Q1Advanced Engineering Mathematics

The Poisson distribution is a fundamental discrete probability distribution that models the probability of a given number of distinct events occurring within a fixed interval of time or space. It is particularly useful when these events occur with a known constant mean rate ($\lambda$) and independently of the time since the last event. Typical applications include modeling the number of phone calls received by a call center per minute or the number of radioactive decays per second.

Derivation as a Limiting Case of Binomial Distribution

The Poisson distribution mathematically arises as a specific limiting case of the Binomial distribution. Let a random variable $X$ follow a Binomial distribution, $X\sim B(n,p)$, where $n$ is the total number of independent trials and $p$ is the probability of success in any single trial. The limiting conditions are:

• The number of trials $n$ becomes extremely large ($n\to \infty$).
• The probability of success $p$ for each trial becomes extremely small ($p\to 0$).
• The expected value or mean number of successes, $np=\lambda$, remains a constant and finite positive number.

The standard probability mass function (PMF) for a Binomial distribution is given by:

Substituting $p=\frac{\lambda }{n}$ into the binomial formula yields:

We can rewrite the factorials and expand the terms:

Now, we apply the limit as $n\to \infty$. As $n$ grows infinitely large, the fraction $\frac{n(n-1)...(n-x+1)}{n^x}$ approaches 1. Furthermore, based on the fundamental limit definition of the natural exponential function, we know that ${\lim }_{n\to \infty }{\left(1-\frac{\lambda }{n}\right)}^n=e^{-\lambda }$. Finally, the term ${\left(1-\frac{\lambda }{n}\right)}^{-x}$ approaches 1 because $x$ is finite and $\frac{\lambda }{n}\to 0$. Substituting these limits yields the final Poisson distribution formula:

Mean and Variance of Poisson Distribution

A unique and defining characteristic of the Poisson distribution is that its Mean and Variance are strictly equal.

• Mean ($E[X]$): The expected value is calculated as $E[X]={\sum }_{x=0}^{\infty }x\frac{e^{-\lambda }{\lambda }^x}{x!}=\lambda e^{-\lambda }{\sum }_{x=1}^{\infty }\frac{{\lambda }^{x-1}}{(x-1)!}=\lambda e^{-\lambda }{e}^{\lambda }=\lambda$.
• Variance ($Var(X)$): Using the formula $Var(X)=E[X^2]-(E[X]{)}^2$, we first find $E[X(X-1)]={\lambda }^2$. Thus, $E[X^2]={\lambda }^2+\lambda$. The variance is $Var(X)=({\lambda }^2+\lambda )-{\lambda }^2=\lambda$.