Barion Pixel Relationship between Poisson and Exponential distribution | mathXplain
 

Contents of this Probability theory episode:

Poisson distribution, Exponential distribution, Average, Density function, Distribution function, Expected value, Standard deviation, Discrete random variable, Continuous random variable.

Text of slideshow

THE RELATIONSHIP BETWEEN THE POISSON DISTRIBUTION AND THE EXPONENTIAL DISTRIBUTION

On average, 12 cars arrive to the gas station in an hour.

1. What is the probability of 3 cars arriving within 10 minutes?

2. What is the probability of at least 10 minutes passing between the arrivals of two cars?

While the first question is about the number of cars, the second question is about the time between their arrivals. The number of cars is a discrete distribution, and since any number of cars can arrive, it is Poisson. The time elapsed is a continuous distribution, and happens to be exponential.

1. number of cars in 10 minutes, cars, POISSON

The expected value is 12 cars per hour, so in one minute it is 12/60=0.2 and in 10 minutes it is cars

[K1]

2. time elapsed between two cars, minutes, EXPONENTIAL

The expected value is 12 cars per hour, so the average elapsed time is 60/12=5 minutes minutes

[K2]

Both distributions describe the same story: one looks at the number of occurrences, while the other looks at the time between those occurrences. So, the mysterious appearance of this at both places is not just a coincidence. The two s are essentially the same.

We need to understand that the expected value of a Poisson distribution depends on the time period checked: a longer period will have more cars, a shorter period will have fewer. Let's say in 10 minutes , but in 15 minutes .

On the other hand, the expected value of the exponential distribution is the expected elapsed time, and that does not depend on the time period being examined. It is because on average, cars arrive every 5 minutes; let it be in a half hour, or in a 20-minutes period. So here will always be the same.

Should it happen that with the Poisson distribution we examine a time period that is exactly the unit of the elapsed time at the exponential distribution, than the two s will be exactly the same. Let's see how this looks for our given example.

If, at the exponential distribution we measure the elapsed time in minutes, then the expected value is 5 minutes, and [K3] . Now let's take a one-minute period, and calculate of the Poisson distribution. There are 12 cars coming every hour, so in a minute it is 12/60=0.2, in other words [K4] : the two s are the same.

If, at the exponential distribution we measure the elapsed time in hours, then the expected value of 5-minutes. Now, 5 minutes = 5/60 hours, so approximately 0.083 hours. Here [K5] . Now let's calculate for the Poisson distribution for a period of one hour. Since the problem said 12 cars are coming every hour, it follows that . So the two s are the same in this case, too.

X = number of occurrences in a given time

Y = elapsed time between two occurrences

In an area, on average, every 16 months there is an earthquake stronger than 5 on the Richter-scale.

a) What is the probability having two such earthquakes in a year?

b) What is the probability of 3 years elapsing between two such earthquakes?

X = number of stronger earthquakes in a year

Let's see how many earthquakes occur per year.

One year is 12 months, and an earthquake occurs every 16 months.

That means there are earthquakes in a year.

Y = elapsed time between earthquakes

X means the number of earthquakes, following the Poisson distribution, while Y means the elapsed time between earthquakes, following the exponential distribution.

So, the expected value here does not mean how many earthquakes are expected, but rather, how many months are expected between them.

3 years = 36 months

Well, this was plenty enough of exponential distribution.

5.1.

An insurance company receives – on average – 5 home insurance claims per day.

a) What is the probability that on a day they receive fewer than the expected number of claims?

b) What is the probability that on three days in a week they receive fewer claims than expected?

X=number of claims

It can happen at any time that aliens attack Earth and then the number of home insurance claims reaches a million per day. That means X has no limit.

Y= number of days

when claims are below average

There are maximum seven days in a week, so Y has a limit.

Let's see what we know:

the probability that the number of claims in a day is below average

 

Relationship between Poisson and Exponential distribution

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