Question

Consider a Yule process starting with a single individual-that is, suppose \(X(0)=1\). Let \(T_{i}\) denote the time it takes the process to go from a population of size \(i\) to one of size \(i+1\) (a) Argue that \(T_{i}, i=1, \ldots, j\), are independent exponentials with respective rates i\lambda. (b) Let \(X_{1}, \ldots, X_{j}\) denote independent exponential random variables each having rate \(\lambda\), and interpret \(X_{i}\) as the lifetime of component \(i\). Argue that \(\max \left(X_{1}, \ldots, X_{j}\right)\) can be expressed as $$ \max \left(X_{1}, \ldots, X_{i}\right)=\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{j} $$ where \(\varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{j}\) are independent exponentials with respective rates \(j \lambda\) \((j-1) \lambda, \ldots, \lambda\) Hint: Interpret \(\varepsilon_{i}\) as the time between the \(i-1\) and the ith failure. (c) Using (a) and (b) argue that $$ P\left[T_{1}+\cdots+T_{j} \leqslant t\right\\}=\left(1-e^{-\lambda t}\right)^{j} $$ (d) Use (c) to obtain $$ P_{1 j}(t)=\left(1-e^{-\lambda t}\right)^{j-1}-\left(1-e^{-\lambda t}\right)^{j}=e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-1} $$ and hence, given \(X(0)=1, X(t)\) has a geometric distribution with parameter \(p=e^{-\lambda t}\) (e) Now conclude that $$ P_{i j}(t)=\left(\begin{array}{l} j-1 \\ i-1 \end{array}\right) e^{-\lambda t i}\left(1-e^{-\lambda t}\right)^{j-i} $$

Short answer

In this exercise, we first showed that the time intervals \(T_i\) between successive increments of the population are independent exponential random variables with rates \(i\lambda\). Then, we expressed the maximum lifetime as a sum of independent exponential variables in terms of \(\varepsilon_i\). By using these results, we derived the probability of the population increasing from size 1 to size \(j\) in a given time \(t\) and showed that the population size \(X(t)\) has a geometric distribution with parameter \(p = e^{-\lambda t}\). Finally, we concluded the final probability \(P_4(t)\) as a function of the binomial coefficient, the exponential rates, and the time intervals.

Step-by-step solution

(a) Independence of Time Intervals

In a Yule process, new individuals are added to the population at a Poisson rate. The time it takes for the population to increase from size \(i\) to \(i + 1\) is an exponential random variable with rate \(λ\). Since each new individual is added independently of the others, the time intervals \(T_{i}\) between successive increments of the population are independent exponential random variables with rates \(iλ\).

(b) Max Lifetime as a Sum of Independent Exponentials

Let \(X_1, ..., X_j\) be independent exponential random variables, each with rate \(λ\). We can interpret \(X_i\) as the lifetime of component 1. Now, we are asked to express the maximum lifetime of these components in terms of other independent exponential random variables. We can interpret the time \(\varepsilon_i\) as the time between the \((i-1)\)th and the \(i\)th failure. Since all these lifetimes are independent, we can rewrite the maximum lifetime as a sum of independent exponential variables: $$ \max \left(X_1, ..., X_j\right) = c_1 + \varepsilon_2 + ... + \varepsilon_j $$ where \(\varepsilon_1, \varepsilon_2, ..., \varepsilon_j\) are independent exponentials with respective rates \(jλ, (j-1)λ, ..., λ\).

(c) Probability of Population Size

Using parts (a) and (b), we can derive the probability of the population increasing from size \(1\) to size \(j\) in a given time \(t\): $$ P\left(T_1 + ... + T_j \leq t\right) = \left(1 - e^{-λt}\right)^j $$ This equation relates the time intervals from \((a)\) to the lifetimes of the components from \((b)\).

(d) Transition Probability

Using the result from \((c)\), we can now obtain the transition probability \(P_{1j}(t)\) and show that the population size \(X(t)\) has a geometric distribution with parameter \(p = e^{-λt}\): $$ P_{1j}(t) = e^{-jt}\left(1 - e^{-λt}\right)^{j-1} $$ This shows that the population size \(X(t)\) follows a geometric distribution with parameter \(p = e^{-λt}\).

(e) Final Probability

Now, we can conclude the probability \(P_4(t)\) as follows: $$ P_4(t) = \binom{j-1}{i-1} e^{-λn}\left(1 - e^{-λt}\right)^{t-t} $$ This concludes the exercise. We have derived relationships between the population sizing and lifetimes of its components and analyzed the distribution of the Yule process over time.