Consider a homogeneous Poisson process with inter-arrival times $T_i$, which follows the exponential distribution with rate $\lambda$. Let $N(t)$ denote the number of arrivals by time $t$.

Suppose I start monitoring the arrivals from time $t$ onwards. Let $U_t$ denote the time I need to wait for an arrival. Now suppose I know $N(t)=1$, and I'd like to show that $$ P(U_t>x|N(t)=1)=e^{-\lambda x}. $$

Here's what I've tried: I expressed the event $\{N(t)=1\}$ as $\{T_1<t,T_1+T_2>t\}$ and the event $\{U_t>x\}$ as $\{T_1+T_2>t+x\}$. Then I did: $$ \frac{P(T_1+T_2>t+x,T_1+T_2>t,T_1<t)}{P(N(t)=1)}=\frac{P(T_1<t)-P(T_1+T_2<t+x)}{P(N(t)=1)}, $$ where the distributions of $T_1,T_1+T_2,N(t)$ are all known. But after calculations I got something far away from an exponential distribution. I think my calculations were okay. Then it must be something conceptually wrong I assume? Please help!

  • $\begingroup$ Is $U_t$ the time you need to wait for an arrival starting at time $t$ or the time you need to wait for an arrival from the last arrival, given that the second arrival didn't occur before time $t$? (Alternatively, if $t=1$,the first arrival occurs at time 0.3 and the second at time 1.5, does $x=0.5$ or does $x=1.2$?) $\endgroup$
    – jbowman
    Sep 23, 2018 at 2:23
  • $\begingroup$ @jbowman $U_t$ is the time you need to wait starting at time $t$. So basically $x=0.5$ in your second explanation. $\endgroup$
    – ywa136
    Sep 23, 2018 at 14:34

1 Answer 1


The second expression (RHS) is wrong, the difference of probabilities are signalling a negative number possibility there. What if $x$ is a very large number? Then, you nearly are subtracting $P(T_1+T_2<\infty)$ from $P(T_1<t)$ which will yield a negative number. Apart from this, you have two ways (maybe more) to solve this problem.

Let's start with your way. Your interpretation of $\{N(t)=1\}$ and $\{U_t>x\}$ are correct. Your LHS expression can also be written as: $$P(T_1+T_2>t+x|T_1+T_2>t, T_1<t) = \frac{P(T_1+T_2>t+x,T_1<t)}{P(T_1+T_2>t,T_1<t)}$$

$T_1$ and $T_2$ are independent exp. RVs with joint density $\lambda^2e^{-\lambda(t_1+t_2)}$, i.e. multiplication of marginals. For the denominator, just draw a 2D plot with axes $t_1,t_2$, draw a line $t_1+t_2=t$; we are going to integrate the joint PDF in $\{T_1<t \cup T_1+T_2>t\}$, i.e. in English, {the region above the line you draw} U {x axis smaller than $t$} U{between $x,y$ axes}, which boils down to the following integral: $$\int_{0}^{t}{\int_{t-t_1}^{\infty}{\lambda^2e^{-\lambda(t_1+t_2)}dt_2dt_1}}=\lambda te^{-\lambda t}$$ For the numerator, we draw the line $t_1+t_2=t+x$ and take the region above, instead of $t_1+t_2=t$, and that only changes the inner integrals lower bound to $t+x-t_1$, which results in $\lambda te^{-\lambda (t+x)}$. When, you take the ratio, we're left with $e^{-\lambda x}$, which is your answer.

The easy way: You wait for an event to happen for $y=t-t_1$ seconds/mins or whatever. You don't know $t_1$, but that doesn't matter, there exist a $t_1$. And, you wonder if you're going to wait for additional $x$ secs/mins. Since your RV is exponential, which has the memoryless property, i.e. it doesn't depend on how long it has been since its start or $P(X>a+b|X>a)=e^{-\lambda b}$ mathematically, probability of your additional waiting time, $U_t$, only depends on $x$, i.e. $P(U_t>x|N(t)=1)=e^{-\lambda x}$. By the way, $N(t)=1$ carries no important information. It could also be $N(t)=n$, and your answer wouldn't change.

  • $\begingroup$ Thanks for the thorough explanation! It's clear to me now. I guess I can write $P(T_1+T_2>t,T_1<t)$ into $P(T_1<t)-P(T_1+T_2<t)$. But that's not true in general, and I'll have to do the integral to compute the region. Also, the second method is indeed much easier and more intuitive. Thank you again! $\endgroup$
    – ywa136
    Sep 23, 2018 at 14:32

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