Poisson Distribution

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Poisson Distribution: Definition, Moments, MGF and Solved Examples

Definition

The Poisson Distribution is a discrete probability distribution used to model the number of times an event occurs in a fixed interval of time, space, area, or volume, when the events occur independently and at a constant average rate.

A random variable \(X\) is said to follow a Poisson distribution with parameter \(\lambda > 0\) if

\[ \boxed{ P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}, \qquad x=0,1,2,\ldots } \]

We write

\[ X\sim\operatorname{Poisson}(\lambda). \]

Here, \(\lambda\) represents the average number of occurrences in the specified interval.

Main Conditions

  1. Events occur independently.
  2. The average rate of occurrence is constant.
  3. The number of occurrences in disjoint intervals is independent.
  4. In a sufficiently small interval, the probability of more than one occurrence is negligible.

Moments of the Poisson Distribution

Let

\[ X\sim\operatorname{Poisson}(\lambda). \]

First Moment: Mean

The mean is

\[ E(X) = \sum_{x=0}^{\infty} x\frac{e^{-\lambda}\lambda^x}{x!}. \]

Using

\[ \frac{x}{x!}=\frac{1}{(x-1)!}, \]

we get

\[ E(X) = \lambda e^{-\lambda} \sum_{x=1}^{\infty} \frac{\lambda^{x-1}}{(x-1)!}. \]

Since

\[ \sum_{r=0}^{\infty}\frac{\lambda^r}{r!} = e^\lambda, \]

therefore

\[ \boxed{E(X)=\lambda}. \]

Second Raw Moment

For the Poisson distribution,

\[ E[X(X-1)]=\lambda^2. \]

Since

\[ X^2=X(X-1)+X, \]

we obtain

\[ E(X^2)=\lambda^2+\lambda. \]

Hence,

\[ \boxed{\mu_2'=\lambda^2+\lambda}. \]

Variance

Using

\[ \operatorname{Var}(X) = E(X^2)-[E(X)]^2, \]

we get

\[ \operatorname{Var}(X) = (\lambda^2+\lambda)-\lambda^2. \]

Therefore,

\[ \boxed{\operatorname{Var}(X)=\lambda}. \]

Thus,

\[ \boxed{\sigma=\sqrt{\lambda}}. \]

A characteristic property of the Poisson distribution is

\[ \boxed{\text{Mean}=\text{Variance}=\lambda}. \]

Higher Raw Moments

\[ \boxed{\mu_1'=\lambda} \]

\[ \boxed{\mu_2'=\lambda^2+\lambda} \]

\[ \boxed{\mu_3'=\lambda^3+3\lambda^2+\lambda} \]

\[ \boxed{ \mu_4' = \lambda^4+6\lambda^3+7\lambda^2+\lambda } \]

Central Moments

\[ \boxed{\mu_2=\lambda} \]

\[ \boxed{\mu_3=\lambda} \]

\[ \boxed{\mu_4=\lambda+3\lambda^2} \]

Skewness

\[ \boxed{ \gamma_1=\frac{1}{\sqrt{\lambda}} } \]

Kurtosis

\[ \boxed{ \beta_2 = 3+\frac{1}{\lambda} } \]

Hence, the excess kurtosis is

\[ \boxed{ \gamma_2=\frac{1}{\lambda} } \]

Moment Generating Function

The moment generating function of \(X\) is defined as

\[ M_X(t)=E(e^{tX}). \]

For a Poisson random variable,

\[ M_X(t) = \sum_{x=0}^{\infty} e^{tx} \frac{e^{-\lambda}\lambda^x}{x!}. \]

Therefore,

\[ M_X(t) = e^{-\lambda} \sum_{x=0}^{\infty} \frac{(\lambda e^t)^x}{x!}. \]

Using the exponential series,

\[ e^a=\sum_{x=0}^{\infty}\frac{a^x}{x!}, \]

we obtain

\[ M_X(t) = e^{-\lambda}e^{\lambda e^t}. \]

Hence,

\[ \boxed{ M_X(t)=e^{\lambda(e^t-1)} } \]

Solved Examples

Example 1: Exact Probability

A call centre receives an average of 4 calls per minute. Assuming a Poisson distribution, find the probability of receiving exactly 6 calls in a minute.

Show Solution

Here, \[ \lambda=4. \]

We require \[ P(X=6). \]

Using the Poisson formula,

\[ P(X=6) = \frac{e^{-4}4^6}{6!}. \]

Since \[ 4^6=4096,\qquad 6!=720, \]

we get

\[ P(X=6) = \frac{4096e^{-4}}{720}. \]

Therefore,

\[ \boxed{P(X=6)\approx0.1042}. \]

Example 2: Probability of No Occurrence

The average number of machine failures in a week is 2. Find the probability that there are no failures during a particular week.

