Find the AutoCorrelation of a Random Process

Question

Below is a problem I did where I suspect I missed something significant because it seemed to easy and my answer is way off.
Problem:
Consider a random process $X(y)$ defined by \begin{eqnarray*} X(t) &=& Y \cos( wt + \Theta) \\ \end{eqnarray*} where $Y$ and $\theta$ are independent r.v's and are uniformly distributed over $(-A,A)$ and $(-\pi,\pi)$, respectively. Find the autocorrelation function $R_x(t,s)$ of $X(t)$.
Answer:
\begin{eqnarray*} R_{X_s\,X_t} &=& \frac{E( (X_t - \mu_t)(X_s - \mu_s ) ) } { \sigma_x \sigma_y } \\ \mu_t &=& E(X(t)) = E( Y \cos( wt + \Theta) ) \\ E(Y) &=& 0 \\ \mu_t &=& 0 \\ \mu_s &=& 0 \\ R_{X_s\,X_t} &=& \frac{E( (X_t - 0)(X_s - 0 ) ) } { \sigma_x \sigma_y } \\ R_{X_s\,X_t} &=& \frac{E( X_t X_s) } { \sigma_x \sigma_y } \\ E( X_t X_s) &=& E( Y \cos( wt + \Theta) Y \cos( ws + \Theta) ) \\ E( X_t X_s) &=& E(Y^2) E(\cos( wt + \Theta) \cos( ws + \Theta) ) \\ \end{eqnarray*} \begin{eqnarray*} E(Y^2) &=& \int_{-A}^{A} y^2 dy = \frac{y^3}{3} \Big|_{-A}^{A} = \frac{2A^3}{3} \\ E(\cos( wt + \Theta) \cos( ws + \Theta) ) &=& \frac{\cos( w(t+s) + 2\Theta ) + \cos(w(t-s))} { 2} \\ \end{eqnarray*} \begin{eqnarray*} E( X_t X_s) &=& \frac{A^3(\cos( w(t+s) + 2\Theta ) + \cos(w(t-s))} {6} \\ {\sigma_x}^2 &=& E(X(t)^2) - mu_t^2 = E(X(t)^2) \\ \end{eqnarray*} \begin{eqnarray*} E(X(t)^2) &=& E( Y^2 ) E( (\cos( wt + \Theta))^2 ) \\ E( (\cos( wt + \Theta))^2 ) &=& \int_{-\pi}^{\pi} (\cos( wt + \Theta))^2 \, dt \\ \text{Now we have the following standard integral to help up.} \\ \int \cos^2(ax) \, dx &=& \frac{x}{2} + \frac{\sin(2ax)}{4} + C \\ \end{eqnarray*} \begin{eqnarray*} \int_{-\pi}^{\pi} (\cos( wt + \Theta))^2 \, dt &=& \\ && \frac{1}{w}\big(\frac{ wt + \Theta}{2} + \frac{\sin( 2( wt + \Theta) ) } {4}\big) \Big|_{-\pi}^{\pi} \\ \end{eqnarray*} \begin{eqnarray*} \int_{-\pi}^{\pi} (\cos( wt + \Theta))^2 \, dt &=& \frac{ t + \frac{\Theta}{w}}{2} + \frac{\sin( 2( wt + \Theta) ) } {4w} \Big|_{-\pi}^{\pi} \\ \frac{ t + \frac{\Theta}{w}}{2} \Big|_{-\pi}^{\pi} &=& \frac{\pi + \frac{\Theta}{2}}{2} - \frac{-\pi + \frac{\Theta}{2} }{2} \\ \frac{ t + \frac{\Theta}{w}}{2} \Big|_{-\pi}^{\pi} &=& \frac{ \pi + \frac{\Theta}{2} + \pi - \frac{\Theta}{2} } { 2 } = \pi \\ \int_{-\pi}^{\pi} (\cos( wt + \Theta))^2 \, dt &=& \pi + \frac{\sin( 2( wt + \Theta) ) } {4w} \Big|_{-\pi}^{\pi} \\ \end{eqnarray*} \begin{eqnarray*} \int_{-\pi}^{\pi} (\cos( wt + \Theta))^2 \, dt &=& \pi + \frac{\sin( 2( \Theta) ) } {4w} - ( \frac{\sin( 2( \Theta) ) } {4w} ) \\ int_{-\pi}^{\pi} (\cos( wt + \Theta))^2 \, dt &=& \pi \\ E( (\cos( wt + \Theta))^2 ) &=& \pi \\ E(X(t)^2) &=& \frac{2\pi A^3}{3} \\ {\sigma_x}^2 &=& \frac{2\pi A^3}{3} \\ {\sigma_x}{\sigma_y} &=& \frac{2\pi A^3}{3} \\ R_{X_s\,X_t} &=& \frac{E( X_t X_s) } { \sigma_x \sigma_y } \\ R_{X_s\,X_t} &=& \cos( w(t+s) + 2 \Theta ) + \cos( w(t-s) ) \\ \end{eqnarray*}

