Stationary independent increments. 1. Random changes of the form X t+h X t fixed h > 0 are called increments of the process.

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1 Stationary independent increments 1. Random changes of the form X t+h X t fixed h > 0 are called increments of the process. 2. If each set of increments, corresponding to non-overlapping collection of time intervals, is mutually independent, then X t is said to be a process with independent increments. That is, for any k and any choice of sampling instants t 1 < t 2 < t k, the random variables are independent random variables. X(t 2 ) X(t 1 ),, X(t k ) X(t k 1 ) 3. If X t+h X t has a distribution that depends only on h, not on t, then X t is said to have stationary independent increments.

2 Markov process A radom process X(t) is said to be Markov if the future of the process given the present is independent of the past, for any k and any choice of sampling instants t 1 < t 2 < < t k and for any x 1, x 2,, x k. (i) continuous-valued X(t) f X(tk ) (x k X(t k 1 ) = x k 1,, X(t 1 ) = x 1 ) = f X(tk ) (x k X(t k 1 ) = x k 1 ) (ii) discrete-valued X(t) P [X(t k ) = x k X(t k 1 ) = x k 1,, X(t 1 ) = x 1 ] = P [X(t k ) = x k X(t k 1 ) = x k 1 ].

3 Sum processes S n = n i=1 X i, X i s are iid random variables It is called binomial counting process if X i s are iid Bernoulli random variables. 1. S n is Markovian: P [S n = α n S n 1 = α n 1 ] = P [S n = α n S n 1 ] = α n 1, S n 2 = α n 2,, S 1 = α 1 ]. This is because S n = S n 1 + X n and the value taken by X n is independent of the values taken by X 1,, X n 1.

4 2. S n has independent increments in non-overlapping time intervals (no X n s are in common). 3. S n has stationary increments P [S n+k S k = β] = P [S n S 0 = β] = P [S n = β], independent of k. This is because S n+k S k = sum of n iid random variables.

5 Example Find P [S n = α, S m = β], n > m. Solution P [S n = α, S m = β] = P [S n S m = α β, S m = β] = P [S n S m = α β]p [S m = β], due to independent increments over non-overlapping intervals. Further, from stationary increments property, we obtain P [S n = α, S m = β] = P [S n m = α β]p [S m = β].

6 Mean, variance and autocovariance of sum processes Let m X and σ 2 X denote the mean and variance of X i, for any i. Since the sum process S n is the sum of n iid random variables, its mean and variance are The autocovariance of S n is m S (n) = E[S n ] = ne[x i ] = nm X VAR[S n ] = nvar[x i ] = nσx 2. C S (n, k) = E[(S n E[S n ])(S k E[S k ])] = E = = n i=1 n k i=1 j=1 min(n,k) i=1 (X i m) k j=1 (X j m) E[(X i m)(x j m)] C X (i, i) = min(n, k)σ 2 since E[(X i m)(x j m)] = σ 2 X δ i,j and only those terms with i = j survive.

7 Alternative method Without loss of generality, we let n k so that n = min(n, k). C S (n, k) = E[(S n nm)(s k km)] = E[(S n nm) {S n nm + (S k km) (S n nm)}] = E[(S n nm) 2 ] + E[(S n nm)(s k S n (k n)m)], and since S n and S k S n are independent, so C S (n, k) = E[(S n nm) 2 ] + E[S n nm]e[s k S n (k n)m] = E[(S n nm) 2 ] = Var[S n ] = nσx 2 = min(n, k)σ2 X.

8 Binomial counting process Let S n be the sum of n independent Bernoulli random variables, that is, S n is a binomial random variable with parameters n and p. P [S n = j] = { nc j p j (1 p) n j for 0 j n 0 otherwise Note that S n has mean np and variance np(1 p)..

9 Example A fair coin is tossed 500 times. Find the probability that the number of heads will not differ from 250 by more than 10. Solution Let X k be the discrete variable which equals 1 if the head turns up and 0 otherwise in the kth toss, k = 1, 2,, 500. These random variables: X 1, X 2,, X 500 are considered to be independent, identically distributed (iid) random variables. Define S 500 = X 1 + X X 500 that gives the number of heads that turn up in N = 500 tosses. The coin tossing experiment is a binomial experiment; and for a fair coin, the probability that a head turns up equals p = 0.5. The mean number of heads showing up in N tosses = Np = 250. The standard deviation of the number of heads showing up in N tosses = Np(1 p) = 500 ( ) ( ) = The sum S 500 tends to the distribution that is Gaussian with mean and standard deviation equal to those of S 500.

10 Define the standardized normal variable: Z = S (a) We require the probability that S 500 lies between 240 and 260, or considering the data as continuous, between and in standard units = in standard units = Let N(y) = y 1 2π e x2 /2 dx = Required probability = area under the standard normal curve between 0.94 and 0.94 = N(0.94) N( 0.94) =

11 Example Let S n be a binomial counting process. Compute P [S n = j, S m = i], n m. Solution (i) n > m P [S n = j, S m = i] = P [S n S m = j i, S m = i] = P [S n m = j i, S m = i] (stationary increments) = P [S n m = j i]p [S m = i] (independent increments) Now, P [S n m = j i] = n m C j i p j i (1 p) (n m) (j i) and P [S m = i] = m C i p i (1 p) m i. (ii) n < m In a similar manner, P [S n = j, S m = i] = P [S n = j, S m n = i j] = P [S n = j]p [S m n = i j].

