Let $X_T$ is a number of events in Poisson process of unitary rate ($\lambda = 1$) within interval of length $T$. It is known that at least one event has been observed in the interval, I want to find probability that there is more events in the interval.

My intuition is that $\Pr(X_T > 1 \mid X_T > 0) = \Pr(X_T > 0)$.

The rationale behind is that

  1. if the observed event was at time $t$ from the beginning of the interval, then it is enough to calculate probability that no event occurred in either $(0, t)$ or $(t, T)$ open intervals: $\Pr(X_T = 1 \mid X_T > 0) = \Pr(X_t = 0) \Pr(X_{T-t} = 0) = e^{-t} e^{t - T} = e^{-T} = \Pr(X_T = 0)$,

  2. $\Pr(X_T > 1 \mid X_T > 0) = 1 - \Pr(X_T = 1 \mid X_T > 0) \\ \Pr(X_T > 0) = 1 - \Pr(X_T = 0) .$


\begin{align} & \Pr(X_T > 1 \mid X_T > 0) = \frac{\Pr(X_T > 1, X_T > 0)}{\Pr(X_T > 0)} = \frac{\Pr(X_T > 1)}{\Pr(X_T > 0)} \\[10pt] = {} & \frac{1 - \Pr(X_T \in \{1, 0\})}{1 - \Pr(X_T = 0)} = \frac{1 - Te^{-T} - e^{-T}}{1 - e^{-T}} , \end{align}

which neither I nor WolframAlpha can prove equal to $\Pr(X_T > 0) = 1 - e^{-T}$.

Since both results can not be true - where is my mistake?

I can see that $X_T > 1$ and $X_T > 0$ are heavily dependent. Does it matter? My intuition is that $X_T > 0$ is just narrowing the sampling space...

[EDIT #1]

I have found one more way to support... both results.

If $t$ is time of the first event in the interval (from the beginning of the interval), its distribution density will be given as $\operatorname{pdf}(t) = \frac{e^{-t}}{1 - e^{-T}}.$ Then $$\Pr(X_T = 1 \mid X_T > 0) = \int_0^T \operatorname{pdf}(t) \Pr(X_{T-t} = 0) \, dt = \int_0^T \frac{e^{-t}e^{t - T}}{1 - e^{-T}} \, dt = T\frac{e^{-T}}{1 - e^{-T}} .$$

However if I repeat similar steps for uniformly distributed ($\operatorname{pdf}(t) = \frac 1 T$) random event in the interval and take into account also events before $t$ I still get $$\Pr(X_T = 1 \mid X_T > 0) = \int_0^T \operatorname{pdf}(t) \Pr(X_t = 0) \Pr(X_{T-t} = 0) \, dt = \int_0^T \frac{e^{-T}} T \, dt = e^{-T} .$$

[EDIT #2]

Followup due to the comment of @combo (about loss of conditioning in the first approach).

I do not understand, why the conditioning is lost.

Imagine a situation where we create an interval of length $T$ with at least one event of unitary Poisson process in it. Let $Y$ is a random event of unitary Poisson process and $t$ is a random variable uniformly distributed in $(0,~T)$. Then $(Y~-~t,~Y~-~t~+~T)$ is an interval of length $T$ containing at least one event, at $t$ (uniformly distributed) from the beginning of the interval. From independence of the events, the probability that there is no more events in the interval is $\Pr(X_t == 0)\Pr(X_{T - t} == 0)$, isn't it? And it is given that there was at least one event in the interval.

Why the situation is different when I have an interval of length $T$ which contains at least one event? Time of a randomly chosen event ($t$; from the beginning of the interval) is uniformly distributed, so I see no difference.

  • $\begingroup$ Why do you state that $P(X_T = 1 \mid X_T > 0) = P(X_t = 0)P(X_{T-t} = 0)$? I do not see why this should be the case, and I this this is the source of your troubles. When I work it out for myself I get the same answer as your approach #2. $\endgroup$
    – combo
    Feb 8, 2017 at 22:08
  • $\begingroup$ @StefanJorgensen I see your point. However shouldn't it be $P(X_t=1)P(X_{T-t}=0) = te^{-t}e^{-(T-t)} = te^{-T}$ ? $\endgroup$
    – abukaj
    Feb 8, 2017 at 22:10
  • $\begingroup$ @StefanJorgensen answering your second comment: it is because I considered the event a point partitioning the interval into two subintervals of unknown number of events. $\endgroup$
    – abukaj
    Feb 8, 2017 at 22:17
  • $\begingroup$ @StefanJorgensen it is like answering a question: given there was an event at time $t$, what is the probability there was no event in either $(0, t)$ or $(t, T)$. $\endgroup$
    – abukaj
    Feb 8, 2017 at 22:28
  • $\begingroup$ Except that the equation $P(X_t = 0)P(X_{T-t}=0)$ is simply the probability that there is no event in either $(0,t)$ or $(t,T)$, and does not express your conditioning at all. This is why your answer ends up being simply the probability that there is at least one event (because conditioning was lost at the outset), when in fact the conditioning does change the result. $\endgroup$
    – combo
    Feb 8, 2017 at 22:47

3 Answers 3


I finally figured it out!

According to @combo's advice, I am going to use term "occurrence".

Surprisingly, the first part of the rationale for my intuition was almost correct.

If the observed occurrence was at time $t$ from the beginning of the interval, then it is enough to calculate probability that no occurrence occurred in either $(0, t)$ or $(t, T)$ open intervals: $\Pr(X_T = 1 \mid t) = \Pr(X_t = 0) \Pr(X_{T-t} = 0) = e^{-t} e^{t - T} = e^{-T} = \Pr(X_T = 0)$.

