# Link between Poisson and Exponential distribution

From my set of notes:
"If a Poisson distribution has an average rate of $r$, then the waiting time is exponential with mean $\frac{1}{r}$. When talking about the Poisson distribution we'd be inclined to say "$\lambda=rt$. while when talking about the exponential distribution we'd be inclined to say $\lambda=\frac{1}{r}$."

My question is: Is this correct? Should the last formula say $\lambda=t\frac{1}{r}$? If not, then why not? I get $\lambda=t\frac{1}{r}$ since $\lambda$ is equal to the rate per time (is this correct?), hence we simply multiply $\frac{1}{r}$ by $t$

• It depends on what $r$ says, considering the alternative parametrization. I believe that the second $\lambda$ in the quote refers to using it as the parameter for an exponential distribution, not a Poisson. Nov 20, 2016 at 21:19
• @O.VonSeckendorff I see what you mean. But if so, is that parameter for an exponential distribution affected by $t$ in any way? Nov 20, 2016 at 21:41
• $t$ is the r.v. for the exponential, unless you do something funky. Nov 20, 2016 at 22:39
• @O.VonSeckendorff I thought $t$ was a constant. Not a r.v. Nov 20, 2016 at 23:19
• Depends on the perspective. How you can you interpret an exponential with a given $t$? Nov 20, 2016 at 23:27

I think you are pretty much on the right track. Here is my favorite way to look at the link between Poisson and exponential distributions:

Suppose the number $N_t$ of random events that occur in time interval $(0, t]$, where $t > 0,$ has the distribution $Pois(\lambda t).$ Thus, the probability of seeing no events by time $t > 0$ is $P(N_t = 0) = e^{-\lambda t}.$ [Notice that when dealing with a Poisson distribution over a time interval, it is necessary to coordinate the rate $\lambda t$ with the length of the interval.]

Another way to specify that there are no events in the interval $(0, t]$ is to let $W$ be the waiting time, starting at time $t = 0,$ until we see the first event. Then $$P(W > t) = 1 - F_W(t) = P(N_t = 0) = e^{-\lambda t},$$ for $t > 0.$

According to the usual notation, $F_W$ is the CDF of $W.$ By differentiation of the CDF we get the density function of $W:$ $$f_W(t) = F_W^\prime(t) = \lambda e^{-\lambda t},$$ for $t > 0$ (and 0 elsewhere). We say that $W$ has an exponential distribution with rate $\lambda.$

Using integration by parts, one can show that $E(W) = \int_0^\infty tf_W(t)\,dt = 1/\lambda.$ [Some texts and software parameterize the exponential distribution in terms of the rate $\lambda$ and others in terms of the mean $\mu = 1/\lambda.$]

Thus starting with the distribution of the discrete Poisson distribution of $N_t,$ we have found the distribution of the continuous random variable $W.$

As a specific example: let $\lambda = 2$ per minute. Then the probability of no events within $(0, 0.5]$ minutes is $P(N = 0),$ where $N \sim Pois(1).$ In R statistical software, this is:

 dpois(0, 1)            # 'dpois' is Poisson PDF
## 0.3678794


It is also $P(W > 0.5),$ where $W \sim Exp(rate\,\, \lambda = 2).$ In R we have:

 1 - pexp(.5, rate=2)   # 'pexp' is exponential CDF
## 0.3678794


Simulating a million realizations of $W$ and averaging them, we have a good estimate of the exponential mean $E(W):$

 w = rexp(10^6, rate=2)  # 'rexp' simulates exponential realizations
mean(w);  sd(w);  mean(w > 0)
## 0.5003201  # aprx E(W) = 1/2
## 0.5000453  # aprx SD(W) = E(W)
## 0.368334   # aprx P(W > 0), fraction of w's exceeding 0


It is not difficult to show that an exponential random variable has variance $1/\lambda^2 = \mu^2,$ so that $SD(W) = 1/\lambda = \mu.$

Acknowledgment: This discussion is much that same as one finds at the beginning of Ch 4 in Suess and Trumbo (2010), Springer.