Show Solution

Here,

\[ X\sim\operatorname{Poisson}(2). \]

We require \(P(X=0)\).

\[ P(X=0) = \frac{e^{-2}2^0}{0!}. \]

Thus,

\[ P(X=0)=e^{-2}. \]

Therefore,

\[ \boxed{P(X=0)\approx0.1353}. \]

Example 3: At Least One Occurrence

A radioactive source emits an average of 3 particles per second. Find the probability that at least one particle is emitted during a particular second.

Show Solution

Here, \[ \lambda=3. \]

We require

\[ P(X\geq1). \]

Using the complement rule,

\[ P(X\geq1) = 1-P(X=0). \]

Now,

\[ P(X=0)=e^{-3}. \]

Therefore,

\[ P(X\geq1)=1-e^{-3}. \]

Hence,

\[ \boxed{P(X\geq1)\approx0.9502}. \]

Example 4: At Most Two Events

A hospital emergency department receives an average of 5 emergency cases per hour. Find the probability of receiving at most 2 cases in an hour.

Show Solution

Let

\[ X\sim\operatorname{Poisson}(5). \]

We require

\[ P(X\leq2) = P(X=0)+P(X=1)+P(X=2). \]

Therefore,

\[ P(X\leq2) = e^{-5} \left[ 1+5+\frac{5^2}{2!} \right]. \]

Thus,

\[ P(X\leq2) = 18.5e^{-5}. \]

Therefore,

\[ \boxed{P(X\leq2)\approx0.1247}. \]

Example 5: Probability Between Two Values

A manufacturing process produces an average of 3 defective components per batch. Find the probability that a batch contains between 2 and 4 defective components, inclusive.

Show Solution

Here,

\[ X\sim\operatorname{Poisson}(3). \]

We require

\[ P(2\leq X\leq4). \]

Therefore,

\[ P(2\leq X\leq4) = P(X=2)+P(X=3)+P(X=4). \]

Thus,

\[ P(2\leq X\leq4) = e^{-3} \left[ \frac{3^2}{2!} + \frac{3^3}{3!} + \frac{3^4}{4!} \right]. \]

Hence,

\[ P(2\leq X\leq4) = e^{-3}(4.5+4.5+3.375). \]

Therefore,

\[ \boxed{P(2\leq X\leq4)\approx0.6161}. \]

Example 6: Changing the Time Interval

A server receives an average of 12 requests per hour. Find the probability that exactly 2 requests arrive during a 10-minute interval.

Show Solution

The average number per hour is 12.

Since 10 minutes is \(\frac{1}{6}\) of an hour,

\[ \lambda = 12\left(\frac16\right) = 2. \]

Therefore,

\[ X\sim\operatorname{Poisson}(2). \]

We need

\[ P(X=2) = \frac{e^{-2}2^2}{2!}. \]

Thus,

\[ P(X=2)=2e^{-2}. \]

Hence,

\[ \boxed{P(X=2)\approx0.2707}. \]

Example 7: Finding the Parameter from a Probability

Suppose \(X\) has a Poisson distribution and

\[ P(X=0)=0.2. \]

Find \(\lambda\), the mean, and the variance.

Show Solution

For a Poisson distribution,

\[ P(X=0)=e^{-\lambda}. \]

Hence,

\[ e^{-\lambda}=0.2. \]

Taking natural logarithms,

\[ -\lambda=\ln(0.2). \]

Therefore,

\[ \lambda=-\ln(0.2)=\ln5. \]

Thus,

\[ \boxed{\lambda\approx1.6094}. \]

For a Poisson distribution,

\[ E(X)=\operatorname{Var}(X)=\lambda. \]

Therefore,

\[ \boxed{ E(X)\approx1.6094,\qquad \operatorname{Var}(X)\approx1.6094 } \]

Example 8: Equal Probabilities

Suppose

\[ X\sim\operatorname{Poisson}(\lambda) \]

and

\[ P(X=2)=P(X=3). \]

Find \(\lambda\).

Show Solution

Using the Poisson formula,

\[ \frac{e^{-\lambda}\lambda^2}{2!} = \frac{e^{-\lambda}\lambda^3}{3!}. \]

Cancel \(e^{-\lambda}\):

\[ \frac{\lambda^2}{2} = \frac{\lambda^3}{6}. \]

Since \(\lambda>0\), divide by \(\lambda^2\):

\[ \frac12=\frac{\lambda}{6}. \]

Hence,

\[ \boxed{\lambda=3}. \]

Example 9: Finding Moments from the MGF

Suppose the moment generating function of \(X\) is

\[ M_X(t)=\exp\left[5(e^t-1)\right]. \]

Identify the distribution and calculate its mean, variance, and third central moment.

