Theorem 8.2.1 Poisson Probability Function
Assume X measures the number of successes in an interval [0,T] within some Poisson process. Then,
f(x)=μxe−μx!
for R={0,1,2,...}.
Section 8.2 Poisson Distribution
Consider a Poisson Process where you start with an interval of fixed length T and where X measures the variable number of successes, or changes, within a that interval. The resulting distribution of X will be called a Poisson distribution.
Proof
For a sufficiently large natural number n, break up the given interval [0,T] into n uniform parts each of width h = T/n. Using the properties of Poisson processes, n very large implies h will be very small and eventually small enough so that
\begin{equation*}
P(\text{exactly one success on a given interval}) = p = \lambda \frac{T}{n}.
\end{equation*}
However, since there are a finite number of independent intervals each with probability p of containing a success then you can use a Binomial distribution to evaluate the corresponding probabilities so long as n is finite. Doing so yields and taking the limit as n approaches infinity gives:
\begin{align*}
f(x) & = P(\text{X changes in [0,T]}) \\
& = \lim_{n \rightarrow \infty} \binom{n}{x} p^x (1-p)^{n-x}\\
& = \lim_{n \rightarrow \infty} \binom{n}{x} (\frac{\lambda T}{n})^x (1-\lambda \frac{T}{n})^{n-x}\\
& = \lim_{n \rightarrow \infty} \frac{n(n-1)...(n-x+1)}{x!} ( \frac{\lambda T}{n})^x (1- \frac{\lambda T}{n})^{n-x}\\
& = \frac{(\lambda T)^x}{x!} \lim_{n \rightarrow \infty} \frac{n(n-1)...(n-x+1)}{n \cdot n \cdot ... \cdot n} (1-\lambda \frac{T}{n})^{n}(1-\lambda \frac{T}{n})^{-x}\\
& = \frac{(\lambda T)^x}{x!} \lim_{n \rightarrow \infty} (1-\frac{1}{n})...(1-\frac{x-1}{n}) (1- \frac{\lambda T}{n})^{n}(1- \frac{\lambda T}{n})^{-x}\\
& = \frac{(\lambda T)^x}{x!}
\lim_{n \rightarrow \infty} (1- \frac{\lambda T}{n})^{n}
\lim_{n \rightarrow \infty} (1- \frac{\lambda T}{n})^{-x}\\
& = \frac{(\lambda T)^x}{x!}
\lim_{n \rightarrow \infty} (1- \frac{\lambda T}{n})^{n} \cdot 1\\
& = \frac{(\lambda T)^x}{x!}
e^{-\lambda T}
\end{align*}
Theorem 8.2.2 Verify Poisson Probability Function
∞∑x=0(λT)xx!e−λT=1
Proof
Using the Power Series expansion for the natural exponential,
\begin{align*}
\sum_{x=0}^{\infty} f(x) & = \sum_{x=0}^{\infty} \frac{(\lambda T)^x}{x!} e^{-\lambda T} \\
& = e^{-\lambda T} \sum_{x=0}^{\infty} \frac{(\lambda T)^x}{x!} \\
& = e^{-\lambda T} e^{\lambda T} \\
& = 1
\end{align*}
Theorem 8.2.3 Statistics for Poisson
μ=λT
σ2=μ
γ1=1√μ
γ2=1μ+3
Proof
Using the f(x) generated in the previous theorem
\begin{align*}
\mu & = E[X] \\
& = \sum_{x=0}^{\infty} x \cdot \frac{(\lambda T)^x}{x!} e^{-\lambda T}\\
& = \lambda T e^{-\lambda T} \sum_{x=1}^{\infty} \frac{(\lambda T)^{x-1}}{(x-1)!} \\
& = \lambda T e^{-\lambda T} \sum_{k=0}^{\infty} \frac{(\lambda T)^k}{k!} \\
& = \lambda T e^{-\lambda T} e^{\lambda T} \\
& = \lambda T
\end{align*}
which confirms the use of \(\mu\) in the original probability formula.
Continuing with \(\mu = \lambda T\text{,}\) the variance is given by
\begin{align*}
\sigma^2 & = E[X(X-1)] + \mu - \mu^2 \\
& = \sum_{x=0}^{\infty} x(x-1) \cdot \frac{\mu^x}{x!} e^{-\mu} + \mu - \mu^2\\
& = e^{-\mu} \mu^2 \sum_{x=2}^{\infty} \frac{\mu^{x-2}}{(x-2)!} + \mu - \mu^2\\
& = e^{-\mu} \mu^2 \sum_{k=0}^{\infty} \frac{\mu^k}{k!} + \mu - \mu^2\\
& = \mu^2 + \mu - \mu^2 \\
& = \mu
\end{align*}
To derive the skewness and kurtosis, you can depend upon Sage...see the live cell below.
xxxxxxxxxxvar('x,mu')assume(x,'integer')f(x) =e^(-mu)*mu^x/factorial(x)mu = sum(x*f,x,0,oo).factor()M2 = sum(x^2*f,x,0,oo).factor()M3 = sum(x^3*f,x,0,oo).factor()M4 = sum(x^4*f,x,0,oo).factor()pretty_print('Mean = ',mu)v = (M2-mu^2).factor()pretty_print('Variance = ',v)stand = sqrt(v)sk = ((M3 - 3*M2*mu + 2*mu^3)).factor()/stand^3pretty_print('Skewness = ',sk)kurt = (M4 - 4*M3*mu + 6*M2*mu^2 -3*mu^4).factor()/stand^4pretty_print('Kurtosis = ',(kurt-3).factor(),'+3')