- Question 1 The Random Telegraph Signal Process Let N T T 0 Denote A Poisson Process With Rate 2 And Let X Be I 1 (160.61 KiB) Viewed 28 times
Question 1: (The random Telegraph Signal Process) Let {N(t),t > 0} denote a Poisson process with rate 2, and let X, be independent of this process and be such that 1 P{X = 1} = P{X = -1) = Defining X(t) = X. (-1)^(t) then {X(t),t 2 0} is called a random telegraph signal process. To see that it is stationary, note first that starting at any time t, no matter what the value of N(t), as X, is equally likely to be either plus or minus 1, it follows that X(t) is equally likely to be either plus or minus 1. Hence, because the continuation of a Poisson process beyond any time remains a Poisson process, it follows that {X(t), t > 0} is a stationary process. 1. Calculate E[X] and Var[Xo). 2. Compute E[X(t)] the mean function of the random telegraph signal. 3. Compute Cov[X(t), X(t + s)] the covariance function of the random telegraph signal. Application: For an application of the random telegraph signal consider a particle moving at a constant unit velocity along a straight line and suppose that collisions involving this particle occur at a Poisson rate 1. Also suppose that each time the particle suffers a collision it reverses direction. Therefore, if X, represents the initial velocity of the particle, then its velocity at time t-call it X(t)-is given by X(t) = X, (-1)N(t), where N(t) denotes the number of collisions involving the particle by time t. Hence, if X, is equally likely to be plus or minus 1, and is independent of{N(t),t > 0}, then {X(t),t > 0} is a random telegraph signal process. If D(t) = = ) X(S) ds Then D(t) represents the displacement of the particle at time t from its position at time 0. With mean E[D(t)] = excs Ś E[X(s) ds = 0