# What is the autocovariance of $x_t = U_1 \sin(2 \pi \omega_0 t) + U_2 \cos(2 \pi \omega_0 t)$

A time series with a periodic component can be constructed from $$x_t=U_1\sin(2\pi \omega_0t)+U_2\cos(2\pi \omega_0t),$$ where $$U_1$$ and $$U_2$$ are independent random variables with zero means and $$E(U_1^2)=E(U_2^2)=\sigma^2$$ Show this series is weakly stationary with autocovariance $$\gamma(h) = \sigma^2 \cos(2 \pi \omega_0 h)$$.

So I set $$p = 2\pi\omega_0$$

I have that the mean function is

$$E(x_t) = E(U_1 \sin(pt)) + E(U_2 \cos(pt)) = 0$$

which is not dependent on $$t$$

What I have for covariance is based on the following property:

If the random variables

$$U=\sum^m_{j=1}a_jX_j$$ and $$V=\sum^r_{k=1}b_kY_k$$

are linear combinations of (finite variance) random variables $${X_j}$$ and $${Y_k}$$, respectively, then

$$cov(U,V)=\sum^m_{j=1}\sum^r_{k=1}a_jb_k\operatorname{cov}((X_j,Y_k)$$

Furthermore, $$\operatorname{var}((U)=\operatorname{cov}((U,U)$$

Then setting

$$U = U_1 \sin(p(t+h)) + U_2 \cos(p(t+h))$$

and

$$V = U_1 \sin(p(t)) + U_2 \cos(p(t))$$

with

$$a_1 = \sin(p(t+h))$$ and $$a_2 = \cos(p(t+h))$$

and

$$b_1 = \sin(pt)$$ and $$b_2 = \cos(pt)$$

and

$$X_1 = Y_1 = U_1$$ and $$X_2 = Y_2 = U_2$$

And noting that $$\operatorname{cov}((U_1, U_2) = 0$$ and $$\operatorname{var}(U_1) = \operatorname{var}(U_2) = \sigma^2$$

we have

$$\gamma(h) = \operatorname{cov}(U, V)$$

$$= \sigma^2 [\sin(p(t+h))\sin(pt) + \cos(p(t+h)) \cos(pt)]$$

see rest of solution below

• You should presumably start by writing out the definition of the autocovariance. That's a crucial part of the problem context, and it's something you should provide. Mar 10, 2021 at 16:35
• You also have that $E(U_1U_2)=0$, since $U_1$ and $U_2$ are statistically independent random variables. Mar 10, 2021 at 16:55
• At this point, remember high school trigonometry... Mar 10, 2021 at 18:17
• ok. thanks. finished answering the question below. Mar 11, 2021 at 14:23

$$\sin(pt + ph)\sin(pt) + \cos(pt +ph)\cos(pt) = \cos(pt + ph -pt) = \cos(ph)$$
and the result is that $$\gamma_x(h) = \sigma^2\cos(2 \pi \omega_0 h)$$