6.262.PS10.sol

# 6.262.PS10.sol - 6.262 Discrete Stochastic Processes Spring...

This preview shows pages 1–3. Sign up to view the full content.

6.262 Discrete Stochastic Processes, Spring 2010 Problem Set 10 — Solutions due: Friday, April 30, 2010 Problem 2 (Voter Problem, Part I) a) For a three-state model, there are three independent Poisson processes each with rate 1. The corresponding merged process with rate ν = 3 yields the times at which voters reconsider their opinions, while the labeling Bernoulli process speciﬁes which of the voters will be the one reconsidering. Denote the opinions at time t by X ( t ) = ( X 1 ( t ) ,X 2 ( t ) 3 ( t )). Given a time t 0 , suppose that the process is in the state X ( t 0 ) = ( x 1 ,x 2 3 ). The wait until the next arrival is exponentially distributed with rate ν = 3, and the arrival is equally likely to correspond to each of the three voters. If the arrival is from the ﬁrst process, then the ﬁrst voter adopts either the second or the third voter’s opinion, where the two choices have equal probability, and similarly for the other arrivals. By the memoryless property of the independent Poisson processes (or equivalently, by the memoryless property of the meged process and that of the labeling process) and by the fact that the opinion choices are independent of everything, the above reasoning holds for any time t and any corresponding state ( x 1 2 3 ). As a result, we obtain the following process: 0 0 0 0 0 1 0 1 1 1 1 0 1 0 1 0 1 0 1 0 0 1 1 1 1/3 1/3 1/6 1/6 1/3 1/6 3 ν = 3 = 3 = 3 = 3 = 3 = 1/3 1/3 1/6 1 1/2 1/6 1/6 1/6 1/3 1/6 1/3 1/6 1/3 1/6 1/3 1/6 1/6 1/3 1/6 1 b) Given any state, wp.1 at most one arrival of the corresponding Poisson process can occur at any given time and therefore at most one person can change his/her opinion. It follows that given a state ( x 1 ,...,x n ) ∈ { 0 , 1 } n , the state can be accessible by a single transition from the following states ( x 1 n ) (0 , 0 ,..., 0), ( x 1 n ) (1 , 0 0), ( x 1 n ) (0 , 1 , 0 0) ( x 1 n ) (0 0 , 1), where the represets element-wise addition over the binary ﬁeld. State x is accessible from x if and only if there is a voter i whose opinion 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
at state x is the same as that of at least one of his/her neighbors. State x + (0 ,..., 1 0), with the 1 in position , is accessible from x if and only if voter has at least one neighbor whose opinion in state x diﬀers from that of x . Given states x ( i ) and x ( j ) , if x ( j ) = x ( i ) + (0 1 0) where the 1 is in some position ∈ { 1 ,...,n } (i.e. voter changes his/her opinion), then the most recent arrival to the merged Poisson process must come from voter , which occurs with probability 1 /n . The rate of transition from x ( i ) to x ( j ) is therefore ± ν n ² ψ ( ‘,i ) d , where d counts the neighbors of voter and ψ ( ) counts the neighbors of voter who have an opinion diﬀerent than that of voter in state x ( i ) . On the other hand, if x ( i ) = x ( j ) , then rate of transition from x ( i ) to x ( j ) equals ν n n X k =1 d k - ψ ( k,i ) d k , where we conditioned on which voter was reconsidering his/her opinion. Note that in both cases, ν = n is the transition rate of the merged Poisson process. (So, when x ( j ) 6 = x ( j ) , the rate of transitions between the two states does not explicitly depend on n .)
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

6.262.PS10.sol - 6.262 Discrete Stochastic Processes Spring...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online