Question

Prove that the total probability is one by using Poisson distribution.

Answer 1

Hint: We will first of all write down the formula used in Poisson distribution, then since Poisson distribution is a discrete probability, we will sum up it for all the values from 0 to infinity and, it will result in 1 on simplification and thus we are done.

Complete step-by-step answer:
Let us discuss Poisson distribution to understand it.
Let X be the discrete random variable that represents the number of events observed over a given time period. Let \[\lambda \] be the expected value (average) of X. If X follows a Poisson distribution, then the probability of observing k events over the time period is:-
$P(X = k) = \dfrac{{{\lambda ^k}{e^{ - \lambda }}}}{{k!}}$
Since, we cannot have the events in negative, therefore, $X \geqslant 0$.
So, if we sum up all the probabilities possible that is upto infinity, we must get 1 to prove the given question.
So, let us now sum it up from 0 to infinity.
$\sum\limits_{k = 0}^\infty {P(X = k)} = \sum\limits_{k = 0}^\infty {\dfrac{{{\lambda ^k}{e^{ - \lambda }}}}{{k!}}} $
Simplifying the RHS by opening up the summation sign, we will have:-
$\sum\limits_{k = 0}^\infty {P(X = k)} = \dfrac{{{\lambda ^0}{e^{ - \lambda }}}}{{0!}} + \dfrac{{{\lambda ^1}{e^{ - \lambda }}}}{{1!}} + \dfrac{{{\lambda ^2}{e^{ - \lambda }}}}{{2!}} + .........$
Taking out ${e^{ - \lambda }}$ common from RHS, we will have with us the following expression:-
$\sum\limits_{k = 0}^\infty {P(X = k)} = {e^{ - \lambda }}\left( {\dfrac{{{\lambda ^0}}}{{0!}} + \dfrac{{{\lambda ^1}}}{{1!}} + \dfrac{{{\lambda ^2}}}{{2!}} + .........} \right)$
Rewriting the expression as follows:-
\[\sum\limits_{k = 0}^\infty {P(X = k)} = {e^{ - \lambda }}\sum\limits_{r = 0}^\infty {\dfrac{{{\lambda ^r}}}{{r!}}} \] ………(1)
We know that the expansion of ${e^x}$ is given by:-
${e^x} = \sum\limits_{r = 0}^\infty {\dfrac{{{x^r}}}{{r!}}} $
Using this we can write the equation (1) as:-
\[\sum\limits_{k = 0}^\infty {P(X = k)} = {e^{ - \lambda }}.{e^\lambda } = 1\].
Hence, we have proved that the total probability is one by using Poisson distribution.

Note: The students must note that the expression just before equation (1), if you open it up by writing the values of factorials, you would not be able to see as a whole combining to form something, which will become an obstacle to reach the required result. So, just try to observe the patterns for once which have factorial in them to turn them in something really simple.
Fun Fact:- The $\lambda $ in Poisson Distribution means both Mean and Variance and you may verify it using its formula given by $P(X = k) = \dfrac{{{\lambda ^k}{e^{ - \lambda }}}}{{k!}}$.