pdf of arrival time I have a hw question that says: Consider a Poisson counting process with arrival rate $λ = 1.$ Suppose it is observed that there is exactly one arrival in the time interval $[0,t_0).$ Find the PDF of that arrival time.
Can you help me understand what the question is asking! Is it just asking what is $P(N(t_0) = 1) = \lambda  t_0 e^{-\lambda t_0}$  (because $P(N(t) = k) = (\lambda  t)^k e^{-\lambda t}/k!$
 A: I think this must mean the conditional p.d.f. of the arrival time given the event that there was exactly one arrival.
Let (capital) $T$ be the time of the first arrival; let $N(t)$ be the number of arrivals at or before time (lower-case) $t.$
Suppose $0\le t\le t_0.$
Then
\begin{align}
& \Pr(T\le t\mid N(t_0)=1) \\[8pt]
= {} & \Pr( N(t_0)- N(t)=0\mid N(t_0)=1) \\[8pt]
= {} &\frac{\Pr( N(t_0)-N(t)=0\ \&\ N(t_0)=1) }{\Pr(N(t_0)=1)} \\[8pt]
= {} & \frac{\Pr( N(t)=1\ \&\ N(t_0) - N(t) =0)}{\Pr(N(t_0)=1)} \\[8pt]
= {} & \frac{\Big(e^{-t\lambda} (t\lambda)^1/1!\Big) \cdot \Big(e^{-\lambda(t_0-t)} (\lambda(t_0-t))^0/0!\Big)}{e^{-\lambda t_0} (\lambda t_0)^1/1!} \\[8pt]
= {} & t/t_0.
\end{align}
This is the c.d.f. Differentiate to get the p.d.f. Then you see that the conditional distribution of the arrival time, given that there was exactly one arrival, is uniform in the interval $[0,t_0).$
PS: Notice that the bottom-line answer is the same regardless of which positive number $\lambda$ is.
A: I agree with Michael Hardy that you are looking for the conditional pdf of the arrival time, given that there was one arrival. Intuitively, this should be uniform. Another way to see this is the following:
Suppose we divide the interval $[0, t_0)$ into $n$ sub-intervals of length $\frac{t_0}{n}$. This makes our interval into a mesh. By the Poisson setup the probability of exactly one arrival in an interval is $\lambda\frac{t_0}{n} + o(\frac{t_0}{n})$. Note that our arrival is equally likely to occur in any sub-interval, and thus is in a discrete uniform distribution. Sending $n \rightarrow \infty$ sends our mesh to a continuum and gives us a uniform distribution.
