What do we mean by rate in the exponential distribution?

When we talk about $X$ being a RV with an exponential distribution $$f(x)= 1-e^{-\theta x} \,\,\,\,\, \text{for}\,\,\,\ \theta>0$$ we say that it describes the time between two events in a Poisson Process, where $\theta$ is the rate.

1. Does $X$ describe the time interval between two specific events? Or does it consider the time interval between any two events. So like if I have 3 events, does it consider the time interval between 1 and 2 and then between 2 and 3?

2. If the rate is constant, then does it mean that we actually know WHEN the event is going to happen? I don't understand how it can be random with a constant rate. If I have a machines that builds cars at a constant rate $\lambda$, then of course after a time $t$ I will have $t\cdot \lambda$.

• In your equation you have $x$ on the left and $t$ on the right. They should agree. – Ross Millikan Feb 11 '17 at 21:35
• @RossMillikan thank you I didn't notice it! – Euler_Salter Feb 11 '17 at 21:42

In a Poisson process, the time between two subsequent events (or arrivals as some call them) has an exponential distribution. It's a random variable, so we can't know exactly when the next event will occur.

The rate parameter $\theta$ tells us how often on average the events come. It's the expected number of arrivals in one unit of time. It doesn't mean that the events happen exactly at a given rate, but in the long run, the average number of events that happened in one unit of time would be $\theta$.

Let's say we model the number of people waiting in line with a Poisson process with the rate $\theta=20$ and let our unit of time be one hour. That means that, on average, 20 people get in the line every hour (according to the model we're using).

The time elapsed from the moment one person got in line to the next person has an exponential distribution with the rate $\theta$. You can't predict when exactly the next person will get in line, but you can expect him to show up in about $3$ minutes ($\frac 1 {20}$ hours).

• thank you for the explanation is very clear. So actually the time between each arrival is not really.. random, or is it? Because for example if I start measuring when there's the $n^{th}$ arrival, at $t=0$, then say the $n+1$ arrival could occur after 1 second, but the $n+2$ could occur after 200 million years. And they should have the same probability, if it is random. So actually when we are thinking about the model, we are already restricting out "randomness" and so using the fact that "the longer you wait, the harder it gets", you are already imposing the exponential distribution – Euler_Salter Feb 11 '17 at 22:01
• Random does not mean that all the possibilities have equal probability. That is specifically a uniform distribution and you can't have one over an infinite interval. There are may probability distributions and depending on how you define the events one may be uniform or not. If I roll a die and regard the outcome as the number showing, it is uniform. If I regard the outcome as either "1" or "not a 1" it is random but one has 1/6 chance and the other has 5/6 chance. – Ross Millikan Feb 11 '17 at 22:14
• @Euler_Salter It isn't 'uniformly random' in a sense that every possible outcome is equally probable. It is random in the sense that we cannot predict the outcome with absolute certainty - it is not deterministic. If you consider 'true' randomness to mean that all events happen with equal probability, then perhaps we do restrict the randomness. – Blaza Feb 11 '17 at 22:17
• @RossMillikan and Blaza , thank you, it does make sense. I understand that, using $f(x) = \frac{1}{b-a}$ we can't have uniform probability over an infinite interval. However is there not another way to define it so that it is possible? Cause it seems plausible to have something like that – Euler_Salter Feb 11 '17 at 23:35
• @Euler_Salter It can be defined on a veeery large interval, but that's also finite. Infinity is a whole different animal. The main reason there isn't a uniform distribution on an infinite interval is that the probabilities won't add up to 1. This has surely been asked here several times, you can do a quick search. This question looks relevant and has many nice answers. – Blaza Feb 11 '17 at 23:43

The rate is the long term average of the number of events divided by the period of time over which they occur. The delay to the next event can be any amount. You are given the cumulative distribution function for the random variable, but that does not tell you when the next event will be.

• So is the rate somewhat empirical? – Euler_Salter Feb 11 '17 at 21:43
• Meaning that, maybe the process itself doesn't really have a constant rate, but we empirically "count" the number of events in a very long period of time and we measure the time in which this happens and decide to use this average? – Euler_Salter Feb 11 '17 at 21:44
• Because if the time between the events is random, then here's my question: The above is a negative exponential distribution right? which means that the bigger $x$, the smaller the probability. Hence it means that the events should occur one next to the other with very high probability and should almost never occur at long intervals of time – Euler_Salter Feb 11 '17 at 21:46
• We might have some theoretical reason to come up with a rate. For example, maybe we are measuring radioactive decay and can look up the rate of decay in the literature. Otherwise, yes we can just wait for some large number of events, measure the amount of time they took, and take that average for the rate. This will be subject to some error, but if the number of events is large the error will be small. It is true that for a given interval, the most likely interval to have the first event is the next one. It is more probable that the next event will be in the next tenth second than – Ross Millikan Feb 11 '17 at 21:52
• between 10 and 10.1 seconds from now. If the rate is low, however, the difference will be small. If the rate is 1/year all the tenth second intervals in the next week are essentially the same probability. – Ross Millikan Feb 11 '17 at 21:53