# Show that $\{X(t), = \cos(t+U)\}$, $U \sim \mathrm{Unif} (0, 2\pi)$ is a wide-sense stationary process

From Introduction to Probability by Pishro-Nik, p. 576, we have the following problem:

"Consider the random problem $$\{X(t), t\in\mathbb{R}\}$$ defined as $$X(t) = \cos(t+U)$$, where $$U \sim \mathrm{Unif} (0, 2\pi)$$. Show that $$X(t)$$ is a wide-sense stationary process."

The first part of this requires checking that $$\mu_X(t) = \mu_X \;\forall \, t\in\mathbb{R}$$. The book walks the reader through it, and I found this part to be simple:

\begin{align} \mu_X(t) &= E[\cos(t+U)]\\ &=\int_0^{2\pi}\frac{1}{2\pi-0} \cos(t+u) \, du\\ &=\frac{1}{2\pi} \sin(t+u)\bigg\rvert_{u=0}^{2\pi}\\ &=0 \end{align}

because $$\sin(t+2\pi) = \sin t$$. So $$\mu_X(t) = 0 = \mu_X$$. Part 1 done.

The second part I know how to start, but there is a step I don't get denoted by $$\overset{??}{=}$$ below:

\begin{align} R_X(t_1, t_2) &= E[X(t_1)X(t_2)] \\ &= E[\cos(t_1+U)\cos(t_2+U)] \\ &= E[\tfrac{1}{2}(\cos(t_1 + t_2 + 2U) + \cos(t_1 - t_2))] \\ &= E[\tfrac{1}{2}\cos(t_1 + t_2 + 2U)] + E[\tfrac{1}{2}\cos(t_1 - t_2)] \\ &\overset{??}{=} \int_0^{2\pi}\frac{1}{2\pi-0} \cos(t_1 + t_2 + u) \, du + \tfrac{1}{2}\cos(t_1 - t_2)\\ &= 0 + \tfrac{1}{2}\cos(t_1 - t_2) \end{align}

Specifically, I have no clue how

$$E[\tfrac{1}{2}\cos(t_1 + t_2 + 2U)] = \int_0^{2\pi}\frac{1}{2\pi-0} \cos(t_1 + t_2 + u) \,du$$

What seems like a conversion of $$2U$$ into just $$u$$ makes absolutely no sense to me. A $$u$$-sub cannot explain this; it would be an integration from $$0$$ to $$\pi$$. What is the hidden step that I am clearly missing?

• It is a typo. The integrand should read $\cos(t_1+t_2+2u)$ May 4, 2020 at 3:43

\begin{aligned} \mathbb{E}(\tfrac{1}{2} \cos(t_1+t_2 + 2U)) &= \int \limits_0^{2\pi} \frac{1}{4 \pi} \cdot \cos(t_1+t_2 + 2u) \ du \\[6pt] &= \Bigg[ \frac{1}{8 \pi} \cdot \sin(t_1+t_2 + 2u) \Bigg]_{u=0}^{u=2\pi} \\[6pt] &= \frac{1}{8 \pi} \Bigg[ \sin(t_1+t_2 + 4 \pi) - \sin(t_1+t_2) \Bigg] \\[6pt] &= \frac{1}{8 \pi} \times 0 = 0. \\[6pt] \end{aligned}