# Confusion about a random process

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1. Jun 4, 2015

### ait.abd

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Let $X(t)$ be a random process such that:
$$X(t) = \begin{cases} t & \text{with probability } \frac{1}{2} \\ 2-at & \text{with probability } \frac{1}{2} \\ \end{cases},$$
where $a$ is a constant.
I need to find the value of $a$ for which $X(t)$ is a wide-sense stationary process. I have made the following definition of the random process:
$$X(t) = \alpha t + (1-\alpha)(2-at),$$
where $\alpha$ is a Bernoulli random variable with $p=q=0.5$.

For mean, we have
$$E[X(t)] = \frac{t + 2-at}{2}.$$
For the autocorrelation function,
\begin{align*} R_X(t_1,t_2) &= E[X(t_1)X(t_2)]\\ &=E[(\alpha t_1 + (1-\alpha)(2-at_1))\times(\alpha t_2 + (1-\alpha)(2-at_2))]\\ &=E[\alpha^2 t_1 t_2 + (1-\alpha)^2(2-at_1)(2-at_2)]\\ &=\frac{t_1 t_2}{2} + \frac{4-a(t_1+at_2)+a^2 t_1 t_2}{2}\\ \end{align*}
As clear from the above equation, there is no value of $a$ for which autocorrelation function is a function of time difference $t_2-t_1$.

My confusion starts from the way we define autocorrelation function as above in so many text books. The above-mentioned definition shows that we sample the ensemble at two time instants $t_1$ and $t_2$ to get two random variables. The two random variables $X(t_1)$ and $X(t_2)$ are (possibly different) functions of the same random variable $\alpha$. My question is why do we need to take the same random variable $\alpha$ at two time instants? It is like saying that if we know $X(t_1)$, we can figure out $X(t_2)$ right away. Shouldn't it be like that at $t_1$ we should take $\alpha_1$ and at $t_2$ we should take $\alpha_2$, where both $\alpha_1$ and $\alpha_2$ are Bernoulli with $p=0.5$.

I can describe the confusion like the following as well. When we sample the random process at two time instants, we get two random variables $A = X(t_1)$ and $B = X(t_2)$, where
$$A = \begin{cases} t_1 & \text{ with probability 0.5} \\ 2-at_1 & \text{ with probability 0.5} \\ \end{cases}$$
and
$$B = \begin{cases} t_2 & \text{ with probability 0.5} \\ 2-at_2 & \text{ with probability 0.5} \\ \end{cases}.$$
To calculate $E[AB]$, we need to take the cases into account where $A=t_1$ and $B=2-at_2$, and $A=2-at_1$ and $B=t_2$. Both of these cases do not appear in the calculation of ensemble autocorrelation function $R_X(t_1,t_2)$. Why do I get to take the two cases into account when I use the formulation in terms of $A$ and $B$, and these two cases do not appear when I calculated $R_X(t_1, t_2)$ using the formulation with $\alpha$ as discussed in the start of the problem?

2. Jun 4, 2015

### mathman

Your covariance is incorrect. You have to subtract the product of the means.

$Cov(X(t_1)X(t_2))=E(X(t_1)X(t_2))-E(X(t_1))E(X(t_2))$

3. Jun 5, 2015

### Stephen Tashi

Yes, that is the correct way to look at it.

The formula $R(\tau) = \frac{ E( X_{\tau} - \mu) E(X_{t + \tau} - \mu)} {\sigma^2}$ does not assert that the random variables $X_{\tau}$ and $X_{t + \tau}$ are functions of the same random variable $\alpha$. In this problem, using two independent random variables $\alpha_1, \alpha_2$ is correct and does not contradict that formula.