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.