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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?

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

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It is a typo, but you should get zero for the integral anyway. You should have:

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

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