Question

Let \(Y_{1}\) and \(Y_{2}\) be independent unit normal random variables and for some constant \(w\) set $$ X(t)=Y_{1} \cos w t+Y_{2} \sin w t, \quad-\infty

Short answer

In summary, \({X(t)}\) is a weakly stationary process because its mean function, \(E[X(t)]\), is constant and its autocovariance function, \(E[(X(t)-E[X(t)])(X(s)-E[X(s)])]\), does not depend on time. Since \({X(t)}\) is also a Gaussian process, weak stationarity implies stationarity. Therefore, \({X(t)}\) is a stationary process.

Step-by-step solution

Part (a): Show that \({X(t)}\) is weakly stationary

First, we need to compute the mean of \(X(t)\): $$ E[X(t)] = E[Y_{1} \cos(wt) + Y_{2} \sin(wt)] = E[Y_{1} \cos(wt)] + E[Y_{2} \sin(wt)] $$ Since each of \(Y_{1}\) and \(Y_{2}\) are standard normal random variables, their expectations are equal to zero: $$ E[X(t)] = 0 + 0 = 0 $$ Therefore, the mean function \(E[X(t)]\) is constant and does not depend on time. Now, we compute the autocovariance function. We want to find: $$ E[(X(t)-E[X(t)])(X(s)-E[X(s)])] = E[X(t)X(s)] $$ As the mean function is 0, we can substitute the given formula for \(X(t)\) and \(X(s)\): $$ E[(Y_{1}\cos(wt)+Y_{2}\sin(wt))(Y_{1}\cos(ws)+Y_{2}\sin(ws))] $$ Next, we expand the expression and apply the expected value operator to each term: $$ E[Y_{1}^2\cos(wt)\cos(ws)]+E[Y_{1}Y_{2}\cos(wt)\sin(ws)]+E[Y_{1}Y_{2}\sin(wt)\cos(ws)]+E[Y_{2}^2\sin(wt)\sin(ws)] $$ Since \(Y_{1}\) and \(Y_{2}\) are independent, \(E(Y_{1}Y_{2}) = E(Y_{1})E(Y_{2}) = 0\). Therefore, the middle two terms are equal to zero. We're left with: $$ E[Y_{1}^2\cos(wt)\cos(ws)]+E[Y_{2}^2\sin(wt)\sin(ws)] $$ Using the property that \(E[kY] = kE[Y]\) (for a constant \(k\)), we have: $$ E[Y_{1}^2]\cdot\cos(wt)\cos(ws) + E[Y_{2}^2]\cdot\sin(wt)\sin(ws) $$ Since \(Y_{1}\) and \(Y_{2}\) are standard normal random variables, their variances are equal to 1, so we have: $$ \cos(wt)\cos(ws) + \sin(wt)\sin(ws) $$ This autocovariance function does not depend on time(the time difference \(t-s\) is not present in the expression), so \({X(t)}\) is weakly stationary.

Part (b): Argue that \({X(t)}\) is stationary

We have shown that \({X(t)}\) is weakly stationary because its mean and autocovariance functions are time-invariant. To make a claim that \({X(t)}\) is stationary, we need to show that its full joint distributions are also time-invariant. Since \({X(t)}\) is constructed as a linear combination of two independent Gaussian random variables, it's also a Gaussian random process. For a Gaussian process, weak stationarity implies stationarity because the full joint distribution of a Gaussian process is determined entirely by the mean and covariance structure, both of which we have shown to be time-invariant. Thus, we can argue that \({X(t)}\) is a stationary process.