STAT 380: Lecture 14
Equivalence of the models
Model 3 implies 1: Fix t, define to be conditional probability of 0 points in (t,t+s] given value of process on [0,t].
Derive differential equation for f. Given process on [0,t] and 0 points in (t,t+s] probability of no points in (t,t+s+h] is
Given the process on [0,t] the probability of no points in (t,t+s] is . Using P(AB|C)=P(A|BC)P(B|C) gives
Now rearrange, divide by h to get
Let and find
Differential equation has solution
Notice: survival function of exponential rv..
General case:
Notation: N(t) =N(0,t).
N(t) is a non-decreasing function of t. Let
Evaluate by conditioning on and N(t)=j.
Given N(t)=j probability that N(t+h) = k is conditional probability of k-j points in (t,t+h].
So, for :
For j=k-1 we have
For j=k we have
N is increasing so only consider .
Rearrange, divide by h and let t get
For k=0 the term is dropped and
Using we get
Put this into the equation for k=1 to get
Multiply by to see
With we get
For general k we have and
Check by induction that
Hence: N(t) has Poisson distribution.
Similar ideas permit proof of
From which (by induction) we can prove that N has independent Poisson increments.
Exponential Interarrival Times
If N is a Poisson Process we define to be the times between 0 and the first point, the first point and the second and so on.
Fact: are iid exponential rvs with mean .
We already did rigorously. The event T> t is exactly the event N(t)=0. So
which is the survival function of an exponential rv.
I do case of . Let be two positive numbers and , . The event
This is almost the same as the intersection of the four events:
which has probability
Divide by and let and go to 0 to get joint density of is
which is the joint density of two independent exponential variates.
More rigor:
First step: Compute
This is just the event of exactly 1 point in each interval for ( ) and at least one point in which has probability
Second step: write this in terms of joint cdf of . I do k=2:
Notice tacit assumption .
Differentiate twice, that is, take
to get
Simplify to
Recall tacit assumption to get
That completes the first part.
Now compute the joint cdf of by
This is
Differentiate twice to get
which is the joint density of two independent exponential random variables.
Summary so far:
Have shown:
Instantaneous rates model implies independent Poisson increments model implies independent exponential interarrivals.
Next: show independent exponential interarrivals implies the instantaneous rates model.
Suppose iid exponential rvs with means . Define by if and only if
Let A be the event . We are to show
and
If n(s) is a possible trajectory consistent with N(t) = k then n has jumps at points
and at no other points in (0,t].
So given with n(t)=k we are essentially being given
and asked the conditional probabilty in the first case of the event B given by
Conditioning on irrelevant (independence).
The numerator may be evaluated by integration:
Let to get the limit
as required.
The computation of
is similar.
Properties of exponential rvs
Convolution: If X and Y independent rvs with densities f and g respectively and Z=X+Y then
Differentiating wrt z we get
This integral is called the convolution of he densities f and g.
If iid Exponential then has a Gamma distribution. Density of is
for s> 0.
Proof:
Then
This telescopes to
Extreme Values: If are independent exponential rvs with means then has an exponential distribution with mean
Proof:
Memoryless Property: conditional distribution of X-x given is exponential if X has an exponential distribution.
Proof:
Hazard Rates
The hazard rate, or instantaneous failure rate for a positive random variable T with density f and cdf F is
This is just
For an exponential random variable with mean this is
The exponential distribution has constant failure rate.
Weibull random variables have density
for t> 0. The corresponding survival function is
and the hazard rate is
which is increasing for , decreasing for . For this is the exponential distribution.
Since
we can integrate to find
so that r determines F and f.