# Probability of Poisson event at time t vs. probability of event by time t

I am modeling three events A, B, and C as Poisson processes with rates $\lambda_A$, $\lambda_B$, and $\lambda_C$ and I would like to calculate the likelihood of observing some data given my model.

A distinction that I have come across is the difference between an event happening at a specific time $t$, $P\left(T\left(E\right) = t \right)$, versus by time $t$, $P\left(T\left(E\right) \leq t \right)$, where $T\left(E\right) = E_n - E_{n-1}$ refers to the inter-arrival time between two consecutive occurrences of the event. For a single event with rate $\lambda$, the probability of these scenarios are the exponential likelihood function

$P\left(T\left(E\right) = t \right) = \lambda e^{-\lambda t}$

and an integration over it

$P\left(T\left(E\right) \leq t \right) = \int_0^t \lambda e^{-\lambda \tau} \, \textrm{d}\tau = 1 - e^{-\lambda t}$

of the exponential distribution, respectively.

It follows that the odds of the event not occurring by $t$ is

$P\left(T\left(E\right)\gt t\right) = 1 - \left(1 - e^{-\lambda t}\right) = e^{-\lambda t}$.

Returning to my three-event model, the probability of observing event A at time $t$ but not events B nor C by that time is

$P\left(T\left(A\right) = t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right) = \lambda_A e^{-\lambda_A t} e^{-\lambda_B t} e^{-\lambda_C t} = \lambda_A e^{-\lambda t}$

where $\lambda = \lambda_A + \lambda_B + \lambda_C$, the overall rate of events.

However, I see two non-equivalent approaches to calculate the probability of observing event A by time $t$ but not events B nor C by that time, $P\left(T\left(A\right) \leq t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right)$.

The first method I see is to explicitly evaluate each expression, as I did previously; that is,

$P\left(T\left(A\right) \leq t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right) = \left(1 - e^{-\lambda_A t}\right) e^{-\lambda_B t} e^{-\lambda_C t} = e^{-(\lambda-\lambda_A)t} - e^{-\lambda t}$.

The second is to integrate over all possible times:

$P\left(T\left(A\right) \leq t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right) = \int_0^t \lambda_A e^{-\lambda \tau} \, \textrm{d}\tau = \frac{\lambda_A - \lambda_A e^{-\lambda t}}{\lambda}$.

Both approaches seem equally valid to me. Which calculation should I be using, if either, and why? Thanks in advance for your help; I really appreciate it.

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First of all, I can't see how the random variable $T(E)$ has $P(T(E)=t)$ non-zero, assuming continues time.

The first method I see nothing wrong with assuming these events are independent. Are they?

I can't see how you got the integral for the second method.

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Another way to consider $P\left(T\left(E\right)=t\right)$ is, what is the probability that $t$ time passes between $E_{n-1}$ and $E_n$? Because the inter-arrival times in a Poisson process are exponentially distributed, this is equivalent to the value of the exponential pdf at $t$. – Arman Aug 11 '13 at 16:51
Yes, the events are independent. The integral in the second method mimics previous calculations. For a single event, we know $P\left(T\left(E\right)=t\right)$, so to find the probability of that event occurring over a period of time we integrate over $t$, $\int_0^tP\left(T\left(E\right)=\tau\right)\,\textrm{d}\tau = 1 - e^{-\lambda t}$. I do the same thing; I know $P\left(T\left(A\right)=t\cap T\left(B\right)\gt t \cap T\left(C\right)\gt t\right)$, so I integrate over all possible times it could occur. See here for an single-event example. – Arman Aug 11 '13 at 17:00
If $T(E)$ is continues in $t$ then $P(T(E)=t)=\int_t^t\lambda e^{-\lambda x}dx=0$. For the second method you would need some joint probability distribution, which would reduce to the first method assuming independence. – Gummi F Aug 11 '13 at 19:33
Yes, the probability of the event happening in the interval $\left(t,t\right]$ is $0$ (and equivalent to the interval $\left(0,0\right]$ because the inter-arrival times are i.i.d.), but I want to calculate the probability that an inter-arrival time is $t$. Let me clarify my notation: I am interested in the $P\left(T\left(E_n\right)-T\left(E_{n-1}\right)=t\right)$, which is the exponential pdf. Regarding the second method, can you expand on why I would need a join probability distribution and how that would reduce to the first method? – Arman Aug 11 '13 at 20:33
If you have random variables $X$ and $Y$, then $P(X\in A)=\int_A f_X(x)dx$ and similarly $P(Y\in B)=\int_B f_Y(y)dy$. If $X$ and $Y$ are independent then $P(X\in A \cap Y\in B)=\int_A f_X(x)dx\int_B f_Y(y)dy$, if not then $P(X\in A \cap Y\in B)=\int_A \int_B f_{X,Y}(x,y)dydx$, where $f_{X,Y}$ is the joint probability distribution. – Gummi F Aug 12 '13 at 0:26

It turns out that the penultimate expression is correct:

$P\left(T\left(A\right) \leq t \cap T\left(B\right) \gt t \cap T\left(C\right) \gt t \right) = \left(1 - e^{-\lambda_A t}\right) e^{-\lambda_B t} e^{-\lambda_C t} = e^{-(\lambda-\lambda_A)t} - e^{-\lambda t}$.