at time $t$,
and where, in any time interval of length $\delta t$ each
individual has, independent of age and other individuals, a
probability $\lambda\delta t+o(\delta t)$ of producing a new
individual, and a probability $\mu\delta t+o(\delta t)$ of dying

$P(X(t+\delta t)=n+1\ |\ X(t)=n)=\lambda n \delta t+o(\delta t)$

$P(X(t+\delta t)=n-1\ |\ X(t)=n)=\mu n \delta t+o(\delta t)\ \
\rm{as}\ \delta t \to 0$

$P(X(t+\delta t)=n\ |\ X(t)=n)=1-(\lambda+\mu)n \delta t+o(\delta
t)$

${}$

\begin{eqnarray*} P_n(t+\delta t) & = & P(X(t+\delta t)=n)\\
& = & P(X(t+\delta t)=n\ |\ X(t)=n-1)P(X(t)=n-1)\\ & &
+P(X(t+\delta t)=n\ |\ X(t)=n+1)P(X(t)=n+1)\\ & & +P(X(t+\delta
t)=n\ |\ X(t)=n)P(X(t)=n)\\ & = & \lambda(n-1)\delta
tp_{n-1}(t)+\mu(n+1)\delta t p_{n+1}(t)\\ & & + (1-(\lambda+\mu)n
\delta t)p_n(t)+o(\delta t) \end{eqnarray*}

${}$

Thus

$\df{p_n(t+\delta t)-p_n(t)}{\delta t}=$

$\lambda(n-1)p_{n-1}(t)+\mu(n+1)p_{n+1}(t)+\mu(n+1)p_{n+1}(t)-(\lambda+\mu)np_n(t)$

Now $M(t)=\ds\sum_{n=0}^\infty n p_n(t)=\ds\sum_{n=1}^\infty n
p_n(t)$

so $M'(t)=\ds\sum_{n=1}^\infty n p_n'(t)$

$=\ds\sum_{n=1}^\infty
\lambda(n-1)np_{n-1}(t)+\ds\sum_{n=1}^\infty\mu(n+1)np_{n+1}(t)
-\ds\sum_{n=1}^\infty(\lambda+\mu)n^2p_n(t)$

$=\ds\sum_{n=0}^\infty \lambda
n(n-1)np_{n}(t)+\ds\sum_{n=0}^\infty\mu n (n-1)np_{n}(t)
-\ds\sum_{n=0}^\infty(\lambda+\mu)n^2p_n(t)$

$=\ds\sum_{n=0}^\infty p_n(t)[\lambda n^2+\lambda n+\mu n^2-\mu
n-\lambda n^2-\mu n^2]$

$=(\lambda-\mu)\ds\sum_{n=0}^\infty n p_n(t)=(\lambda-\mu)M(t)$

so $M'(t)=(\lambda-\mu)M(t)$

The general solution is $M(t)=Ae^{(\lambda-\mu)t}$

$X(0)=b$ so $M(0)=b$

Thus $M(t)=be^{(\lambda-\mu)t}$

Now $W(t)=\ds\sum_{n=1}^\infty n^2 p_n(t)$ so

$\begin{array}{rcl} W'(t) & = & \ds\sum_{n=1}^\infty n^2 p_n'(t)\\
& = & \ds\sum_{n=1}^\infty \lambda
n^2(n-1)p_{n-1}(t)+\ds\sum_{n=1}^\infty \mu n^2 (n+1)p_{n+1}(t)\\
& & -\ds\sum_{n=1}^\infty (\lambda+\mu)n^3 p_n(t)\\ & = &
\ds\sum_{n=0}^\infty \lambda
(n+1)^2n(n-1)p_{n}(t)+\ds\sum_{n=0}^\infty \mu (n-1)^2 np_{n}(t)\\
& &  -\ds\sum_{n=0}^\infty (\lambda+\mu)n^3 p_n(t)\\ & = &
\ds\sum_{n=0}^\infty p_n(t)[\lambda n^3+2\lambda n^2+\lambda n+\mu
n^3\\ & & -2\lambda \mu^2+\mu n-\lambda n^3-\mu n^3]\\ & = &
2(\lambda-\mu)\ds\sum_{n=0}^\infty
n^2p_n(t)+(\lambda+\mu)\ds\sum_{n=0}^\infty n p -n(t)\end{array}$

Thus $W'(t)=2(\lambda-\mu)W(t)+(\lambda+\mu)M(t)$

Since $M(t)$ is known, this is a linear 1st order equation which
can be solved for $W(t)$.

Then $Var(t)=W(t)-M9t)^2$.




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