Poisson Process as counting process distribution

Exponential distribution


This means that the condition on the random variable does not really matter in such cases. So, if you go to a barber whose service time follows an exponential distribution. Suppose you ask him what the expected time for a haircut is and he says 30 mins. Now during the haircut, after say 15 min, if you will ask him how much expected time is left, he will still say 30 mins! This is why the above property is known as memoryless property. Let’s plot a histogram to see how the density looks in Poisson distribution.
par(mfrow =c(2,2))
x <-rpois(1000, lambda =2)
hist(x, main ="Rate = 2" )
x <-rpois(1000, lambda =4)
hist(x, main ="Rate = 4" )
x <-rpois(1000, lambda =0.5)
hist(x, main ="Rate = 0.5" )
x <-rpois(1000, lambda =10)
hist(x, main ="Rate = 10" )

Poisson process modeling
The Poisson process is defined as follows: Let λ>0 be fixed. The process {N(t),t∈R^+} is called the Poisson process if it holds below conditions:
* N(0)=0
* N(t) has independent increments * N(t)∼poisson(λt), i.e., arrival count is distributed as Poisson distribution.
From the above definition, it can be proved that the inter-arrival time is exponential.
Let’s look at an example from Poisson distribution.
Q. If the number of the earthquake in Florida follows a Poisson distribution with rate 2/year, then find the probability that:
a. There arethree earthquakes in the year 2004.b. Given there are three earthquakes in 2004, there arefour earthquakes in 2005.c. Expected number of earthquakes in 2010-2012.
Ans: a. The probability that there are three earthquakes in 2004 is 0.18. We can calculate the same in r
dpois(3 , 2)
## [1] 0.180447
The number of events is independent at a disjoint time interval. So the condition has no effect on probability. We can simply calculate the probability for the 4 events in a year, which is 0.09
dpois(4 , 2)
## [1] 0.09022352
The number of earthquakes in the 3 years follows poisson (3* 2 = 6). Hence, the expected number of earthquakes is with variance, also being 6.
Example:
Q. The sequencing error in a genome sequencing project is, on average 1 wrong base pair in 100,000. A genome has length of 50,000,000 base pairs.
(a) Explain what a Poisson distribution is.
(b) (i) Using a Poisson distribution, calculate the expected number of sequencing errors in the genome.
(ii) If the genome were sequenced multiple times, how would the number of errors fluctuate around this expected value?
Ans:
- Poisson distribution: X follows poisson if
, for a real positive parameter , It has mean and variance both .
- (i) Since the Poisson distribution here has mean
, we have expected number of errors