Answer 1

Your definition of autocorrelation is wrong, i.e. $R_x(t,s) = E(X(t))E(X(s))$ is incorrect.

Hints:

$1:$ Use $$R_{X_t,X_s}(s,t)=\frac{E[(X_t-\mu_t)(X_s-\mu_s)]}{\sigma_{t}\sigma_{s}}$$

$2:$ Apply the fact that $Y$ and $\Theta$ are independent

$3:$ Conclude with $$\cos(x)\cos(y)=\left(\cos(x+y)+\cos(x-y)\right)/2.$$

Edit: Now your calculation for $E[cos(w(s+t)+2\Theta)]$ is not there. I just wrote it on a paper. See:

Above I just omitted the scaling factor coming from the uniform density, i.e.

$$\int_{w(t+s)-2\pi}^{w(t+s)+2\pi} 1 dx=4\pi.$$ So you may want to replace $1/2$ terms with $1/8\pi$ but the result would again be $0$.

And dont forget: expectation is a linear operator so $$E[f(X)+f(Y)]=E[f(X)]+E[f(Y)]$$

In your solution this part is wrong: $$E( (\cos( wt + \Theta))^2 ) = \int_{-\pi}^{\pi} (\cos( wt + \Theta))^2 \, dt $$

You take the expectation with respect to a random variable not a time variable. If you sum something over time, you will get the average of that something over a certain set or lets say an interval. You can correct this also via having a look at the JPG file that I put in to this answer.

$$E( X_t X_s) = E(Y^2) E(\cos( wt + \Theta) \cos( ws + \Theta) )$$

$$E(\cos( wt + \Theta) \cos( ws + \Theta) ) =E\left[ \frac{\cos( w(t+s) + 2\Theta ) + \cos(w(t-s))} { 2} \right]\\=E\left[ \frac{\cos(w(t-s))} { 2} \right]=\frac{\cos(w(t-s))} { 2} $$ Because, $$E\left[ \frac{\cos( w(t+s) + 2\Theta )} { 2} \right]=0$$ See the JPEG file I uploaded for how you get it.

Since $$E(Y^2) = \frac{1}{2A}\int_{-A}^{A} y^2 dy = \frac{1}{2A}\frac{y^3}{3} \Big|_{-A}^{A} = \frac{1}{2A}\frac{2A^3}{3}=\frac{A^2}{3}$$

We have

$$E( X_t X_s) = \frac{A^2}{3} \frac{\cos(w(t-s))} { 2}=\ \frac{A^2\cos(w(t-s))} {6}$$

Normally, to find the autocorrelation function one needs to normalize $E( X_t X_s)$ by $\sigma^2=\sigma_t\sigma_s.$ But from the answer that you had provided before I suspect that they are actually after $E( X_t X_s)$

You can find $$\sigma^2=E[Y^2]E[\cos(wt+\Theta)^2]$$

We already found $E[Y^2]$ and to find $E[\cos(wt+\Theta)^2]$ you will use the same tricks as in the JPEG file above. In particular $z=wt+\Theta\sim U[-\pi+wt,\pi+wt]$. Then,

$$E[\cos(wt+\Theta)^2]=E[\cos(z)^2]=\frac{1}{2\pi}\int_{-\pi+wt}^{\pi+wt} \cos(z)^2 \mathrm{d}z$$

I leave the rest to you. The solution was already there and you said "I give up"

Dont forget $$E[X]=\int x f_X(x)\mathrm{d}x$$ where $f_X$ is the density function of the random variable $X$.

In your solution you repeatedly did this wrong.