Suppose someone catches fish according to a Poisson process with rate 2 fish per hour. He quits fishing for the day whenever an hour has gone by since his last catch, or after 8 hours, whichever comes first. For example, if he catches his first fish after exactly 14 minutes but then no other fish during the first 74 minutes, then he goes home after 74 minutes with one fish. What is the probability that he has caught exactly 3 fish after 2 hours?
Since in order to catch 3 fish, he must catch the first and second fish so that the first two interarrivals are less than 1. Assuming the first two interarrivals are truncated exponentials with rate 2 that range from 0 to 1, I found the PDF of the sum of the two, which happens to be a piecewise function. The integrands in the two WolframAlpha links are the PDFs from 0 to 1 and then 1 to 2, and we see the integral is 1 for the sum of both parts.
If the time "a" of the second fish is between 0 and 1, we want probability of one fish coming in the next hour, and then no fish in the remaining time "1-a". If "a" is between 1 and 2, we want the probability of one fish coming in the remaining time "2-a". These probabilities can be found quickly using the Poisson distribution with rate 2t, where t is the length of the time interval. The integrals could be found below.
However, I simulated this question and find the answer to be around 0.142, while adding those two integrals and then even multiplying by the probability the first two interarrivals are less than 1, which is (1-e^(-2))^2, gives me an answer that is different.
How would I go about doing this more elegantly, since I would imagine there is a better solution that is actually correct? Sorry for the long post but wanted to provide my approach in case someone can tell me what is incorrect.
You need to evaluate the following probability: $$P(X_2 \leq 1, X_3 \leq 1,T_3\leq2,T_4>2)$$ Here, $X_1$ is the arrival time of the first fish, $X_2$ is the interarrival time between the first and second fish, $X_3$ is the interarrival time between the second and third fish, $T_3=X_1+X_2+X_3$ is the arrival time of the third fish, and $T_4$ is the arrival time of the fourth fish. Notice here we're allowing this Poisson process to continue even in the event that the fisherman leaves so we can discuss the arrival of the $4^{\text{th}}$ fish. Using the total law of probability we can express this probability as $$\int_0^2P(X_2\leq1,X_3\leq1,X_4>2-t|T_3=t)f_{T_3}(t)dt$$ where $X_4$ is the interarrival time between the third and forth fish and $f_{T_3}(t)=4t^2e^{-2t}$ is the pdf of $T_3$. Using independence we can simplify this expression a little bit: $$\int_0^2P(X_2 \leq 1,X_3 \leq 1|T_3=t)P(X_4>2-t)f_{T_3}(t)dt$$ We know $X_4\sim \text{Exponential}(2)$ so $P(X_4>2-t)=\int_{2-t}^{\infty}2e^{-2x}dx=e^{2t-4}$ and this probability becomes $$\frac{4}{e^4}\int_0^2t^2P(X_2\leq 1,X_3 \leq 1|X_1+X_2+X_3=t)dt$$ Clearly $P(X_2\leq 1,X_3\leq 1|X_1+X_2+X_3=t)=1$ whenever $t\leq 1$, but how do we handle the case when $t\in(1,2]$? Here's how I did it. Suppose $(X,Y,Z)\sim f_{XYZ}$ is a trivariate random vector supported on $(0,\infty)^3$ with pdf $$f_{XYZ}(x,y,z)=8e^{-2(x+y+z)}$$ Observe how $$f_{XYZ}(x,y,z)=f_{X_1}(x)f_{X_2}(y)f_{X_3}(z)$$ We can evaluate $P(Y\leq 1,Z\leq 1|X+Y+Z=t)$ using surface integration. When $t\in(1,2]$ this equals $$\frac{I_1 + I_2}{\int_0^t \int_0^{t-x}\sqrt{3}f(x,y,t-x-y)dydx}$$ where $$I_1=\int_0^{t-1} \int_{-x+t-1}^1\sqrt{3}f(x,y,t-x-y)dydx \\ I_2=\int_{t-1}^t \int_0^{-x+t}\sqrt{3}f(x,y,t-x-y)dydx$$ This evaluates to $\frac{4}{t}-\frac{2}{t^2}-1$ so the probability you catch three fish in two hours equals $$\frac{4}{e^4}\Bigg[\int_0^1 t^2dt + \int_1^2 (-t^2+4t-2)dt\Bigg]=\frac{8}{e^4}\approx 0.147$$