Lambda - Exponential vs. Poisson Interpretation

Question

I'm trying to understand $\lambda$'s role in both the Poisson and Exponential Distributions and how it is used to find probabilities (yes, I have read the other post regarding this topic, didn't quite do it for me).

What (I think) I understand:

1. Poisson Distribution -

   • discrete

   • $\lambda$ is defined as the average number of successes (however "success" is defined, given problem context) per unit of time or space

   • PMF: $~~P(X = k;\lambda) = \frac{ \lambda^ke^{-\lambda} }{k!} $

   • $P(X\leq k) = P(X = 0)~+~P(X = 1)~+~P(X = 2)~+~\ldots~+~P(X = k)$

   • $P(X< k) = P(X = 0)~+~P(X = 1)~+~P(X = 2)~+~\ldots~+~P(X = k~-~1)$

   • $P(X\geq k) = 1~-~P(X<k)$

   • $P(X > k) = 1~-~P(X\leq k)$

2. Exponential Distribution -

   • continuous

   • $\lambda$ is defined as the average time/space between events (successes) that follow a Poisson Distribution

   • Where my understanding begins to fade:

   • PDF: $~~f(x; \lambda)~=~ \lambda e^{-\lambda x} $

   • CDF: $ P(X \leq k; \lambda)~=~1~-~e^{-\lambda x} $

   • $ P(X > k; \lambda) ~=~ 1 ~-~P(X \leq k; \lambda)~=~e^{-\lambda x}$

Where I think the misunderstanding lies:

As of now I'm assuming $\lambda$ can be interchanged between the two distributions. Is this the case? I have briefly read about, "re-parameterizing" and I think that might be the key, but I don't know what that process is referring to. How do I do this and how does it affect the PMF and CDF of the exponential distribution?

This all came from a problem asking: Given a random variable X that follows an exponential distribution with lambda = 3, find P(X > 8). My approach was $ e^{-3*8} $, which gives a probability that seems far too low.

Answer 1

Suppose I am waiting for a bus at a stop. And suppose that a bus usually arrives at the stop in every 10 mins. Now I define λ to be the rate of arrival of a bus per minute. So, λ = (1/10).

Now I want to compute the probability that no bus arrives in the next minute. I can do it using both the poisson and the exponential distribution.

Poisson

λ = 1/10

Probability of 0 arrivals in the next minute: P(X = 0) = 0.9048

Exponential

λ = 1/10

Probability of my having to wait for more than 1 minute: P(X>1) = 0.9048

Note: Look at the expected values of both the distributions. For Poisson, we get that the average number of buses arriving per minute E(X) = λ = 0.10 buses. For exponential, the average waiting time for a bus to arrive E(X) = (1/λ) = 10 mins

Answer 2

The fact that both of these distributions use the same parameter is probably a coincidence stemming from notational convention. After all, one random variable is discrete, and the other is continuous. They should never model the exact same thing.

However, sometimes it isn't a coincidence. One instance where these parameters can mean similar things, and where you can use either of these distributions in the same modeling task is when you are using a Poisson process. This would be useful in a situation where you're modeling things arriving randomly in time (e.g. you receiving texts on your cell phone). Say you start measuring at time $0$. Fixing any time $t > 0$, you could call the total received by this time $N_t$ and assume $$ N_t \sim \text{Poisson}(\lambda t); $$ $\lambda$ here is a rate measured in calls per unit time.

You could also look at the waiting times between texts $X_1, X_2, \ldots$ and assume that $$ X_i \overset{iid}{\sim} \text{Exponential}(\lambda); $$ here $\lambda$ is also a rate (if you're writing the density the way you are in this question). The arrival times of each text message are the partial sums $S_n = \sum_{i=1}^n X_i$.

The relationship between these two distributions in this particular situation is as follows: given an integer $n$, and a time $t$, $$ P(S_n \le t) = P(N_t \ge n). $$ In the special case where $n = t = 1$, both of these expressions are equal to $1 - e^{-\lambda}$.

