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Stochastic Processes 1 Monday, November 14, 11 Definition and Classification X( , t): stochastic process: X :⌦⇥T ! ( , t) R X( , t) where ⌦ is a sample space and T is time. {X( , t) is a family of r.v. defined on {⌦, A, P and indexed by t 2 T . • For a fixed 0, X(( 0 , t) is a function of time or a sample function • For a fixed t0 , X(( , t0 ) is a random variable or a function of Discrete-time random process: n 2 T , X( , n) Continuous-time random process: t 2 T X( , n) 2 Monday, November 14, 11 2⌦ Probabilistic characterization of X( , t) {X( , t)} is an infinite family of r.v., we then need to know: 1. First-order p.d. or distribution F (x, t) = P [X( , t) x] for all t @F (x, t) f (x, t) = @x 2. Second-order p.d. or distribution F (x1 , x2 , t1 , t2 ) = P [X( , t1 ) x1 , X( , t2 ) x2 ] for all t1 , t2 2 T @ 2 F (x, t) f (x1 , x2 , t1 , t2 ) = @x1 @x2 3. n-th order p.d. or distribution F (x1 , · · · , xn , t1 , · · · , tn ) = P [X( , t1 ) x1 , · · · , X( , tn ) xn ] for all t1 , · · · , tn 2 T @ n F (x, t) f (x1 , · · · , xn , t1 , · · · , tn ) = @x1 · · · @xn Remark The n-th order p.d./p.D must also satisfy • Symmetry F (x1 , · · · , xn , t1 , · · · , tn ) = F (xi1 , · · · , xin , ti1 , · · · , tin ) for any permutation of {1, 2, · · · , n}. • Consistency in marginals F (x1 , · · · , xn Monday, November 14, 11 1 , t1 , · · · , tn 1) = F (x1 , · · · , xn 3 1 , 1, t1 , · · · , tn ) Kolmogorov’s Theorem If F (x1 , · · · , xn , t1 , · · · , tn ) for all sets of {ti } 2 T satisfy the symmetry and consistency conditions, there corresponds a stochastic process. Second-order Properties F (x, t), f (x, t), first-order p.d./p.D, and F (x1 , x2 , t1 , t2 ), f (x1 , x2 , t1 , t2 ) are not sufficient to characterize X( , t) but provide important information: • Mean function ⌘(t) ⌘X (t) = E(X(t)) = Z 1 xf (x, t)dx 1 function of t that depends on the first-order p.d. • Auto-covariance function CX (t1 , t2 ) CX (t1 , t2 ) = E[(X(t1 ) ⌘(t1 ))(X(t2 ) = E[X(t1 )X(t2 )] ⌘(t1 )⌘(t2 ) | {z } ⌘(t2 ))] RX (t1 ,t2 ) RX (t1 , t2 ) is the autocorrelation function, a function of t1 , t2 and secondorder probability density ZZ RX (t1 , t2 ) = E[X(t1 )X(t2 )] = x1 x2 f (x1 , x2 , t1 , t2 )dx1 dx2 4 Monday, November 14, 11 Remarks • Note that CX (t, t) = E[(X(t) ⌘(t))2 ] = 2 X (t) or the variance of X(t). • CX (t1 , t2 ) relates r.v.’s at times t1 and t2 , i.e., in time, while CX (t, t) relates the r.v. X(t) with itself, i.e., in space. Correlation coefficient CX (t1 , t2 ) X (t1 ) X (t2 ) rX (t1 , t2 ) = Cross-covariance X(t), Y (t) real processes CXY (t1 , t2 ) = E[(X(t1 ) ⌘X (t1 ))(Y (t2 ) ⌘Y (t2 ))] = E[X(t1 )Y (t2 )] ⌘X (t1 )⌘Y (t2 ) | {z } RXY (t1 ,t2 ) RXY (t1 , t2 ) is the cross-correlation function. X(t), Y (t) are uncorrelated i↵ CXY (t1 , t2 ) = 0 for all t1 and t2 2 T . 