Extracting time interval from poisson distribution given a probability

Question

Let's say I have a Poisson distribution with average of 3 events per hour. X ~ poisson(3). Consider this scenario:

Determine the amount of time it would take for no events to occur, and also have the probability during that time be less than .5.

My initial thoughts: convert the Poisson distribution to an exponential distribution, then solve for k such that P(X < k) = .5. That would give me the time up to P = .5.

Am I on the right track?

Answer 1

Your question highlights an important connection between Poisson and exponential distributions.

Let $X_t \sim Pois(3t)$ be the number of events seen in $t$ hours. Now let $Y \sim Exp(rate = 3).$ Then the following probabilities are equal:

$$P(\text{No event in } t \text{ hours}) = P(X_t = 0) = P(Y > t).$$

You seek $t$ so that they are all equal to 1/2. One way to do this is to find the median of $Y$. You can do this by setting the CDF $F_Y(t) = .5$ and solving for $t$.

In R statistical software, the inverse CDF, called the quantile function is denoted qexp. Thus the computation:

 qexp(.5, 3)
 ## 0.2310491

Note that if $X \sim Pois(3(0.321)) = Pois(0.693),$ then $P(X = 0) =. 500.$ In R dpois is the Poisson PDF:

 dpois(0, .693)
 ## 0.5000736