Transcript Discrete Random Variables - Indian Institute of Technology

EE255/CPS226
Discrete Time Markov Chain (DTMC)
Dept. of Electrical & Computer engineering
Duke University
Email: [email protected], [email protected]
7/17/2015
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Discrete Time Markov Chain
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Markov process: dynamic evolution is such that future state
depends only on the present (past is irrelevant).
Markov Chain  Discrete state (or sample) space.
DTMC : time (index) is also discrete i.e. system is observed
only at discrete intervals of time.
X0, X1, .., Xn, .. :observed state (of a particular ensemble}
member (of the sample space) at discrete times, t0, t1,..,tn, ..
{X0, X1, .., Xn , ..} describes the states of a DTMC
Xn = j  system state at time step n is j. Then for a DTMC,
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P(Xn = in| X0 = i0, X1 = i1, …, Xn-1 = in-1) = P(Xn = in| Xn-1 = in-1)
pj(n)  P(Xn = j) (pmf), or,
pjk(m,n)  P(Xn = k | Xm = j ),
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Transition Probability
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pjk(m,n): probability transition function of a DTMC.
Homogeneous DTMC: pjk(m,n) = pjk(m-n) i.e., transition
probabilities exhibit stationary property. For such a DTMC,
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1-step transition prob, pjk = pjk(1) = P(Xn = k| Xn-1 = j) ,
Assuming 0-step transition prob as:
P(X0 = i0, X1 = i1, …, Xn = in)
= P(X0 = i0, X1 = i1, …, Xn-1 = in-1). P(Xn = in| X0 = i0, X1 = i1, …, Xn-1 = in-1)
= P(X0 = i0, X1 = i1, …, Xn-1 = in-1). P(Xn = in| Xn-1 = in-1) (due to Markov
Joint pmf is given by,
prop)
= P(X0 = i0, X1 = i1, …, Xn-1 = in-1).pin-1, in
:
= pi0(0)pi0, i1 (1) …pin-1, in (1) = pi0(0)pi0, i1 …pin-1, in
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Transition Probability Matrix
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The initial prob. is, pi0(0) = P(X0 = i0 ). In general,
p0(0) = P(X0 = 0 ), …, pk(0) = P(X0 = k ) etc, or,
p(0) = [p0(0), p1(0), … ,pk(0), ….] (initial prob. vector)
This allows us to define transition prob. matrix as,
Sum of ith row elements pi,0(0)+ pi,1(0)+ …
?
Any such sq. matrix with non-negative entries whose row
sum =1 is called a stochastic matrix.
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
State Transition Diagram
pij : describes random state value evolution from i to j
Node with labels i, j etc. and an arc labeled pij
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i
j
pij
Concept of ri reward (cost or penalty) for each state I allows
evaluation of various interesting performance measures.
Example: 2-state DTMC for a cascade binary comm.
channel. Signal values: ‘0’ or ‘1’ form the state values.
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xn-1 = 0
xn = 0
1-a
1-a
a
0
b
xn-1 = 1
th
(n-1) stage
1-b
a
1-b
xn = 1
1
b
th
n stage
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Total Probability
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Finding total pmf:
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
n-Step Transition Probability
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For a DTMC, find
Events: state reaches k (from i) & reaches j (from k) are
independent due to the Markov property (i.e. no history)
pik(m)
k
pkj(n)
j
i
0
m
m+n
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Invoking total probability theorem:
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Let P(n) : n-step prob. transition matrix (i,j) entry is pij(n).
Making m=1, n=n-1 in the above equation,
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Marginal pmf
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j, in general can assume countable values, 0,1,2, …. Defining,
pj(n) for j=0,1,2,..,j,… can be written in the vector form as,
Or,
Pn can be easily computed if n is finite. However, if n is countably
infinite, it may not be possible to compute Pn (and p(n) ).
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Marginal pmf Example
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For a 2-state DTMC described by its 1-step transition prob. matrix,
the n-step transition prob. Matrix is given by,
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Proof follows easily by using induction, that is, assuming that the above
is true for Pn-1. Then, Pn = P. Pn-1
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Computing Marginal pmf
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Previous example of a cascade digital comm channels: each stage
described by a 2-state DTMC, We want to find p(n) (a=0.25 & b=0.5),
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The ’11’ element for n=2 and n=3 are,
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Assuming initial pmf as, p(0) = [p0(0) p1(0)] = [1/3 2/3] gives,
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What happens to Pn as n becomes very large ( infinity)?
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
DTMC State Classification
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From the previous example, as n becomes infinity, pij(n) becomes
independent of n and i ! Specifically,
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Not all Markov chains may exhibit this behavior.
State classification may be based on the distinction that asymtotically:
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some states may be visited infinitely many times. Whereas, some other
states may be visited only a small number of times
Transient state: iff there is non-zero probability that the system will
NOT return to this state.
Define Xji to be the # of visits to state i, starting from state j, then,
For a transient state (i), visit count needs to finite, which requires pji(n)
 0 as n  infinity. Eventually, the system will always leave state i.
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
DTMC State Classification (contd.)
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State i is a said to be recurrent iff, starting from state i, the process
eventually returns to the state i with probability 1.
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For a recurrent state, time-to-return is a relevant measure. Define fij(n) as
the cond. prob. that the first visit to j from i occurs in exactly n steps.
If j = i, then fii(n) denotes the prob. of returning to i in exactly n steps.
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Known result:
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Let,
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Mean recurrence time for state i is
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Recurrent state
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Let i be recurrent and pii(n) > 0, for some n > 0.
For state i, define period di as GCD of all such +ve n’s that result in
pii(n) > 0
If di=1,  aperiodic and if di>1, then periodic.
