Question

What is the moment generating function of a Poisson distribution?

Answer 1

In order to solve this, we will first let $X$ be a discrete random variable having Poisson distribution with parameter $\lambda$ . Then using the formula of a moment generating function i.e., ${M_X}\left( t \right) = \sum\limits_{x = 0}^{} {\Pr \left( {X = x} \right){e^{tx}}}$ we will generate the moment generating function of a Poisson distribution.
Here $\Pr \left( {X = x} \right)$ is the probability mass function or discrete density function
and the probability mass function of the Poisson distribution is defined as: $\Pr \left( {X = x} \right) = \dfrac{{{\lambda ^x}{e^{ - \lambda }}}}{{x!}}$

Complete solution:
The moment generating function of a discrete random variable $X$ is a function ${M_x}\left( t \right)$ defined as ${M_x}\left( t \right) = \sum\limits_{x = 0}^\infty {\Pr \left( {X = x} \right){e^{tx}}}$
where $\Pr \left( {X = x} \right)$ is the probability mass function or discrete density function.
Now we will see what is Poisson distribution,
A random variable $X$ is said to have a Poisson distribution if discrete density function is defined as $\Pr \left( {X = x} \right) = \dfrac{{{\lambda ^x}{e^{ - \lambda }}}}{{x!}}$ where $\lambda$ is the parameter of the Poisson distribution.
And it is represented as $X \sim P\left( \lambda \right)$
Now we will find the moment generating function of a Poisson distribution
From the definition of the Poisson distribution, $X$ has probability mass function:
$\Pr \left( {X = x} \right) = \dfrac{{{\lambda ^x}{e^{ - \lambda }}}}{{x!}}$
And from the definition of a moment generating function, we have
${M_x}\left( t \right) = \sum\limits_{x = 0}^\infty {\Pr \left( {X = x} \right){e^{tx}}}$
Therefore, on substituting the value we get
${M_x}\left( t \right) = \sum\limits_{x = 0}^\infty {\dfrac{{{\lambda ^x}{e^{ - \lambda }}}}{{x!}}{e^{tx}}}$

${e^{ - \lambda }}$ is a constant term, so we can take it out from the summation
Therefore, we get
$\Rightarrow {M_x}\left( t \right) = {e^{ - \lambda }}\sum\limits_{x = 0}^\infty {\dfrac{{{\lambda ^x}}}{{x!}}{e^{tx}}}$
On combining the numerator, we get
$\Rightarrow {M_x}\left( t \right) = {e^{ - \lambda }}\sum\limits_{x = 0}^\infty {\dfrac{{{{\left( {\lambda {e^t}} \right)}^x}}}{{x!}}}$
Now we know that
Power series expansion for exponential function is
${e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + .....$
Therefore, we get
$\Rightarrow {M_x}\left( t \right) = {e^{ - \lambda }}{e^{\lambda {e^t}}}$
$\Rightarrow {M_x}\left( t \right) = {e^{\lambda \left( {{e^t} - 1} \right)}}$
which is the required moment generating function of a Poisson distribution.

Note:
Poisson distribution can also be defined as a limiting case of binomial distribution where $n \to \infty$ and $p \to 0$ .
Also remember the mean of a Poisson distribution, $E\left( X \right) = \lambda$ and variance of a Poisson distribution, $VAR\left( X \right) = \lambda$ which means Poisson distribution is a discrete distribution in which the value of mean and variance is equal.