It's been quite a while since I did any statistics so I am struggling with the definitions of a Poisson distribution. What I understand by the "rate is constant" is that if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?

Where I believe I am confused is with the final sentence. Is this saying that the time between a customers purchases would grow exponentially as time goes on? Doesn't this contradict the idea that we have a constant rate of purchase?

The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.

Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money.

The exponential distribution is widely used in the field of reliability. Reliability deals with the amount of time a product lasts.

Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:

λe−λt

where λ is Poisson intensity, i.e. average number of events in unit of time, and t is the waiting time. The average waiting time is obviously 1/λ.

Let XtXt be the number of arrivals in the Poisson process with rate λλ between time 00 and time t≥0.t≥0.Then we have

Pr(Xt=x)=(λt)xe−λtx! for x=0,1,2,3,…Pr(Xt=x)=(λt)xe−λtx! for x=0,1,2,3,…
Let TT be the time until the first arrival.

Then the following two events are really both the same event:

[Xt=0].[T>t].[Xt=0].[T>t].
So they both have the same probability. Thus

Pr(T>t)=Pr(Xt=0)=(λt)0e−λt0!=e−λt.Pr(T>t)=Pr(Xt=0)=(λt)0e−λt0!=e−λt.
So

Pr(T>t)=e−λt for t≥0.Pr(T>t)=e−λt for t≥0.
That says TT is exponentially distributed.