I got a task and I have no idea how to solve it:
Task: During the time 'T' number of events 'A' is a random variable distributed according to Poisson's law with the parameters $$a*T$$
Find a probability, that the time between two events 'A' will be not more than 't'.
Having: a = 0.13; t = 1.37
My solution:
I know that Poisson distribution is commonly used to model rate of random events that occur in some fixed time interval. In my opinion, we can use a 'exponential distribution' with a parameter 'lambda' = a is less than or equal to t:
P(time between two events 'A'<= t) = P(Exponential(lambda = a) <= t)
Thus,
$$P(Exponential(lambda = a) <= t) = 1 – e^{(-at)}$$
Then, substituting the values of a and t, we get:
$$P(time between two events 'A' <= 1.37) = 1 – e^{(-0.13 * 1.37)} = 0.163141$$
I am not sure is my solution correct.
Best Answer
The way to properly argue is as below:-
$$\{\text{Time between two events is less than}\, t\}=\{\text{No events in time period }\, t\}^{c} $$ .
Now if the number of events in time period $T$ (call it $N(T)$) is Poisson distributed with mean $a\cdot T$ .
Then the probability that there is no event in time period $t$ is $P(N(t)=0)=\exp({-a t})$
Thus the $$P(\{\text{Time between two events is less than}\, t\})=1-P(N(t)=0)=1-\exp(-a t)$$