The difference is replacement of $\Pr(X_T = 1 \mid X_T > 0)$ by $\Pr(X_T = 1 \mid t)$, which may be viewed as $\Pr(X_T = 1 \mid X_{dt} = 1)$, where $dt$ is a length of infinitysmall interval $[t - \frac{dt}{2}, t + \frac{dt}{2}]$. Since $\Pr(X_T = 1, t) = e^{-T} dt$, we may see $\operatorname{pdf}(t) = \Pr(X_T = 1 \mid t)$ as a density in point $t$ of the only occurrence within the $(0, T)$ interval.

Since the value of $t$ is unknown, we integrate $\int_0^T \operatorname{pdf}(t) dt = Te^{-T} = \Pr(X_T = 1)$. So far so good - we know the unconditional probability of having exactly one occurrence in the interval. However by conditioning on $X_T > 0$ the $e^{-T}$ fraction of the sampling space has been discarded - thus it is necessary to normalize every remaining probability/density by a factor of $\frac{1}{1 - e^{-T}}$.

Thus, the intuition was fallacy of mistaking probability with density.


A quick note on terminology to avoid confusing ourselves: I will refer to something happening at a particular time as an "occurence" rather than an "event" in order to avoid confusion with the more rigorous definition of an event as an element of the sample space.

Let's start from the definition of the Poisson counting process. Let $N(t)$ be the number of occurrences that happen by time $T$. It has the following properties:

  1. $N(0) = 0$
  2. $N(T_1), (N(T_2)- N(T_1))$ are independent for $T_1 < T_2$ (independent increments property)
  3. The number of occurrences in any interval of length $t$ is a Poisson random variable with parameter $\lambda t$ (for our purposes, $\lambda = 1$).

The independent increments property is what is tripping you up - specifically the implications of the strict inequality.

We are asking the question, given that an occurrence happens at time $t \in [0,T]$, what is the probability that $N(T) = 1$? Following your approach, let's break the process into three segments. For some $\epsilon < \min\{t, T-t\}$, we have: $$N(T) = \left(N(T) - N(t+\epsilon) \right) + \left(N(t+\epsilon) - N(t-\epsilon) \right) + \left(N(t-\epsilon) - N(0)\right) $$ and now we are looking at the probability \begin{align*} &\lim_{\epsilon \to 0} P\left( N(T) = 1 \mid N(t+\epsilon) - N(t-\epsilon) = 1 \right)\\ &=\lim_{\epsilon\to 0} P\left( N(t-\epsilon) - N(0) = 0, N(T) - N(t+\epsilon) = 0 \mid N(t+\epsilon) - N(t-\epsilon) = 1\right) \end{align*} Note that this is not quite the same as what you wrote. There are two key differences:

  1. The intervals are not disjoint, so while $N(t-\epsilon) - N(0)$ is independent from $N(T) - N(t+\epsilon)$, neither are independent of $N(t+\epsilon) - N(t-\epsilon)$.
  2. The intervals all have positive measure. In your approach you break the interval $[0,T]$ into three pieces, $[0,t), [t], (t,T]$. The problem is that now you are conditioning on the event that an occurrence happens in the interval $[t]$, which has measure zero (for more details on why this is an issue, see this post).
  • $\begingroup$ 1. I am a bit confused about $\epsilon$ and $s$ in the last paragraph. Do you mean there is an interval overlap of size $\epsilon - s$? 2. It may be an effect of the confusion, but I fail to see a difference between $\lim_{s\to 0} P\left( N(t-s) - N(0) = 0, N(T) - N(t+s) = 0 \mid N(t+s) - N(t-s) = 1\right)$ and $\lim_{s\to 0} P\left( N(t-s) - N(0) = 0, N(T) - N(t+s) = 0\right)$ given that the occurrence at $t$ is given: $P\left( N(t+s) - N(t-s) = 1\right) = 1$. $\endgroup$
    – abukaj
    Feb 9, 2017 at 19:18
  • $\begingroup$ Sorry about that, I meant for them to be the same but my computer died while editing. The difference is precisely the source of your confusion: the event $\{N(t-\epsilon)-N(0) = 0, N(T)-N(t+\epsilon) = 0\}$ is not independent of the event $\{N(t+\epsilon)-N(t-\epsilon) = 1\}$ for $\epsilon > 0$. If you try to make $\epsilon = 0$ (with equality, not the limit), then the interval is a singleton and you are conditioning on a set with measure zero. $\endgroup$
    – combo
    Feb 10, 2017 at 0:58
  • 1
    $\begingroup$ Well, I do not see the dependence. However I managed to find the fallacy (see my answer) - and since interval $[t - \epsilon, t + \epsilon]$ was important, thus +1. $\endgroup$
    – abukaj
    Feb 10, 2017 at 4:35

Of your two ways of approaching the problem, the second one seems to be spot on. It is much more rigorous and makes complete sense to me. However, I had a little more difficulty in understanding the first approach, which gave reason to believe that this is the source of your mistake.

First of all, out of curiosity,why are you calculating $P(X_{T}=1|X_{T}>0)$? Given the very same approach you used below, this probability (as far as it is relevant) should equal $\frac{Te^{-T}}{1-e^{-T}}$ which is not by definition equal to $e^{-T}$. Hope this helps!

  • $\begingroup$ I still can not see my mistake in the first approach. I am calculating it since $P(X_T > 1| X_T > 0) = 1 - P(X_T = 1| X_T > 0)$ - see the edit. $\endgroup$
    – abukaj
    Feb 8, 2017 at 16:35

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