Is there any relationship between λ (or μ) in Poisson and Exponential distribution? In other words, if I know λ (or μ) for one of the distributions, is it possible to calculate the corresponding value in the other distribution?



The exponential distribution models the time between events, while the Poisson is used to represent the number of events within a unit of time. Both distributions are a function of the rate parameter, $\lambda$.

The mean of the exponential distribution is $\frac{1}{\lambda}$ and can be expressed in time units (e.g. $\text {sec}$). $\lambda$ corresponds to the mean in the matching Poisson distribution, and is the expected number of events per unit of time, which would be expressed in inverse time units (e.g. $1/\text {sec}$).

Provided that the Poisson distribution makes reference to $1$ time unit, the rate parameter $\lambda$ is identical in both distributions.

So, if the pdf of the exponential is $f(X=t;\lambda)=\lambda \, e^{-\lambda x}$, the pmf of the Poisson modeling the same experiment will be $f(X=k;\lambda)=\frac{\lambda^k}{k!}\,e^{-\lambda}$. And if the question makes reference to the number of events in other than the time unit, $f(X=k;\lambda)=\frac{(\lambda t)^k}{k!}\,e^{-\lambda t}$.

  • $\begingroup$ All of this is correct in the context of a Poisson process. The exponential and Poisson distributions also arise in other contexts where there's no particular connection between them. $\endgroup$ – Brian Borchers Jan 4 '16 at 19:13

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