Homogeneous Poisson Process (original) (raw)

If we denote number of occurrences during a time interval of length t as X(t) then

P(X(t)=n) = \frac{e^{-\lambda t}(\lambda t)^n}{n!}

**Examples -
Many real life situations can be modelled using Poisson Process. Suppose we consider number of accidents in a road. We can easily understand that the three above conditions are satisfied. For two disjoint time intervals number of accidents in a given road are independent, Again it is quite improbable that two or more accidents occur at a small interval of time. Intuitively we can also assume that probability that an accident occurs during a small interval of time is proportional to the length of the time interval. Number of earthquakes in a place can also be modelled using Poisson process.

**Derivation -
Now we prove our claim that if X(t) be the number of occurrence in an interval of length t, then

P(X(t)=n) = \frac{e^{-\lambda t}(\lambda t)^n}{n!}

where

\lambda

is the rate of occurrence.

We will use mathematical induction to prove the statement. First we write the assumptions written above in mathematical terms. According to assumption 3 in a small time interval h

P(X(h) >1) = o(h)

where

\frac{o(h)}{h}

tends to zero as h tends to zero or

1-P(X(h)=0) -P(X(h)=1) = o(h)

.
Again if

\lambda

be the rate of occurrence then according to assumption 2 we get,

P(X(h)=1) =\lambda h

.

Let us take an interval (0, t) and a small interval (t, t+h). We will denote P(X(t)=n) as

P_n(t)

. So the above equations can be written as,

1-P_0(h) -P_1(h) = o(h)

or

P_0(h) = 1-\lambda h -o(h)

So we have to prove that

P_n(t) = \frac{e^{-\lambda t}(\lambda t)^n}{n!}

.
First we will prove the result for n=0 and n=1. Then we will show that if the result is true for n=m then it will be true for n=m+1.

Take the interval (0, t+h). Now,

P_0(t+h)=P_0(t)P_0(h)

(since the occurrences in the interval (0, t) and (t, t+h) are independent) or ,

P_0(t+h) = P_0(t)(1-\lambda (h) -o(h)

or

\frac{P_0(t+h)-P_0(t)}{h} = -\lambda P_0(t) -\frac{o(h)}{h}

.

Taking limit as h tends to zero we get,

P_{0}'(t) =- \lambda P_0(t).

.
The solution of above differential equation is,

P_0(t)=ce^{-\lambda t}

,
taking the initial condition

P_0(0)=1

we evaluate c=0. Hence,

P_0(t)=e^{- \lambda t}

, so our claim is true for n=0.

Now we try to prove it for n=1.

P_1(t+h)=P_1(t)P_0(h) + P_0(t)P_1(h)

(We use the fact that the occurrence must be in either of the interval (0, t) and (t, t+h)), or

P_1(t+h) =P_1(t)(1-\lambda h-o(h))+ e^{-\lambda t}(\lambda h)

,
or

\frac{P_1(t+h) -P_1(t)}{h}= -\lambda P_1(t) - \lambda e^{-\lambda t}- \frac{o(h)}{h}

.

Again taking limit as h tends to zero,

P_{1}'(t)=-\lambda P_1(t) - \lambda e^{-\lambda t}

.
This is a first order linear differential equation and the solution is,

P_1(t)=\lambda t e^{-\lambda t}+c_1

where

c_1

is a constant. Since,

P_1(0)=0

. We get,

c_1=0

. Hence

P_1(t)=\lambda t e^{-\lambda t}

, or

P_1(t)=\frac{(\lambda t)^1 e^{-\lambda t}}{1!}

.
So our claim is true for n=1. We assume that our claim is true for n=m.

We will show that it is true for n=m+1. So,

P_{m+1}(t+h) =P_{m+1}(t)P_0(h) +P_m(t)P_1(h) +\sum_{j=1}^m {P_{m-j}(t)P_{j+1}(h)}

,
(We assume that the m+1 occurrence can happen in different ways such as m+1 occurrences in (0, t) and no occurrence in (t, t+h) or m occurrences in (0, t) and 1. occurrence in (t, t+h), or m-j occurrences in (0, t) and j+1 occurrence in (t, t+h) for j=1 to m). So,

P_{m+1}(t+h) =P_{m+1}(t)(1-\lambda h-o(h)) + \frac{e^{-\lambda t}(\lambda t)^m}{m!}\lambda h +\sum_{j=1}^m P_{m-j}(t)o(h)

Since,

P_{j+1}(h)=o(h)

for j>=1.
Or,

\frac{P_{m+1}(t+h)-P_{m+1}(t)}{h}=-\lambda P_{m+1}(t)+\frac{{\lambda}^{m+1}t^m}{m!}e^{-\lambda t}+\frac{o(h)}{h}\sum_{j=1}^{m}P_{m-j}(t)-\frac{o_(h)}{h}P_{m+!}(t)

Taking limit as h goes to zero we have,

P'_{m+1}(t)=-\lambda P_{m+1}(t) + \frac{{\lambda}^{m+1}t^m}{m!}e^{-\lambda t}

.
This is again a first order differential equation whose solution so,

P_{m+1}(t)=\frac{(\lambda t)^{m+1}e^{-\lambda t}}{(m+1)!}+c_2

.
If e assume that

P_{m+1}(0)=1

we get

c_2=0

.

So the final result is,

P_{m+1}(t)=\frac{(\lambda t)^{m+1}e^{-\lambda t}}{(m+1)!}

.
Hence the result is proved.

Thus we have derived the pmf of no. of occurrences in a Poisson Process which is a Poisson Distribution with parameter

\lambda

. Now if this

\lambda

is a function of time we call the process as non-homogeneous Poisson process.