User:Stuart.winter02/sandbox

The Poisson Distribution can also be derived from the Exponential Distribution.

Using the football analogy, let $$f(t)=\lambda e^{-\lambda t} ,t\ge 0$$ be the probability density function for scoring a goal at time t.

A match is a time interval of unit length (0, 1]. A team scores on average λ goals per match.

Then let $$p(k;\lambda )$$ be the probability of scoring exactly k goals in a match.

To score no goals in a match means you don’t score in the interval (0, 1]. Therefore,

$$p(0;\lambda )=\int _{1}^{\infty }\lambda e^{-\lambda t} dt=-\left[e^{-\lambda t} \right]_{1}^{\infty } =e^{-\lambda }$$

To score exactly one goal at time x, where 0 < x ≤ 1, means you cannot then score again in the interval (x, 1]. Because the probability density function is “memoryless”, this is calculated as not scoring in the interval (0, 1-x]. Therefore,

$$p(1;\lambda )=\int _{0}^{1}\lambda e^{-\lambda x} \left(\int _{1-x}^{\infty }\lambda e^{-dt}  dt\right)dx=\int _{0}^{1}\lambda e^{-\lambda x}  e^{-\lambda (1-x)} dx=\lambda e^{-\lambda }$$

To score the first goal at time y, where 0 < y ≤ 1, and the second goal at time x + y, where x > 0 and x + y ≤ 1, means you cannot then score again in the interval (x + y, 1]. Therefore,

$$p(2;\lambda )=\int _{0}^{1}\lambda e^{-\lambda y} \left\{\int _{0}^{1-y}\lambda e^{-\lambda x} \left(\int _{1-x-y}^{\infty }\lambda e^{-\lambda t}  dt\right)dx \right\}dy=\int _{0}^{1}\lambda e^{-\lambda y}  \left\{\int _{0}^{1-y}\lambda e^{-\lambda x} e^{-\lambda (1-x-y)} dx \right\}dy=\int _{0}^{1}\lambda ^{2} e^{-\lambda } (1-y) dy=\frac{\lambda ^{2} }{2} e^{-\lambda } $$

To score the first goal at time z, where 0 < z ≤ 1, and the second goal at time y + z, where y > 0 and y + z ≤ 1, and the third goal at time x + y + z, where x > 0 and x + y + z ≤ 1, means you cannot then score again in the interval (x + y + z, 1]. Therefore,

$$p(3;\lambda )=\int _{0}^{1}\lambda e^{-\lambda z} \left\langle \int _{0}^{1-z}\lambda e^{-\lambda y}  \left\{\int _{0}^{1-y-z}\lambda e^{-\lambda x} \left(\int _{1-x-y-z}^{\infty }\lambda e^{-\lambda t}  dt\right)dx \right\}dy\right\rangle dz=...=\frac{\lambda ^{3} }{3!} e^{-\lambda }$$

The above formulae can be seen to be following the generic pattern:

$$p(k,\lambda )=\frac{\lambda ^{k} }{k!} e^{-\lambda }$$

Stuart.winter02 (talk) 17:02, 12 March 2012 (UTC)