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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

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    $\begingroup$ 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. $\endgroup$ Commented Mar 10, 2021 at 16:35
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    $\begingroup$ You also have that $E(U_1U_2)=0$, since $U_1$ and $U_2$ are statistically independent random variables. $\endgroup$
    – minmax
    Commented Mar 10, 2021 at 16:55
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    $\begingroup$ At this point, remember high school trigonometry... $\endgroup$ Commented Mar 10, 2021 at 18:17
  • $\begingroup$ ok. thanks. finished answering the question below. $\endgroup$
    – Hasselhoff
    Commented Mar 11, 2021 at 14:23

1 Answer 1

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By trig identities

$\sin(pt + ph)\sin(pt) + \cos(pt +ph)\cos(pt) = \cos(pt + ph -pt) = \cos(ph)$

and the result is that $\gamma_x(h) = \sigma^2\cos(2 \pi \omega_0 h)$

so neither the mean nor the autocovariance is dep on t and the series is stationary.

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