In this problem the random variable is theta and according to the formula there should be two integrations but in the solution there is only one . Nor am i able to understand the meaning of x1 and x2 in the formula. this problem is driving me nuts.. in this problem the random variable is theta this is the solution here is the formula


2 Answers 2


You have a single random variable $\Theta$ in this example, so taking the expectation with respect to that random variable results in a single integral. The formula for $R_X(\tau)$ in your question is rather messy and full of mistakes; it should be as follows:

$$\begin{align}R_X(\tau)&=E[X(t+\tau)X(t)]\\&=E[a\sin(\omega_0t+\omega_0\tau+\Theta)\cdot a\sin(\omega_0t+\Theta)]\\&=\frac{a^2}{2}E[\cos(\omega_0\tau)-\cos(2\omega_0t+\omega_0\tau+2\Theta)]\\&=\frac{a^2}{2}\frac{1}{2\pi}\int_{0}^{2\pi}[\cos(\omega_0\tau)-\cos(2\omega_0t+\omega_0\tau+2\Theta)]d\Theta\\&=\frac{a^2}{2}\cos(\omega_0\tau)\end{align}$$

  • $\begingroup$ What would be the formula then ..if there is only one random variable in the autocorrelation function...?? $\endgroup$ Aug 4, 2019 at 16:12
  • 1
    $\begingroup$ @VishalPoddar: It's not about autocorrelation, it's about how to evaluate the expectation operator: $E[g(\theta,\ldots)]=\int p(\theta)g(\theta,\ldots)d\theta$, where $\ldots$ stands for other variables that the function $g(\cdot)$ may depend on, as long as they're not random. The variable $\theta$ is the (only) random variable here, so you have to integrate over $g(\cdot)$ multiplied by the PDF $p(\theta)$. $\endgroup$
    – Matt L.
    Aug 4, 2019 at 16:16

You have the definition of the random process as :

$$ X(t) = a \sin(\omega_0 t + \Theta) \tag{1} $$

where $a$ and $\omega_0$ are deterministic constants and $\Theta$ is a continuous R.V. uniformly distributed in $[0,2\pi]$.

According to the indexed-set of RV interpretation of a R.P., for each index $t$ you have a new R.V denoted as $X_t$ which has its own density function $f_{X_t}(x)$ for each $t$. Then the following expectation is actually looking for

$$ E\{ X(t) \} = \int x f_{X_t}(x) dx \tag{2} $$

However, based on a well known relation in probability theory, the following helps. If a RV $Y$ is a function of another RV $X$, such as given by $Y = g(X)$, then the following expectations are equivalent:

$$ E\{ Y \} = \int y f_Y(y) dy \tag{3}$$ $$ E\{ g(X) \} = \int g(x) f_X(x) dx \tag{4}$$

Now if you look at the indexed RV $X_t$ of the R.P. $X(t)$ you can see that it's a function of the R.V. $\Theta$ :

$$X_t = g(\Theta) = a \sin( \omega_0 t + \Theta) $$

Hence the Expectation in Eq.2 can be computed based on the reasoning of Eq.4 as

$$ E\{ X(t) \} = E\{ g(\Theta) \} = \int g(\theta) f_{\Theta}(\theta) d\theta \tag{5} $$

$$ E\{ X(t) \} = \int a \sin(\omega_0 t + \theta) f_{\Theta}(\theta) d\theta \tag{6} $$

A very similar reasoning shows that the expectation of the product $X(t+\tau) X(t)$ can be computed by treating the product as a function of the RV $\Theta$, instead of using the joint PDF of the RVs $X_{t}$ and $X_{t+\tau}$ as:

$$ X(t+\tau) X(t) = h( \Theta) \tag{7} $$

where $h(\cdot)$ is given by:

$$ h(\theta) = a^2 \sin(\omega_0 (t+\tau) + \theta) \sin(\omega_0 t + \theta) \tag{8}$$

Then the expectation will be evaluated as

$$ E\{ X(t+\tau) X(t) \} = E\{ h(\Theta) \} = \int h(\theta) f_{\Theta} (\theta) d\theta \tag{9} $$

$$ E\{ X(t+\tau) X(t) \} = \int a^2 \sin(\omega_0 (t+\tau) + \theta) \sin(\omega_0 t + \theta) f_{\Theta} (\theta) d\theta \tag{10} $$


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