3.9 Branching Process

Consider a population which consists of individuals that are able to produce offsprings of the same kind. Assume that each individual will by the end of its life time have produce \(j\) new offsprings with probability \(P_j, \, j\geq 0\) independently of the numbers produced by other individuals. Further assume that \(P_j < 1\, \, \forall \, j > 0\). Let \(X_0\) denote the size of the \(0^{\text {th}}\) generation. i. e number of individuals present at \(t = 0\). All offsprings of the \(0^{\text {th}}\) generation constitute the first generation. Let \(X_n, \, n = 0, \, 1, \, 2, \, \cdots \) denote the size of the \(n^{\text {th}}\) generation. Hence \(\{X_n , \, n = 0, \, 1, \, 2, \, \cdots \}\) is a Markov chain with state space \(\{0, \, 1, \, 2, \, \cdots \}\).

Useful Derivation For The Markov Chain \(\{X_n, \, n = 0, \, 1, \, 2, \, \cdots \}\)
Let \(P_{ij}\) be the transition probability from state \(i\) to state \(j\)

(i).: \(P_{00} = 1\)
(ii).: Let \(P_0 > 0\), then there is a positive probability that no offspring exist to an individual. Hence all states become transient because they transit to a recurrent (absorbing) state \(E_0\). If \(P_0 > 0\), and there are \(i\) individuals the \(n^{\text {th}}\) generation, then the probability \(\left (P_0\right )^i\) that non of the \(i\) individual produced and offspring.
Hence \(P_{i0} > 0\) implying state \(E_i, \hspace {0.2cm} i = 0, \, 1, \, \cdots \) are transient.
(iii).: Expected size of the \(n^{\text {th}}\) generation. Let \[\mu = \sum ^{\infty }_{j = 0} j\, P_j\] denote the expected number of offsprings of an individual. Let \(X_0 = 1.\)
Let \(Z_i\) denote the number of offsprings of the \(i^{\text {th}}\) individual from the \((n - 1)^{\text {th}}\) generation. \[X_n = Z_1 + Z_2 + \cdots + Z_{n - 1}\] \begin {align*} E(X_n) & = E\left (E(X_n/X_{n - 1})\right )\\ & = E\left (E\left (Z_1 + Z_2 + \, \cdots \, + Z_{n - 1} \,/\, X_{n - 1}\right )\right )\\ & = E\left (E(Z_1) + E(Z_2) + \, \cdots \, + E(Z_{n - 1}) \, / X_{n - 1}\right )\\ & = E\left (\mu \, X_{n - 1}\right )\\ & = \mu \, E(X_{n - 1}). \end {align*}
\[\therefore \hspace {0.3cm} E(X_n) = \mu \, E(X_{n - 1}).\] Since \(X_0 = 1\) then \(\hspace {0.3cm}\begin {aligned}[t] E(X_1) & = \mu \\ E(X_2) & = \mu ^2\\ \vdots & \\ E(X_n) & = \mu ^n. \end {aligned}\)
(iv).: If \(0< \mu < 1\), then the population will die out eventually (i.e the population will become extinct). \[\lim _{n\rightarrow \infty } E(X_n) = \lim _{n \rightarrow \infty } \mu ^n = 0.\]
(v).: The probability of extinction \begin {align*} \mu ^n & = E(X_n) \\ & = \sum ^{\infty }_{j = 0} j\, P(X_n = j)\\ & = \sum ^{\infty }_{j = 1} j\, P(X_n = j)\\ & > \sum ^{\infty }_{j = 1} P(X_n = j)\\ & = 1 - P(X_n = 0). \end {align*}
If \(\mu < 1\), \[\lim _{n\rightarrow \infty } \mu ^n \geq \lim _{n\rightarrow \infty } (1 - P(X_n = 0))\] \[\lim _{n\rightarrow \infty } P(X_n = 0) = 1.\] Hence population will become extinct eventually.

Theorem 3.9.1 (Probability of Extinction given \(X_0 = 1\)). \(X_0 = 1\) implies that the offsprings of the individuals of the \(i^{\text {th}}\) generation constitute the \(1^{\text {st}}\) generation. Hence \[P(X_1 = j) = P_j\hspace {0.2cm} , \hspace {0.5cm} j = 0, \, 1, \, 2, \, \cdots \]

Let \(\pi _0\) denote the probability that the population eventually die out \begin {align*} \pi _0 = & P\left (\text {population eventualy dies out}\right )\\ = & \sum ^{\infty }_{j =0}P\left (\text {population eventualy dies out}\, |\, X_1 = j\right )\, P(X_i = j)\\ & \text {using}\hspace {0.3cm} P(E) = \sum _y P(E/Y = y)\, P(Y = y)\\ & = \sum ^{\infty }_{j = 0} P\left (\text {population eventualy dies out}\, |\, X_1 = j\right )\, P_j\\ & = \sum _{j = 0}^{\infty } \pi _0^j\, P_j \end {align*}

since each of the \(j\) individuals (population) of the \(1^{\text {st}}\) generation is dying independent with probability \(\pi _0\).

Example 3.9.2. Find \(\pi _0\) given that \[P_0 = \frac {1}{2}\, , \hspace {0.3cm} P_1 = \frac {1}{4}\, , \hspace {0.4cm} P_2 = \frac {1}{4}\]

Solution. \(\pi _0 = \pi _0^0\, (P_0) + \pi ^1_0\, (P_1) + \pi _0^2\, (P_2)\) \[\pi _0 = \frac {1}{2} + \pi _0 \left (\frac {1}{4}\right ) + \pi _0^2\left (\frac {1}{4}\right )\] \[4 \pi _0 = 2 + \pi _0 + \pi _0^2\] \[\pi _0^2 - 3\pi _0 + 2 = 0\] \[(\pi _0 - 2)(\pi _0 - 1) = 0\] \[\pi _0 = 1.\] □

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