Answer 3

Does this, from Wikipedia sum up the relationship in a simple way?

If for every t > 0 the number of arrivals in the time interval [0, t] follows the Poisson distribution with mean λt, then the sequence of inter-arrival times are independent and identically distributed exponential random variables having mean 1/λ.

Reference: S. M. Ross (2007). Introduction to Probability Models (ninth ed.). Boston: Academic Press. ISBN 978-0-12-598062-3. pp. 307–308.

Answer 4

The lambdas are interchangeable in certain contexts. Suppose that I'm measuring radioactivity with a Geyger counter. In one case, $\lambda=2$ means that on average I get 2 clicks per second, and the average time between clicks is $1/2$ seconds. The number of clicks per second is from a Poisson distribution, and the time between clicks is from an Exponential distribution, with both of these having $\lambda=2$.

Answer 5

As stated in Taylor's answer, there is a connection between the exponential distribution and the Poisson distribution through the equivalence of certain probability statements relating these in a Poisson process. Mathematically, this equivalence comes from a recurrence property of the incomplete gamma function, which can be used to show the probabilistic equivalence at issue.

In a Poisson process we have events occurring at some specified rate $\lambda > 0$ and we can analyse the process by looking either at the time between events, or the number of events in a given time. To do the former, let $X_1, X_2, X_3, ... \sim \text{IID Exp} (\lambda)$ be the time between events in the process, and define the partial sums $S_n \equiv \sum_{i=1}^n X_i$, which represent the time taken for the first $n$ events. Then we have $S_n \sim \text{Ga} (n, \lambda)$ so that:

$$\begin{equation} \begin{aligned} \mathbb{P}(S_n \leqslant t) = 1 - \mathbb{P}(S_n >t) &= 1 - \int\limits_t^{\infty} \text{Ga} (s|n, \lambda) ds \\ &= 1-\frac{\lambda^n}{\Gamma (n) } \int\limits_t^{\infty} s^{n-1} \exp (- \lambda s) ds \\ &= 1-\frac{1}{\Gamma (n)} \int\limits_t^{\infty} (\lambda s) ^{n-1} \exp (- \lambda s) \lambda ds \\ &= 1-\frac{1}{\Gamma (n)} \int\limits_{\lambda t}^{\infty} r^{n-1} \exp \left( - r \right) dr \\ &= 1-\frac{\Gamma(n, \lambda t)}{\Gamma (n)}. \\ \end{aligned} \end{equation}$$

Using integration by parts, the upper incomplete gamma function follows the recurrence:

$$\begin{matrix} \Gamma (n, x) = (n-1) \Gamma(n-1, x) + x^{n-1} \exp (-x) & & \Gamma (1, x) = \exp(-x). \end{matrix}$$

For integer $n$, repeated application of this recurrence yields:

$$\Gamma (n, x) = \Gamma (n) \exp (-x) \sum_{k=0}^{n-1} \frac{x^k}{k!}.$$

So, letting $N_t \sim \text{Pois} (\lambda t)$ we have:

$$\begin{equation} \begin{aligned} \mathbb{P}(S_n \leqslant t) &= 1- \exp (-\lambda t) \sum_{k=0}^{n-1} \frac{(\lambda t)^k}{k!} \\ &= 1 - \sum_{k=0}^{n-1} \text{Pois} (k|\lambda t) \\ &= \sum_{k=n}^{\infty} \text{Pois} (k|\lambda t) \\ &= \mathbb{P} (N_t \geqslant n). \end{aligned} \end{equation}$$

This gives us a basic intuitive result for the Poisson process. If the time taken for the first $n$ events is no greater than $t$ then the number of events that have occurred at time $t$ is at least $n$. If the time between events follows an exponential distribution then the number of events at a given time follows a Poisson distribution.