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?
Thanks!
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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, $f(X=k;\lambda)=\frac{(\lambda t)^k}{k!}\,e^{-\lambda t}$.
answered Jan 4 '16 at 16:44
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$\begingroup$ 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
