Question

Consider a discrete time branching process \(\left\\{X_{n}\right\\}\) with probability generating function $$ \varphi(s)=\frac{1-(b+c)}{1-c}+\frac{b s}{1-c s}, \quad 01\). Assume \(X_{0}=1\). Determine the conditional limit distribution: $$ \lim \operatorname{Pr}\left\\{X_{n}=k \mid X_{n}>0\right\\} $$

Short answer

The conditional limit distribution for the given branching process is given by: \[ \lim_{n \to \infty} \operatorname{Pr}\left\{X_{n} = k \mid X_{n} > 0\right\} = \frac{\lim_{n \to \infty} \operatorname{Pr}\left\{X_{n} = k\right\}}{\lim_{n \to \infty} (1 - q_n)} \] where the numerator represents the \(k\)th coefficient of \(\varphi_{+}^{n}(s)\), and the denominator represents the extinction probability at the \(n\)-th step.

Step-by-step solution

Calculate the probability generating function for all non-zero outcomes

To calculate the probability generating function for all non-zero outcomes, first define the generating function for non-zero offspring in a single step: \[ \varphi_{+}(s) = \frac{\varphi(s) - (1 - b - c)}{b} \] Notice that this generating function represents the original generating function but without the probability of having \(0\) offspring. #Step 2: Define the extinction probability and the probability of extinction at step n#

Define the extinction probability and the probability of extinction at step n

The extinction probability, denoted by \(q\), is the probability that the branching process dies out eventually. Also, define the extinction probability at step \(n\), which is the probability of having no offspring at step \(n\), denoted by \(q_n\). For example, \(q_1 = 1 - b - c\). #Step 3: Calculate the extinction probability using the probability generating function#

Calculate the extinction probability using the probability generating function

The extinction probability \(q\) satisfies the equation \(\varphi(q) = q\). We can find the extinction probability by substituting \(\varphi(s)\) in the equation: \[ q = \frac{1 - (b + c)}{1 - c} + \frac{bq}{1 - cq} \] Solve for \(q\). #Step 4: Calculate the conditional limit distribution denominator#

Calculate the conditional limit distribution denominator

To calculate the required probability from the exercise, we need to condition on the event that \(X_{n} > 0\). Hence, the denominator for the conditional probability is \(1 - q_n\). #Step 5: Calculate the conditional limit distribution numerator#

Calculate the conditional limit distribution numerator

To calculate the numerator of the conditional probability, note that the probability generating function for the branching process conditioned on \(X_{n} > 0\) is given by: \[ \varphi_{+}^{n}(s) \] So, the numerator of the conditional limit distribution is the \(k\)th coefficient of \(\varphi_{+}^{n}(s)\), denoted by \(\operatorname{Pr}\left\{X_{n} = k\right\}\). #Step 6: Calculate the conditional limit distribution and the limit#

Calculate the conditional limit distribution and the limit

The conditional limit distribution, as requested in the exercise, will be the ratio of the numerator (step 5) and the denominator(step 4): \[ \operatorname{Pr}\left\{X_{n} = k \mid X_{n} > 0\right\} = \frac{\operatorname{Pr}\left\{X_{n} = k\right\}}{1 - q_n} \] Finally, finding the limit \(\lim n \to \infty\) of this expression will yield the required answer. This may be further investigated by considering the properties of \(\varphi_{+}(s)\) and taking the limit as \(n \to \infty\).