Show Solution

The MGF of a Poisson random variable is

\[ M_X(t)=e^{\lambda(e^t-1)}. \]

Comparing,

\[ \lambda=5. \]

Therefore,

\[ \boxed{X\sim\operatorname{Poisson}(5)}. \]

For a Poisson distribution,

\[ E(X)=\lambda, \qquad \operatorname{Var}(X)=\lambda, \qquad \mu_3=\lambda. \]

Hence,

\[ \boxed{ E(X)=5,\qquad \operatorname{Var}(X)=5,\qquad \mu_3=5 } \]

Example 10: Sum of Independent Poisson Variables

Suppose

\[ X\sim\operatorname{Poisson}(3), \qquad Y\sim\operatorname{Poisson}(5), \]

where \(X\) and \(Y\) are independent. Find

\[ P(X+Y=6). \]

Show Solution

The sum of independent Poisson random variables is also Poisson.

Let

\[ Z=X+Y. \]

Then,

\[ Z\sim\operatorname{Poisson}(3+5) = \operatorname{Poisson}(8). \]

Therefore,

\[ P(X+Y=6) = P(Z=6) = \frac{e^{-8}8^6}{6!}. \]

Hence,

\[ \boxed{ P(X+Y=6)\approx0.1221 } \]

Example 11: Conditional Distribution

Let

\[ X\sim\operatorname{Poisson}(2), \qquad Y\sim\operatorname{Poisson}(3), \]

independently. Find

\[ P(X=2\mid X+Y=5). \]

Show Solution

For independent Poisson random variables,

\[ X\mid(X+Y=n) \sim \operatorname{Bin} \left( n, \frac{\lambda_X}{\lambda_X+\lambda_Y} \right). \]

Here,

\[ n=5, \qquad \frac{\lambda_X}{\lambda_X+\lambda_Y} = \frac{2}{2+3} = \frac25. \]

Therefore,

\[ P(X=2\mid X+Y=5) = \binom52 \left(\frac25\right)^2 \left(\frac35\right)^3. \]

Thus,

\[ = 10 \left(\frac4{25}\right) \left(\frac{27}{125}\right). \]

Therefore,

\[ \boxed{ P(X=2\mid X+Y=5)=0.3456 } \]

Example 12: Poisson Approximation to Binomial Distribution

A factory produces components with probability \(0.002\) of being defective. A random sample of 1000 components is selected. Use the Poisson approximation to find the probability that exactly 3 components are defective.

Show Solution

Originally,

\[ X\sim\operatorname{Bin}(1000,0.002). \]

Since \(n\) is large and \(p\) is small, we use the Poisson approximation with

\[ \lambda=np. \]

Thus,

\[ \lambda = 1000(0.002) = 2. \]

Therefore,

\[ X\approx\operatorname{Poisson}(2). \]

Hence,

\[ P(X=3) \approx \frac{e^{-2}2^3}{3!}. \]

Thus,

\[ P(X=3) \approx \frac{8e^{-2}}{6}. \]

Therefore,

\[ \boxed{P(X=3)\approx0.1804}. \]

Important Formula Summary

Quantity Formula
Probability Mass Function \(\displaystyle P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}\)
Mean \(\displaystyle E(X)=\lambda\)
Variance \(\displaystyle \operatorname{Var}(X)=\lambda\)
Standard Deviation \(\displaystyle \sqrt{\lambda}\)
MGF \(\displaystyle e^{\lambda(e^t-1)}\)
Second Raw Moment \(\displaystyle \lambda^2+\lambda\)
Third Raw Moment \(\displaystyle \lambda^3+3\lambda^2+\lambda\)
Fourth Raw Moment \(\displaystyle \lambda^4+6\lambda^3+7\lambda^2+\lambda\)
Third Central Moment \(\displaystyle \lambda\)
Fourth Central Moment \(\displaystyle \lambda+3\lambda^2\)
Skewness \(\displaystyle \frac{1}{\sqrt{\lambda}}\)
Kurtosis \(\displaystyle 3+\frac{1}{\lambda}\)
Excess Kurtosis \(\displaystyle \frac{1}{\lambda}\)

Additive Property

If \(X_1,X_2,\ldots,X_k\) are independent Poisson random variables with parameters \(\lambda_1,\lambda_2,\ldots,\lambda_k\), then

\[ \boxed{ X_1+X_2+\cdots+X_k \sim \operatorname{Poisson} (\lambda_1+\lambda_2+\cdots+\lambda_k) } \]

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