Question

What is the moment generating function of a Poisson distribution?

Answer 1

Hint: 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.