The distribution inherited by Y is called the Cauchy distribution. Using that. d dy ln(1 + y2 ) = 1 arctan(y)

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1 Stochastic Processes - MM3 - Solutions MM3 - Review Exercise Let X N (0, ), i.e. X is a standard Gaussian/normal random variable, and denote by f X the pdf of X. Consider also a continuous random variable Y with pdf f Y (y) = π( + y 2 ), y R. We know that E[X] = 0 and we are not surprised by this fact since f X is symmetric around zero. Now, since f Y is symmetric around zero as well, what is your immediate and intuitive guess of E[Y ]? - Our intuitive guess of E[Y ] would be zero as well. However, this is wrong! The distribution inherited by Y is called the Cauchy distribution. Using that d dy arctan(y) = + y 2 and d dy ln( + y2 ) = 2y + y 2, y R, you should verify that f Y is indeed a probability density function and afterwards calculate E[Y ] using the definition of expectation. - The pdf integrates to unity, since f Y (y)dy = π + y 2 dy = [ ] arctan(y) π = π π =. However, when we try to calculate the mean, we find that E[Y ] = yf Y (y)dy = 2π 2y + y 2 dy = [ ln ( + y 2)] 2π, which is an expression of the form. Such expressions are not welldefined and we say that E[Y ] does not exist. The tails of the Cauchy density are too heavy for the integral in E[Y ] to converge. Surprised? Verify it yourself by a small experiment in Matlab. To draw observations of Y, let Z U ( π 2, π 2 ) and set Y = tan(z). Draw an i.i.d. sample Y,..., Y N and estimate the mean. Repeat it over and over. Increase also N sequentially. Does the mean stabilize when N gets larger and larger?

2 MM3 - Exercise Let X(t), t T be a random process, either discrete-time (T Z) or continuoustime (T R). a) State the conditions that must be satisfied for the random process X(t) to be wide-sense stationary (WSS). Write the definition on the blackboard. - Its mean must be constant (no dependency on time t) and also its autocorrelation function must depend only on the time lag, i.e. t 2 t (or t t 2 ). b) State the conditions that must be satisfied for the random process X(t) to be strict-sense stationary (SSS). Write the definition on the blackboard. - All n th-order joint cdf s must be invariant under arbitrary time shifts and it must hold for all n N. c) Let c and k be constants and suppose that E[X(t)] = c and that Var[X(t)] = k for all time instances t T. Are these two assumptions sufficient to conclude that X(t) is WSS? - No, they are not. They are necessary conditions only. If X(t) is WSS then the first two moments are constant, but assuming that the first two moments are constant does not (in general) imply that the autocorrelation function depends on the time lag only. Try to design a random process for which the first two moments are constant but such that the autocorrelation function depends directly on absolute time instances and not just the time lag. MM3 - Exercise 2 Let Y (t), t T be a random processes and suppose we wish to show that Y (t) is indeed not strict-sense stationary (SSS). Will it be sufficient to show that: a) P ( Y (t ) 0 ) P ( Y (t 2 ) 0 ) for some time instances t and t 2 with t t 2? b) E [ Y n (t) ] exists and depends on t for some n N? c) Y (t) is not WSS? - Yes, is the answer in all three cases. Showing either of the three conditions will immediately imply that Y (t) is not SSS. Showing a) or b) implies that the first-order cdf s of Y (t) are not invariant under arbitrary time shifts. Showing c) will be sufficient as well, since SSS WSS. The contrapositive statement reads: not WSS not SSS. Hence, if the process is not stationary in the weaker sense, then it is definitely not stationary in the stronger sense either. 2

3 MM3 - Exercise 3 Consider the random process X(t) = A sin(2πt), t R, () where A is some real-valued random variable. Consider also the random process Y (t) = cos(2πt + Θ), t R, (2) where Θ U(0, 2π). Recall Exercise from MM and compare it with (2). Finally, consider the random process Z(t) = A sin(2πt + Θ), t R, (3) where A and Θ are as above and assume that they are independent. a) Is the process X(t) in () WSS? Why or why not? - We calculate the mean m X (t) := E [ X(t) ] = E [ A sin(2πt) ] = E[A] sin(2πt), and we realize that X(t) can be WSS only if E[A] = 0. However, if so, we still have to check its autocorrelation function and since R X (t, t 2 ) := E [ X(t )X(t 2 ) ] = E [ A 2] sin(2πt ) sin(2πt 2 ) = 2 E[ A 2]( cos ( 2π(t t 2 ) ) cos ( 2π(t + t 2 ) )), we realize that the process X(t) cannot be WSS (no matter if E[A] = 0). b) Is the process Y (t) in (2) WSS (take a look at your calculations from MM)? - Indeed, the process Y (t) is WSS since m Y (t) = 0 for all t R and the autocorrelation function reads R Y (t, t 2 ) = 2 cos ( 2π(t t 2 ) ). c) Is the process Z(t) in (3) WSS? Justify your answer. - Again, we calculate the mean and since A and Θ are independent we get m Z (t) := E [ Z(t) ] = E [ A sin(2πt + Θ) ] = E[A] E [ sin(2πt + Θ) ] = 0, and we realize that no restrictions apply to E[A] this time (only that it is finite!). The autocorrelation function reads R Z (t, t 2 ) := E [ Z(t )Z(t 2 ) ] = E [ A 2] E [ sin(2πt + Θ) sin(2πt 2 + Θ) ] = 2 E[ A 2] cos ( 2π(t t 2 ) ), and in conclusion we find that Z(t) is WSS whenever E[A] and E [ A 2] are both finite. 3

