Calculating distribution of time until first event in a poisson process

Question

Let crashes of a stock market be modelled as a Poisson process with rate 0.1 per year and let $N_t$ denote the number of crashes by time t. Then $N_t$ has a Poisson distribution with mean $0.1t$.

For $0<s<t$, calculate $P(N_s=0 \vert N_t=1)$. What is the distribution of the time until the first crash given $N_t=1$.

My solution for this is as follows:

\begin{align} P(N_s=0 \vert N_t=1) &= \frac{P(\{N_s=0\} \cap \{N_t=1\} )}{P(\{N_t=1\})} \\ &=\frac{P(\{N_{t-s}=0\})}{P(\{N_t=1\})} \\ &= \frac{e^{-0.1(t-s)}}{0.1t\cdot e^{-0.1t}}\\ &= \frac{e^{0.1s}}{0.1t} \end{align}

I do not know what to do past this point, consequently I cannot find the distribution required.

Answer 1

Hints:

• $P(\{N_s=0\} \cap \{N_t=1\} )$ is the probability is the probability of zero events from time $0$ to $s$ and one event from time $0$ to $t$; these are not independent
• Instead, try saying $P(\{N_s=0\} \cap \{N_t=1\} )$ is the probability is the probability of zero events from time $0$ to $s$ and one event from time $s$ to $t$ (a time period of $t-s$; these are independent so you can find the probability of each and multiply them
• $P(\{N_t=1\} )$ is the probability is the probability of zero events from time $0$ to $t$

You can calculate these from knowing over a time period $t$ saying "$N_t$ has a Poisson distribution with mean $0.1t$" and so the result from $P(N_s=0 \vert N_t=1) = \dfrac{P(\{N_s=0\} \cap \{N_t=1\} )}{P(\{N_t=1\})}$.

With tidying and cancellation, the final answer is remarkably simple

Your calculation is incorrect. $\dfrac{P(\{N_s=0\} \cap \{N_t=1\} )}{P(\{N_t=1\})} \not =\dfrac{P(\{N_{t-s}=0\})}{P(\{N_t=1\})}$. Instead $\dfrac{P(\{N_s=0\} \cap \{N_t=1\} )}{P(\{N_t=1\})} =\dfrac{P(\{N_s=0\}\,P(\{N_{t-s}=1\})}{P(\{N_t=1\})}$