Absorbing state: state i absorbing iff pii=1.
Communicating states: i and j are said to be communicating if there
exits directed paths from i and j and from j and i.
Closed set of states: A commutating set of states C forms a closed set,
if no state outside of C can be reached from any state in C.
Closed set c1
Closed set c2
Closed set ck
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Irreducible Markov Chains
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Markov chain states partitioned into two distinct subsets: c1,
c2, .., ck-1, ck , such that
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ci, i=1,2,..k-1 are closed set of recurrent nun-null.
ck transient states.
If ci contain only one state, then ci’s form a set absorbing states
If k=2 and ck empty, then c1 forms an irreducible Markov chain
Irreducible Markov chain: is one in which every state can be
reached from every other state in a finite no. of steps, i.e.,
for all i,j ε I, for some integer n > 0, pij(n) > 0. Examples:
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Cascade of digital comm. channels is
0
1
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Irreducible Markov Chains (contd.)
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If one state is recurrent aperiodic, then so are all the other
states. Same result if periodic or transient.
For a finite aperiodic irreducible Markov chain, pij(n)
becomes independent of i and n as n goes to infinity.
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All rows of Pn become identical
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Irreducible Markov Chains (contd.)
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Law of total probability gives,
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Therefore, 1st eq. can be rewritten as,
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In the matrix form,
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v is a probability vector, therefore,
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Self reading exercise (theorems on pp. 351)
For an aperiodic, irreducible, finite state DTMC,
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Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Irreducible Markov Chain Example
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Typical computer program: continuous cycle of compute & I/O
q0
0
1
q1
q2
2
1
1
1
q3
m
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The resulting DTMC is irreducible with period =1. Therefore,
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Sojourn Time
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If Xn = i, then Xn+1 = j should depend only on the current
state i, and not on the time spent in state i.
Let Ti be the time spent in state i, before moving to state j
DTMC will remain in state i in the next step with prob. pii
and,
Next step (n+1), toss a coin, ‘H’ Xn+1 = i, ‘T’Xn+1 # i
At each step, we perform a Bernoulli trial. Then,
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Markov Modulated Bernoulli Process
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Generalization of a Bernoulli process: the Bernoulli
process parameter is controlled by a DTMC.
Simplest case is Binary state (on-off) modulation
‘On’ Bernoulli param = c1; ‘Off’  c2’ (or =0)
1-a
1-b
a
0
1
b
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Controlling process is an irreducible DTMC, and,
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Reward assignment, r0 = c1 and r1=c2. This gives,
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Examples of Irreducible DTMCs
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Example 7.14: non-homogeneous DTMC for s/w reliability
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Slotted ALOHA wireless multi-access protocol
Terminal #5
Terminal #4
Terminal #3
Terminal #2
Terminal #1
Time slot
Pure ALOHA
Advantages:
• 2X more efficient than pure Aloha.
Terminal #5
• Automatically adapts to changes in station population
Terminal #4
Disadvantages:
Terminal #3
• Throughput maximum of 36.8% theoretical limit.
Terminal #2
• Requires queuing (buffering) for re-transmission
Terminal #1
Time slot
• Synchronization.
Slotted ALOHA
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Slotted ALOHA DTMC
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New and backlogged requests
Successful channel access iff :
1.
2.
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Backlogged Requests
New Requests
+
Exactly one new req. and no backlogged req.
Exactly one backlogged req. and no new req.
DTMC state: # of backlogged requests.
new
m-n
backlogged
n
++
+ +
x
xx
Σ
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Channel
Slloted Aloha contd.
• In a particular state n, successful contention occurs with prob. rn
• rn may be assigned as a reward for state n.
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Discrete-time Birth-Death Processes
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Special type of DTMC in which P has a tri-diagonal form,
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
DTMC solution steps
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Solving for v = vP, gives the steady state probabilities.
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Finite DTMCs with Absorbing States
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Example: Program having a set of interacting modules.
Absorbing state: failure state ( ps5 : unreliability)
s1
0.6
0.4
0.2
s2
s3
0.4
0.6
s4
0.6
0.4
0.4
0.4
s5
1
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Finite DTMCs, Absorbing States (contd.)
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M contains useful information.
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si
Xij : rv denoting random number to visits to j starting from i
E[Xij] = mij (for i, j = 1,2,…, n-1) . Need to prove this statement.
There are three distinct situations that can be enumerated
sk
si n
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sj
Xij =
{
δij ,
occurs with prob. pij
Xkj + δij , occurs with prob. Pik
k=1,2,..n
(δij : term accounts for i=j case)
Let rv Y denote the state at step #2
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(initial state: i)
E[Xij| y = n] = δij
E[Xij| y = k] = E[Xkj + δij]= E[Xkj]+ δij
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Finite DTMCs, Absorbing States (contd.)
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Since, P(Y=k) = pik , k=1,2,..n, total expectation rule gives,
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Over all (i,j) values, we need to work with the matrix,
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Therefore, fundamental matrix M elements give the
expected # of visit to state j (from i) before absorption.
If the process starts in state “1”, then m1j gives the average
# of visits to state j (from the start state) before absorption.
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Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University
Performance Analysis, Absorbing States
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By assigning rewards values to different state, a variety of
performance measures may be computed.
Average time to execute a program
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s1 is the start state; rjk : (fractional) execution time/visit for sj
Vj = m1j is the average # times statement block sjis executed
We need to calculate total expected execution time, I.e. until the
process gets absorbed into stop state (s5 )
Software reliability: jth reward = Rj: Reliability of sj .Then,
Bharat B. Madan, Department of Electrical and Computer Engineering, Duke University