STAT 380 Lecture 19
End of Summary for Continuous Time Markov Chains:
In this case we write for the instantaneous ``birth'' rate:
and for the instantaneous ``death'' rate:
Necessary condition for existence of is
In this case
Suppose X a Markov Chain with stationary transitions. Then
which is the Chapman-Kolmogorov equation.
Now consider the chain starting from i and let be the first t for which . Then is a stopping time.
for each t.] Then
by the Markov property.
Conclusion: given X(0)=i, has memoryless porperty so has an exponential distribution. Let be the rate parameter.
Embedded Chain: Skeleton
Let be the stopping times at which tranisitons occur. Then . The sequence is a Markov chain by the Markov property. That reflects the fact that by design.
As before we say if for some t. It is fairly clear that for the X(t) if and only if for the embedded chain .
We say (i and j communicate) if and .
Suppose the chain has made n transitions so far so that . Then the event X(t+h)=j is, except for possibilities of probability o(h) the event that
The probability of this is
The Chapman-Kolmogorov equations are
Subtract from both sides, divide by h and let . Remember that is the identity. We find
The Chapman-Kolmogorov equations can also be written
Now subtracting from both sides, dividing by h and letting gives
Look at these equations in component form:
For our calculations of instantaneous transition rates gives
For i = k we have
(X(h)=i either means which has probability or there have been two or more transitions in [0,h], a possibility of probability o(h).) Thus
Let be the matrix with entries
is the infinitesimal generator of the chain.
Called Kolmogorov's backward equations.
On the other hand
These are Kolmogorov's forward equations.
Remark: When the state space is infinite the forward equations may not be justified. In deriving them we interchanged a limit with an infinite sum; the interchange is always justified for the backward equations but not for forward.
Example: . Then
and the chain is otherwise specivied by and . The matrix is
The backward equations become
while the forward equations are
Add first and third backward equations to get
Put t=0 to get . This gives
Plug this back in to the first equation and get
Multiply by and get
which can be integrated to get
can be written as
so we get
Notice: rows of are a stationary initial distribution. If rows are then
Fact: is long run fraction of time in state 0.
Ergodic Theorem in continuous time.
Birth and Death Processes
Consider a population of X(t) individuals. Suppose in next time interval (t,t+h) probability of population increase of 1 (called a birth) is and probability of decrease of 1 (death) is .
Jargon: X is a birth and death process.
All ; called a pure birth process.
All (0 is absorbing): pure death process.
and is a linear birth and death process.
, : Poisson Process.
and is a linear birth and death process with immigration.
Ingredients of Queuing Problem:
1: Queue input process.
2: Number of servers
3: Queue discipline: first come first serve? last in first out? pre-emptive priorities?
4: Service time distribution.
Example: Imagine customers arriving at a facility at times of a Poisson Process N with rate . This is the input process, denoted M (for Markov) in queuing literature.
Single server case:
Service distribution: exponential service times, rate .
Queue discipline: first come first serve.
X(t) = number of customers in line at time t.
X is a Markov process called M/M/1 queue:
Customers arrive according to PP rate . Each customer begins service immediately. X(t) is number being served at time t. X is a birth and death process with