5 Monday, November 14, 11 Remark A zero-mean process ⌫(t) is called strictly white noise if ⌫(ti ) and ⌫(tj ) are independent for ti 6= tj . It is called white noise if it is uncorrelated for di↵erent values of t, i.e., ⇢ 2 ti = tj ⌫ C⌫ (ti , tj ) = ⌫2 (ti tj ) = 0 ti 6= tj Gaussian Processes X(t) is Gaussian if {X(t1 ), · · · , X(tn ) are jointly Gaussian for any value of n and times {t1 , · · · , tn }. Remark The statistics of a Gaussian process are completely determines by the mean ⌘(t) and the covariance CX (t1 , t2 ) functions. Thus f (x, t) requires ⌘(t), C(t, t) = 2 X (t) f (x1 , x2 , t1 , t2 ) requires ⌘(t1 ), ⌘(t2 ), C(t1 , t2 ), C(t1 , t1 ), C(t2 , t2 ) and in general the nth -order characteristic function 2 3 X X 4 !i ⌘X (ti ) 0.5 !i !k CX (ti , tj )5 X (bf !, t) = exp j i i,k 6 Monday, November 14, 11 Stationarity Strict Stationarity X(t) is strictly stationary i↵ F (x1 , · · · , xn , t1 , · · · , tn ) = F (x1 , · · · , xn , t1 + ⌧, · · · , tn + ⌧ ) for all n and ⌧ values.i.e., statistics do not change with time. If X(t) is strictly stationary then • its k th -moments are constant, i.e., E[X k (t)] = constant for all t. • RX (t, t + ⌧ ) = RX (0, ⌧ ), or it only depends on the lag ⌧ Proof • Using strict stationarity Z 1 Z E[X k (t)] = xk f (x, t)dx = 1 1 xk f (x, t + ⌧ )dx = E[X k (t + ⌧ )] 1 for any ⌧ , which can only happen when it is a constant. • By strict stationarity ZZ ZZ RX (t, t + ⌧ ) = x1 x2 f (x1 , x2 , t, t + ⌧ )dx1 dx2 = x1 x2 f (x1 , x2 , 0, ⌧ )dx1 dx2 = RX (0, ⌧ ) Wide-sense Stationarity (wss) X(t) is wss if 1. E[X(t)] = constant , Var[X(t)] = constant, for all t. 2. RX (t1 , t2 ) = RX (t2 t1 ) 7 Monday, November 14, 11 Examples of random processes Discrete-time Binomial process Consider the Bernoulli trials where ⇢ 1 event occurs at time n X(n) = 0 otherwise with P [X(n) = 1] = p, P [X(n) = 0] = 1 p = q. The discrete-time Binomial process counts the number of times the event occurred (successes) in a total of n trials, or n X Y (n) = X(i), Y (0) = 0, n 0 i=1 Since Y (n) = n X1 X(i) + X(n) = Y (n 1) + X(n) i=1 the process is also represented by the di↵erence equation Y (n) = Y (n 1) + X(n) 8 Monday, November 14, 11 Y (0) = 0, n 0 Characterization of Y (n) First-order p.d. f (y, n) = n X P [Y (n) = k] (y n ✓ ◆ X n k) = k=0 k=0 k (y k), 0kn Second and higher-order p.d. are difficult to find given the dependence of the Y (n)’s. Statistical averages E[Y (n)] = n X i=1 Var[Y (n)] E[X(i)] = np | {z } p = E[(Y (n) = X np)2 ] = E[( n X i=1 E[(X(i) p)(X(j) p)] = i,j = n X i=1 = npq p)2 ] (X(i) X i,j E[(X(i) | {z p)2 ] } Var(X(i))=(12 ⇥p+0⇥q) E[X(i) | p]E[X(j) {z =0 if i6=j p] independence of X(i) } p2 =pq Notice that E[Y (n)] = np, so it depends on time n, and Var[Y (n)] = npq also a function of time n, so the discrete binomial process is non-stationary. 9 Monday, November 14, 11 Random Walk Process Let the discrete binomial process be defined by Bernoulli trials ⇢ s event occurs at time n X(n) = s otherwise so that Y (n) = n X X(i), Y (0) = 0, n 0 i=1 Possible values at times bigger than zero n=1 n=2 n=3 .. . Y (1) = 1, 1 Y (2) = 2, 0, 2 Y (3) = 3, 1, 1, 3 .. . In general, at step n = n0 Y (n) can take values {2k n0 , 0 k n0 }, so for instance for n = 2, Y (2) can take values 2k 2, 0 k 2 or 2, 0, 2. 