Discussion
Which of the following is true ? Suppose \(\{X(t), t \ge 0\}\) is a Yule-Furry process with birth rate \(\lambda\) and X(0) = 1. Then the distribution of X(t) is geometric with success probability p, where :
Answer: B
X(t) is geometric with \(p = e^{-\lambda t}\) and support {1, 2, 3, ....}.
For the Yule-Furry (pure birth) process with rate \(n\lambda\) in state n and \(X(0) = 1\), the forward equations give
\[ \begin{aligned} P[X(t) = n] &= e^{-\lambda t}\left(1 - e^{-\lambda t}\right)^{n-1}, \\ n &= 1, 2, \ldots \end{aligned} \]
which is the geometric distribution counting trials up to the first success, with success probability \(e^{-\lambda t}\). The mean is \(e^{\lambda t}\), the familiar exponential growth.
The population never drops below its starting size 1, so a support starting at 0 is wrong, and the options with \(p = 1 - e^{-\lambda t}\) swap success and failure.
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