I am trying to prove a time series is stationary and I think I am misunderstanding something. My question is

$$\ X_t=U_1 sin(yt) + U_2cos(yt) $$ with $E[U_1]=E[U_2]=0$ and $E[U_1^2]=E[U_2^2]=\sigma^2$ and $y=(2 \pi w_0)$ is a constant.

it was pretty easy to show that the mean does not depend on t, but then my book gave me the autocovariance = $ \gamma (h)=\sigma^2 cos(yh)$ and I am struggling to reach this.

my thoughts... let $s=t+h$ $$cov(x_t, x_s) = E[(x_s - \mu_s)(x_t - \mu_t)]=E[x_s x_t] - 0 =E[x_s x_t]$$

so we need the expectation at two different time periods. I tried to $$\ E[x_t x_s]=E[(U_1 sin(yt) + U_2cos(yt) )(U_1 sin(y(t+h)) + U_2cos(y(t+h)) )]$$

but after continuing with this line, nothing came of it. I don't think we need to use trigonometric identities, so can anyone point me in the right direction?

  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$
    – Silverfish
    Oct 1, 2015 at 2:05
  • $\begingroup$ You're missing some information in the question; were you told that $U_1$ and $U_2$ were independent? Or instead, were you given $\mathbb{E}(U_1 U_2)$ or $\operatorname{Cov}(U_1,U_2)$? Reconsider "I don't think we need to use trigonometric identities" - the angle in the trig function in the answer does not match the angle in the any of the trig functions in your working, so you'll have to use a trig identity. Can you identify why "nothing came of it"? Keep going. And, as early as you can, use that missing info. The result will pop out. The trig identity is a simple one, nothing fiendish. $\endgroup$
    – Silverfish
    Oct 1, 2015 at 2:06
  • $\begingroup$ If you are willing to add the restriction that $U_1$ and $U_2$ are independent random variables (as @Silverfish suggests), then you can find an answer to your question on math.SE $\endgroup$ Oct 1, 2015 at 2:46
  • $\begingroup$ I had actually gotten to this step in my own proof, $=E[A2]cos(t)cos(t+τ)+E[B2]sin(t)sin(t+τ)$ but wasnt aware this was an identity equal to σ2cos(τ), anyone know what the specific identity is? im not that great at trig $\endgroup$ Oct 1, 2015 at 19:12
  • $\begingroup$ Ummm... $$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)$$ applied bass ackwards with $\alpha = t+\tau$ and $\beta=t$, maybe? See Equation 4.3.17 here and use the fact that $\cos(-\beta)=\cos(\beta)$ while $\sin(-\beta) = -\sin(\beta)$. $\endgroup$ Oct 2, 2015 at 3:00