10 Monday, November 14, 11 Characterization of Y (n) First-order p. mass d. P [Y (n0 ) = 2k n0 , 0 k n0 ] Convert the random walk process into a binomial process by (s = 1) ⇢ X(n) + 1 1 event occurs at time n Z(n) = = 0 otherwise 2 n n n X X X Z(i) n = 2Ỹ (n) (2Z(i) 1) = 2 Y (n) = X(i) = i=1 i=1 i=1 where Ỹ (n) is the binomial process in the previous example. Letting m = 2k n0 , 0 k n0 then we have for 0 n0 n P [Y (n0 ) = m] = P [2Ỹ (n0 ) n0 = 2k n0 ] = P [Ỹ (n0 ) = k] = ✓ ◆ n 0 k n0 p q k Mean and variance E[Y (n)] = E[2Ỹ (n) n] = 2E[Ỹ (n)] Var[Y (n)] = 4Var[Ỹ (n)] = 4npq n = 2np Both of which depend on n, so that the process is nonstationary. 11 Monday, November 14, 11 n n k Sinusoidal processes X(t) = A cos(⌦t + ) A constant,frequency ⌦ and phase are random and independent. Assume ⇠ U[ ⇡, ⇡]. The sinusoidal process X(t) is w.s.s. E[X(t)] = AE[cos(⌦t + )] = AE[cos(⌦t) sin( ) = A[E[cos(⌦t)] E[sin( )] E[sin(⌦t)] E[cos( )]] independence | {z } | {z } 0 = sin(⌦t) cos( )] 0 0 The E[sin( )] = Z ⇡ ⇡ 1 sin( )d = 0 2⇡ area under a period of the sinusoid T0 = 2⇡, likewise for E[cos( )] = 0. The autocorrelation RX (t + ⌧, t) = = = E[X(t + ⌧ )X(t)] = A2 E[cos(⌦(t + ⌧ ) + ) cos(⌦t + )] A2 E[cos(⌦⌧ ) + cos(2⌦t + ⌦⌧ + 2 )] 2 ⇢ 2 A2 A A2 E[cos(⌦⌧ )] + E[cos(2⌦t + ⌦⌧ ) sin(2 )] E[sin(2⌦t + ⌦⌧ ) cos(2 )] 2 2 2 | {z } 0 = A2 E[cos(⌦⌧ )] 2 where the zero term is obtained because of the independence and that the expected values E[sin(2 )] = E[cos(2 )] = 0 by similar reasons as above. 12 Monday, November 14, 11 X(t) = A cos(!0 t), A ⇠ U[0, 1], !0 constant. X(t) is non-stationary. t 0 X(t) A cos(0) = A ⇠ U[0, 1] ⇡ 4!0 A cos( ⇡/4 ) ⇡ !o = p A 2 2 ⇠ U[0, p 2/2] A ⇠ U[ 1, 0] For each time the first-order p.d. is di↵erent so the process is not strictly stationary. The process can be shown not to be wide-sense stationary: E[X(t)] = cos(!0 t)E[A] = 0.5 cos(!0 t) RX (t + ⌧, t) = E[A2 ] cos(!0 (t + ⌧ )) cos(!0 t) RX (t, t) = = 2 A [cos(!0 ⌧ ) Var[X(t)] = + cos(!0 (2t + ⌧ ))] 2 A (1 + cos(2!0 t) which gives that the mean and the variance are not constant, and the autocorrelation is not a function of the lag, therefore the systems is not wide-sense stationary. 13 Monday, November 14, 11 Gaussian processes Let X(t) = A + Bt, where A and B are Gaussian. Consider 2 3 2 X(t1 ) 1 6 7 6 .. .. 4 5=4 . . 1 X(tn ) jointly Gaussian. Determine if X(t) is 3 t1 .. 7 A . 5 B tn for any n and times {tk }, 1 k n, the above is a linear combination of A and B which are Gaussian then {X(tk )} are jointly Gaussian, and so the process X(t) is Gaussian. Gaussian processes Consider a Gaussian process W (n), 1 < n < 1 such that E[W (n)] = 0 for all n R(k, `) = 2 (k `) = ⇢ 2 0 k=` k 6= ` such a process is a discrete white noise, determine its nth -order p.d. The covariance is C(k, `) = R(k, `) E[W (k)]E[W (`)] = R(k, `) which is zero when k 6= `, so W (k) and W (`) are uncorrelated and by being Gaussian they are independent. So Y f (wn1 , · · · , wnm ) = f (wni ) i 14 Monday, November 14, 11