Poisson Process

In the last post, we know what is Poisson distribution. If we give a restriction, we can get the Poisson Process (we don't discuss the details here): \[ P(X=k, t)=\frac{(\lambda t)^{k}}{k !} e^{-\lambda t}\ . \tag{1} \]

Exponential Distribution

Let say \(Y=k_1-k_2\), then according to formula \((1)\) \[ P(Y>t)=P(X=0, t)=\frac{(\lambda t)^{0}}{0 !} e^{-\lambda t}=e^{-\lambda t}, \quad t \geqslant 0\ , \] then \[ P(Y \leq t)=1-P(Y>t)=1-e^{-\lambda t}\ . \] That is the probability distribution function of exponential distribution, \[ F(y)=P(Y \leq y)=\left\{\begin{array}{ll}1-e^{-\lambda y}, & y \geq 0 \\ 0, & y<0\end{array}\right. \ . \tag{2} \] and the PDF is \[ p(y)=\left\{\begin{array}{ll}\lambda e^{-\lambda y}, & y \geq 0 \\ 0, & y<0\end{array}\right. \ . \tag{3} \] 　The MGF is \[ \psi(t)=E\left[e^{t x}\right]=\int_{0}^{\infty} e^{t x} \lambda e^{-\lambda x} \mathrm{d} x=\frac{\lambda}{\lambda-t}\ . \tag{4} \]

　Expected Value \(E[X]=1 / \lambda\)

　Variance \(\operatorname{Var}(X)=1 / \lambda^{2}\)

Relationship

Because Poisson and exponential both generate from binomial distribution, they are all memoryless distribution. They are all belong to exponential function as well, relating to normal distribution and so on.

　Let say a case where selling hamburgers.

  • Poisson distribution describes the probability of how many hamburgers will be sold over a period of time.
  • Exponential distribution describes the probability distribution of the next time someone buys a hamburger. The expected value of Poisson distribution is \(\lambda\) , thus the expected value of exponential distribution, meaning the average interval between each hamburger sold, is \(1/ \lambda\).
  • Gamma distribution describes the probability distribution of how long it takes to sell \(n\) hamburgers.

　All kinds of distributions have countless ties.

distribution

The figure comes from Univariate Distribution Relationships

References