4 MM3 - Exercise 4 Consider the random process W (t) = A cos(ωt) + B sin(ωt), t R, where A and B are i.i.d. real-valued random variables. a) How are the mean and the variance of A related to the mean and the variance of B (respectively)? - When A and B are identically distributed all their moments must coincide, i.e. E [ A n] = E [ B n] for all n N. Hence, it follows that E[A] = E[B] and Var[A] = Var[B]. In general, if E [ A n] E [ B n] for some particular integer n, then certainly A and B cannot be identically distributed. b) For A and B, state the condition(s) that must be satisfied in order for W (t) to be WSS. - We calculate the mean and find that m W (t) := E [ W (t) ] = E [ A cos(ωt) + B sin(ωt) ] = E[A] cos(ωt) + E[B] sin(ωt), so in this case it is also required that E[A] = 0, and hence E[B] = 0 as well, in order for W (t) to be WSS. Furthermore, when assuming that E[A] = 0, we must check the autocorrelation function R W (t, t 2 ) := E [ W (t )W (t 2 ) ] [ (A = E cos(ωt ) + B sin(ωt ) )( A cos(ωt 2 ) + B sin(ωt 2 ) )] = E [ A 2] cos(ωt ) cos(ωt 2 ) + E [ B 2] sin(ωt ) sin(ωt 2 ) + E [ AB ]( ) cos(ωt ) sin(ωt 2 ) + cos(ωt 2 ) sin(ωt ), and since A and B are independent it follows that E [ AB ] = E[A]E[B] = 0. Using also the fact that E [ A 2] = E [ B 2], we get R W (t, t 2 ) = E [ A 2]( ) cos(ωt ) cos(ωt 2 ) + sin(ωt ) sin(ωt 2 ) = E [ A 2] cos ( ω(t t 2 ) ), and in conclusion, we realize that when E[A] = E[B] = 0 the random process W (t) is WSS. 4

5 c) Assume the necessary condition(s) from part b) such that W (t) is a WSS process. For technical reasons, assume furthermore that the moments E[A], E [ A 2], E [ A 3] and E [ A 4] exist. Show that W (t) is not SSS. - We assume that E[A] = 0 and with this assumption we know from part b) that W (t) is WSS and in particular we know that E [ W (t) ] = 0. Since W (t) is WSS we also know that E [ W 2 (t) ] is constant for all t (what is the constant?). You should verify that the third order moment of W (t) cannot help us either, unless we assume that E [ A 3] = 0. Accordingly, we proceed by calculating the fourth order moment E [ W 4 (t) ] = E [ W 2 (t)w 2 (t) ] [( ) 2 ] = E A 2 cos 2 (ωt) + B 2 sin 2 (ωt) + 2AB cos(ωt) sin(ωt), and by completing the square and applying the expectation operator we get E [ W 4 (t) ] = E [ A 4] cos 4 (ωt) + E [ B 4] sin 4 (ωt) + 6E [ A 2] E [ B 2] cos 2 (ωt) sin 2 (ωt) + 4E [ A 3] E[B] cos 3 (ωt) sin(ωt) + 4 E[A] E [ B 3] cos(ωt) sin 3 (ωt) = E [ A 4] ( cos 4 (ωt) + sin 4 (ωt) >0 ) + 6 E [ A 2] 2 cos 2 (ωt) sin 2 (ωt), >0 and this is a non-constant function of t. Hence, the first-order cdf s of W (t) are not invariant under arbitrary time shifts, i.e. W (t) is not SSS. Consider carefully the way we argue here. We are dealing with a process which we happen to know to be WSS. We ask ourselves whether this process is also SSS? Then we show that some moment of the process exists and depends on time. From this we conclude that the process is not SSS since if it had been, then all of its first-order cdf s would have been invariant under arbitrary time shifts and hence no moment (of any order) would depend on time. 5