Variance of a transformed uniform random variable

Given a sequence of random variables $(Z_t)_{t\in\mathbb{Z}}$ defined as $Z_t = \cos(tu)$ where $u \sim \text{Uniform}(0,2\pi)$. I want to calculate the mean, covariance and variance to show that this is a White noise process which is not strictly stationary. I have calculated the mean and variance in the following way: The cumulative dist function is given by $$F_{Z_t}(z) = P(Z_t \leq z) = P(\cos(tu) \leq z) = \frac{\cos^{-1}(z)}{2\pi t}.$$ Then the density is given by $f_{z_t}(z) = \frac{1}{2 \pi t \sqrt{1-z^2}}$ by using the derivate of the arccos. Thus for the mean we get \begin{align} EZ_t &= \int^1_{-1} z f_{Z_t}(z)dz = \frac{1}{2 \pi t} \int^1_{-1} \frac{z}{\sqrt{1-z^2}}dz\\ &= -\frac{1}{2(2 \pi t)} \int^1_{-1} -\frac{2z}{\sqrt{1-z^2}}dz = -\frac{1}{2(2 \pi t)} \left[\sqrt{1-z^2}\right]^1_{-1} = 0. \end{align}

Now for the variance I already know that it should be $\text{Var}(Z_t)=\frac{1}{2}$. But I get the following: \begin{align} \text{Var}(Z_t) &= EZ_t^2 = \int_{-1}^1 z^2 f_{Z_t}(z)dz = \int_{-1}^1 \frac{z^2}{2\pi t \sqrt{1-z^2}}dz \\ &= \frac{1}{2\pi t} \int^1_{-1} \frac{z^2}{\sqrt{1-z^2}}dz = \frac{1}{4t} \end{align} As $\int^1_{-1} \frac{z^2}{\sqrt{1-z^2}}dz = \frac{\pi}{2}$. Where am I going wrong and how does it get independent of $t$?

Is there any easy way to derive the covariance for some $Z_t$ and $Z_j$ with $t\neq j$? Thanks for any help.

• 1. First of all why do you think that the variance should not depend on $t$ ? 2. The covariance calculation is not correct as the $Z_t$ and $Z_j$ are not independent. In fact you can calculate the covariance function and check whether it depends only on $(j-t)$. If it is not, then it is not even Wide Sense Stationary, so strict sense would definitely not hold. – rajatsen91 Feb 4 '15 at 20:03
• I know that for this $Z_t$ for all $t \in \mathbb{Z}$ the mean is 0, variance is 1/2 and the covariance is 0. This example of a White noise process is used to prove that a there exists a white noise process that is not strictly stationary. These results can be easily found. Yet the derivation of the variance and covariance I cannot find and am trying to compute myself. – user155670 Feb 4 '15 at 20:08

I think that your mistake is that the cosine is not invertible in $(0,2\pi)$ so that your density formula is wrong. Anyway, your problem can be solved a lot easier:

Let $U \sim U[0,2\pi]$, then it follows:

$$\mathbb E [Z_t] = \mathbb E[\cos(tU)] = \frac{1}{2\pi}\int_0^{2\pi} cos(tu) du =0$$

\begin{aligned} \mathbb E[Z_t^2] &= \mathbb E[\cos^2(tU)] = \frac{1}{2\pi}\int_0^{2\pi} \cos^2(tu) du \\ &= \frac{1}{2\pi}\int_0^{2\pi} \frac{1 + \cos(2tu)}{2} du \\ &= \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{2}du = \frac{1}{2} \end{aligned}

Finally for $t \neq s$:

$$E[Z_t Z_s] = \mathbb E[\cos(tU)\cos(sU)] = \frac{1}{2\pi}\int_0^{2\pi} \cos(tu)\cos(su) du =0$$

The final equality can be shown e.g. by two repeated applications of partial integration.

• Thanks alot. I didn't notice I could write out the expectation in alot cleaner way. The density of the uniform distribution is of course $\frac{1}{2\pi}$ – user155670 Feb 4 '15 at 21:57
• Oh this factor is just the density of U, i.e.$\mathbb E[\cos(tU)]= \int cos(tu)d \mathbb P_U = \int_{0}^{2\pi} cos(tu) f_U(u) du$, where $f_U$ is the Lebesgue density of $U$. – air Feb 4 '15 at 21:59
• One more comment for people trying to calculate the covariance without PI. We can actually use $\cos(\alpha x)\cos(\beta x) = \frac{1}{2}\left(\cos((\alpha -\beta)x) + \cos(\alpha +\beta)x)\right)$. We then easily obtain that it is zero. – user155670 Feb 5 '15 at 22:15