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  • Poisson Random Variables
  • Poisson Experiment
  • Poisson Distribution
  • Problem 1
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  1. math
  2. Statistics (hackerrank)

Poisson Distribution

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Last updated 6 years ago

P(X=k)=lim⁡n→∞(nk)(λn)k(1−λn)n−k=λkk!e−λ P(X =k) = \lim_{n \rightarrow \infty}\binom{n}{k}(\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{n-k} \newline = \frac{\lambda ^k}{k!} e^{-\lambda}P(X=k)=limn→∞​(kn​)(nλ​)k(1−nλ​)n−k=k!λk​e−λ

λ⟹is just any given number\lambda \Longrightarrow \text{is just any given number}λ⟹is just any given number

Poisson Random Variables

We've already learned that we can break many problems down into terms of nnn, xxx, and ppp and use the following formula for binomial random variables:

p(x)=(nx)⋅px⋅(1−p)n−xp(x)=\binom{n}{x} \cdot p^x \cdot (1-p)^{n-x}p(x)=(xn​)⋅px⋅(1−p)n−x

But what do we do when cannot be calculated using that formula? Enter the Poisson random variable.

Poisson Experiment

A Poisson experiment is a statistical experiment that has the following properties:

  • The outcome of each trial is either success or failure.

  • The average number of successes (λ)(\lambda)(λ) that occurs in a specified region is known.

  • The probability that a success will occur is proportional to the size of the region.

  • The probability that a success will occur in an extremely small region is virtually zero.

A Poisson random variable is the number of successes that result from a Poisson experiment. The probability distribution of a Poisson random variable is called a Poisson distribution:

P(k,λ)=λke−λk!\huge P(k, \lambda) = \frac{\lambda^k e^{-\lambda}}{k!}P(k,λ)=k!λke−λ​

Here,

  • e=2.71828e=2.71828e=2.71828

  • λ\lambdaλ is the average number of successes that occur in a specified region.

  • kkk is the actual number of successes that occur in a specified region.

  • P(k,λ)P(k, \lambda)P(k,λ) is the Poisson probability, which is the probability of getting exactly kkk successes when the average number of successes is λ\lambdaλ.

Example

Acme Realty company sells an average of 222 homes per day. What is the probability that exactly 333 homes will be sold tomorrow?

Here, λ=2\lambda=2λ=2 and k=3k=3k=3, so P(k=3,λ=2)=λke−λk!=0.180P(k=3, \lambda =2)=\frac{\lambda^ke^{-\lambda}}{k!}=0.180P(k=3,λ=2)=k!λke−λ​=0.180

Example

Suppose the average number of lions seen by tourists on a one-day safari is 555. What is the probability that tourists will see fewer than lions on the next one-day safari? P(k≤3,λ=5)=∑r=03λre−λr!=0.2650P(k\leq3,\lambda=5)=\sum_{r=0}^{3}\frac{\lambda^re^{-\lambda}}{r!}=0.2650P(k≤3,λ=5)=∑r=03​r!λre−λ​=0.2650

Special Case

Consider some Poisson random variable, XXX. Let E[X]E[X]E[X] be the expectation of XXX. Find the value of E[X2]E[X^2]E[X2].

Let Var(X)Var(X)Var(X) be the variance of XXX. Recall that if a random variable has a Poisson distribution, then:

  • E[X]=λE[X] = \lambdaE[X]=λ

  • Var(X)=λVar(X) = \lambdaVar(X)=λ

Now, we'll use the following property of expectation and variance for any random variable, XXX:

Var(X)=E[X2]−(E[X])2⇒E[X2]=Var(X)+(E[X])2Var(X)=E[X^2]-(E[X])^2 \Rightarrow E[X^2] = Var(X) +(E[X])^2Var(X)=E[X2]−(E[X])2⇒E[X2]=Var(X)+(E[X])2

So, for any random variable XXX having a Poisson distribution, the above result can be rewritten as:

⇒E[X2]=λ+λ2\Rightarrow E[X^2]=\lambda + \lambda^2⇒E[X2]=λ+λ2

Problem 1

A random variable, XXX, follows Poisson distribution with mean of 2.52.52.5. Find the probability with which the random variable XXX is equal to 555.

def factorial(n):
    if n == 0:
        return 1
    else:
        return n * factorial(n-1)

k = 5
l = 2.5
e = 2.71828

dist = factorial(k)
p = ((l**k)*(e**(l*-1)))/dist
print(round(p, 3))

Problem 2

The manager of a industrial plant is planning to buy a machine of either type AAA or type BBB. For each day’s operation:

  • The number of repairs, XXX, that machine AAA needs is a Poisson random variable with mean 0.880.880.88. The daily cost of operating AAA is CA=160+40X2C_A = 160 + 40X^2CA​=160+40X2.

  • The number of repairs, YYY, that machine BBB needs is a Poisson random variable with mean 1.551.551.55. The daily cost of operating BBB is CB=128+40Y2C_B=128+40Y^2CB​=128+40Y2.

Assume that the repairs take a negligible amount of time and the machines are maintained nightly to ensure that they operate like new at the start of each day. Find and print the expected daily cost for each machine.

X = 0.88
Y = 1.55

print(round((160 + (40*(X + X**2))), 3))  # => 226.176
print(round((128 + (40*(Y + Y**2))), 3))  # => 286.1
🪶
Poisson Distribution