Question

Consider a stochastic process Z(t)=X(t)Y(t) where X(t), Y(t) are independent and
wide-sense stationary(WSS) . Is Z(t) wide-sense stationary? Why?

Answer 1

Given

Z(t)=X(t)Y(t), where X(t), Y(t) are independent and
wide-sense stationary(WSS)

consider

E[Z(t)] = E[X(t)Y(t)]

= E[X(t)] E[Y(t)]

= Constant . Constant

= Constant

Hense Z(t) is a 1st order stationary random process. ------- (1)

Consider

RZZ (t,t+T) = E[Z(t) Z(t+T)]

  = E[X(t) Y(t) X(t+T) Y(t+T)]

  = E[X(t) X(t+T)] E[Y(t) Y(t+T)]

  = RXX(T).Ryy(T)

  = Rzz(T)

Hense Z(t) is a 2nd order stationary random process. ------ (2)

From equations (1) and (2) we can clearly say that if,Z(t)=X(t)Y(t) then Z(t) wide-sense stationary.