[Reading Notes - > statistics] 07-03 introduction to the concept of discrete probability distribution Poisson distribution

Poisson distribution

Suppose a situation ： There will be a big promotion at the cinema next week , The cinema manager wants everything to be perfect . The average number of failures of the popcorn machine per week is 3.4, Or the failure rate of popcorn machine is 3.4.

What is the probability that the popcorn machine will not fail next week ？（ If too many failures are expected , I'm going to buy a new popcorn machine .）

Different from the front , This time there is no series of experiments , contrary , This time the situation is like this ： The probability of failure is known , And the fault occurs randomly .

The difficulty of this kind of problem is , Although we know the average number of failures , But the actual number of failures is not fixed . On the whole , The number of failures we can expect is... Per week 3 or 4 Time , But on a bad week , There will be many more faults , And on a good week , The failure will not happen at all .

Poisson distribution Is specifically designed to deal with this situation .

The condition of Poisson distribution

1. The individual time is random in a given interval 、 Occur independently , A given interval can be time or space , For example, it can be a week , Or a mile .
2. The average number of events in this interval is known （ Or incidence ）, And is a finite value . The average number of occurrences of this event is usually in the Greek alphabet $\lambda$（lambda） Express .

Let's use it X Indicates the number of events in a given interval , For example, the number of damages in a week . If X It fits the poisson distribution , And the average occurrence of... In each interval λ Time , Or the incidence is λ, Then writing ：
$$X\sim Po(\lambda )$$
We won't deduce here . Occurs in a given interval r The probability of this event , Please use the following formula to calculate ：
$$P(X=r)=\frac{e^{-\lambda }{\lambda }^r}{r!}$$
e Is a mathematical constant , It's usually 2.718, Just substitute this number into the Poisson distribution formula .

If X～Po(2), be ：
$$P(X=3)=\frac{e^{-2}\ast 2^3}{3!}=0.180$$

Expectation and variance of Poisson distribution

If X～Po(λ), be E(X) For the number of events we can expect in a given interval , For the popcorn machine , Is the number of machine damage we can expect in an ordinary week , in other words ,E(X) Is the average number of events in a given interval .

Now? , If X～Po(λ), Then the average number of events is expressed in λ Express , be E(X) be equal to λ. meanwhile , Variance is also λ.
$$\begin{gathered} E(X)=\lambda \\ Var(X)=\lambda \end{gathered}$$

The shape of the Poisson distribution

The shape of Poisson distribution varies with λ The value of has changed .λ Small , The distribution is skewed to the right , With λ Bigger , The distribution gradually becomes symmetrical .

If λ It's an integer , There are two modes λ and λ-1, If λ Is not an integer , Then the mode is λ.

Example of popcorn machine ：

ask ： How did the formula of Poisson distribution come from ？

answer ： In fact, it can be deduced from other formulas , But it will involve a lot of mathematical knowledge . in application , The best way is to remember this formula and its application conditions .

ask ： What is the difference between Poisson distribution and other probability distributions ？

answer ： The main difference is that Poisson distribution does not require a series of tests , But it describes the number of times an event occurs in a specific interval .

Combined Poisson variables

Suppose a situation ： Now there are popcorn machines and beverage machines , The average number of times a popcorn machine breaks down per week is 3.4（X～Po(3.4)）, The average number of beverage machine failures per week is 2.3（Y~Po(2.3)）. Calculate the probability that the machine will not fail next week .

If X Represents the number of failures of the popcorn machine per week ,Y Represents the number of failures of the beverage machine per week , be X and Y All conform to Poisson distribution , in addition ,X and Y It's independent of each other , That is, whether one party fails has no effect on the probability of failure of the other party .

We need to find out the total number of failures next week is 0 Probability , namely ：
$$P(X+Y=0)$$

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because X and Y Is an independent random variable , be
$$\begin{gathered} P(X+Y)=P(X)+P(Y) \\ E(X+Y)=E(X)+E(Y) \end{gathered}$$
That is, if X～Po(${\lambda }_x$) And X～Po(${\lambda }_y$), be ：
$$X+Y\sim Po({\lambda }_x+{\lambda }_y)$$
namely , If X and Y All conform to Poisson distribution , be X+Y It also conforms to Poisson distribution . in other words , You can use X+Y The distribution of X+Y Probability .

Example solution ：

Poisson distribution is used to approximate binomial distribution

Poisson distribution has another use ： It can be used to approximately replace the binomial distribution under certain conditions . This makes it easy to calculate , You don't have to calculate factorial （ There are combinations in the formula of binomial distribution $C_n^r$,n It's not easy to calculate if it's too big ）.

When is it close to ？

Suppose we have a variable X, And X～B(n,p), Such a condition is required ：B(n,p) Approximately equal to Po(λ).

Let's first study the expectation and variance of the two distributions . Our goal is the case where the expectation and variance of Poisson distribution at Zhao are approximately equal to the expectation and variance of binomial distribution , Hope ：

When q Approximately equal to 1 And n When a large ,np and npq Approximately equal . namely ：

 When n Large and p Very hour , It can be used X~Po(np) Approximate instead of X～B(n,p).

When n Greater than 50 And p Less than 0.1 when , Is a typical approximation .

Example ：

Poisson distribution summary

End the example ：

I just don't understand why there is one more question P(X=0)？

Chapter summary