12 Example One dimensional random walk (frog hopping) The initial position of a frog is taken to be at position 0. Let p and q = 1 p be the probabilities that the frog will choose to move to the right and to the left, respectively. Let X n be the position of the frog after n moves. P [X 1 = 1] = p, P [X 1 = 1] = q P [X 2 = 2] = p 2, P [X 2 = 0] = 2pq, P [X 2 = 2] = q 2 P [X 3 = 3] = P 3, P [X 3 = 1] = 3p 2 q, P [X 3 = 1] = 3pq 2, P [X 3 = 3] = q 3 {X 1,, X n } is an indexed family of random variables. The random walk process is a discrete-valued discrete time random process.

13 Questions 1. Is this random process Markovian? Check whether only the information of the present position is relevant for predicting the future movements. Suppose the frog is at position 4 after 10 moves, does the probability that it will be in position 8 after 16 moves (6 more moves) depend on how it moves to position 4 within the first 10 moves? 2. Are X 10 X 4 and X 16 X 12 independent? How about X 10 X 4 and X 12 X 8?

14 Each move is an independent Bernoulli trial. P X3 (3) = p 3, P X3 (1) = 3 C 1 p 2 q, P X3 ( 1) = 3 C 2 pq 2, P X3 ( 3) = q 3. In general, P Xk (j) = k C k+j p k+j 2 q k j 2, k j k, 2 and when k is odd, j must be odd. How to find the joint pmf s? For example, P [X 2 = 0, X 3 = 3] = P [X 3 = 3 X 2 = 0]P [X 2 = 0] = 0; P [X 2 = 2, X 3 = 1] = P [X 3 = 1 X 2 = 2]P [X 2 = 2] = qp 2.

15 Wiener process Let the random walk be symmetric, that is, p = 1/2. We attempt to obtain a continuous-time process by letting X(t; t) be the accumulated sum up to time t, where t is the time interval between successive jumps. Let the jump distance be h and t = n t. Here, n is the number of jumps within time interval [0, t]. Let D n be the iid process of ±1 random variables, and let S n denote the corresponding sum process. Then, X(t) = hs n, whose mean and variance are E[X(t)] = he[s n ] = 0 VAR[X(t)] = h 2 nvar[d n ] = h 2 n, since VAR[D n ] = 1.

16 Suppose we shrink t 0 and h 0, but h 2 = α t, where α is a finite constant. Then VAR[X(t)] = (α t) = αt. t We call X(t) the Wiener random process. It is a continuous-time process that begins at the origin, has zero mean for all times, and has a variance that increases linearly with time. ( t ) The pdf of X(t) approaches that of a Gaussian random variable with mean zero and variance αt: f X(t) (x) = 1 2παt e x2 /2αt,

17 Example Continuous processes with stationary independent increments Suppose X(t) is a continuous process with stationary independent increments, show that Solution VAR[X t ] = αt for some constant α. For s 0 and t 0, we write X t+s = (X t+s X s ) + (X s X 0 ). Since X t+s X s and X s X 0 are independent, we have VAR[X t+s ] = VAR[X t+s X s ] + VAR[X s X 0 ] = VAR[X t ] + VAR[X s ].

18 Write f(u) = VAR[X u ], then f(t + s) = f(t) + f(s), s 0, t 0. How to find f(t)? Assume f(t) to be differentiatiable df(t) dt so that f(t) = αt. = lim t 0 = lim t 0 f(t + t) f(t) t f( t) f(0) t = lim t 0 = f (0) α f(t) + f( t) f(t) t

19 Covariance From VAR[X t ] = VAR[X t+s X s ] = VAR[X t+s ] + Var[X s ] 2COV[X t+s, X s ] so that COV[X t+s, X s ] = 1 2 {VAR[X s] + Var[X t+s ] Var[X t ]} = 1 {as + a(t + s) at} = as. 2 Hence, COV[X t1, X t2 ] = a min(t 1, t 2 ).

20 Example Let X n consist of an iid sequence of Poisson random variables with mean α. (a) Find the pmf of the sum process S n. (b) Find the joint pmf of S n and S n+k. (c) Find COV(S n+k, S n ). Solution (a) Recall that the sum of independent Poisson variables remains to be Poisson distributed. Let µ i be the mean of the ith Poisson variable, i = 1, 2,, n, among this group of n independent Poisson variables. The sum of these n variables is a Poisson variable with mean µ = µ 1 + µ µ n. Now, S n is a Poisson variable with mean nα. P [S n = m] = (nα)m e nα. m!

21 (b) P [S n = m, S n+k = l] = P [S n+k = l S n = m]p [S n = m], l m = P [S k = l n]p [S n = m] (stationary increments) = (kα)l n e kα (l n)! (nα) m e nα. m! (c) VAR(S n ) = VAR(S n+k S k ) = VAR(S n+k ) + VAR(S k ) 2COV(S k, S n+k ) so that COV(S k, S n+k ) = 1 2 [VAR(S n+k) + VAR(S k ) VAR(S n )] = 1 [(n + k)α + kα nα] = kα. 2