Question comes from “Introduction to Probability” by Charles M. Grinstead.
Problem Description:
Prove that the values of the Poisson Distribution given by:
\[ \begin{equation} P(X=k) = \frac{\lambda^k}{k!}e^{-\lambda} \end{equation} \] sum together to 1.
Solution:
To solve this problem, we need simply to take the sum of \(P(X=k)\) for all values of \(k\) from 0 to \(\infty\) (\(k\) cannot be negative) and show:
\[ \begin{equation} \sum_{k=0}^{\infty} P(X=k) = \sum_{k=0}^{\infty} \frac{\lambda^k}{k!}e^{-\lambda} = 1 \end{equation} \]
Thus we have (note: for clarity’s sake I remove the limits of integration in the work below…):
\[ \begin{align} \sum P(X=k) &= \sum\frac{\lambda^k}{k!}e^{-\lambda} \\ &=e^{-\lambda}\sum\frac{\lambda^k}{k!}\\ &=e^{-\lambda}\left(\frac{\lambda^0}{0!} + \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!}+ ... \right) \end{align} \]
Given that the summation expansion of \(e^x\) is written as:
\[ \begin{align} e^x &= \sum_{x=0}^{\infty} \frac{x^n}{n!} \\ & = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!}+ ... \end{align} \]
We conclude that:
\[ \begin{align} \sum_{k=0}^{\infty} P(X=k) &= e^{-\lambda}e^\lambda \\ &= e^{-\lambda+\lambda} \\ &=e^0 \\ &=1 \end{align} \]
We can also confirm this computationally in R using different \(\lambda\) values and summing to a reasonably high \(k\):
k_max <- 100
for (l in 1:10){
tot <- 0
for (k in 0:k_max){
tot <- tot + (((l ** k) / factorial(k)) * exp(-l))
}
output <- paste('Using k_max =', k_max, 'and lambda =', l, ': sum =', tot)
print(output)
}## [1] "Using k_max = 100 and lambda = 1 : sum = 1"
## [1] "Using k_max = 100 and lambda = 2 : sum = 1"
## [1] "Using k_max = 100 and lambda = 3 : sum = 1"
## [1] "Using k_max = 100 and lambda = 4 : sum = 1"
## [1] "Using k_max = 100 and lambda = 5 : sum = 1"
## [1] "Using k_max = 100 and lambda = 6 : sum = 1"
## [1] "Using k_max = 100 and lambda = 7 : sum = 1"
## [1] "Using k_max = 100 and lambda = 8 : sum = 1"
## [1] "Using k_max = 100 and lambda = 9 : sum = 1"
## [1] "Using k_max = 100 and lambda = 10 : sum = 1"