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Confusion about a random process

  1. Jun 4, 2015 #1
    Question already asked on http://math.stackexchange.com/quest...m-process?noredirect=1#comment2661260_1310194, but couldn't get an answer so reposting here
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    Let [itex]X(t)[/itex] 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 [itex]a[/itex] is a constant.
    I need to find the value of [itex]a[/itex] for which [itex]X(t)[/itex] is a wide-sense stationary process. I have made the following definition of the random process:
    $$
    \begin{equation}
    X(t) = \alpha t + (1-\alpha)(2-at),
    \end{equation}
    $$
    where [itex]\alpha[/itex] is a Bernoulli random variable with [itex]p=q=0.5[/itex].

    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 [itex]a[/itex] for which autocorrelation function is a function of time difference [itex]t_2-t_1[/itex].

    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 [itex]t_1[/itex] and [itex]t_2[/itex] to get two random variables. The two random variables [itex]X(t_1)[/itex] and [itex]X(t_2)[/itex] are (possibly different) functions of the same random variable [itex]\alpha[/itex]. My question is why do we need to take the same random variable [itex]\alpha[/itex] at two time instants? It is like saying that if we know [itex]X(t_1)[/itex], we can figure out [itex]X(t_2)[/itex] right away. Shouldn't it be like that at [itex]t_1[/itex] we should take [itex]\alpha_1[/itex] and at [itex]t_2[/itex] we should take [itex]\alpha_2[/itex], where both [itex]\alpha_1[/itex] and [itex]\alpha_2[/itex] are Bernoulli with [itex]p=0.5[/itex].

    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 [itex]A = X(t_1)[/itex] and [itex]B = X(t_2)[/itex], 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 [itex]E[AB][/itex], we need to take the cases into account where [itex]A=t_1[/itex] and [itex]B=2-at_2[/itex], and [itex]A=2-at_1[/itex] and [itex]B=t_2[/itex]. Both of these cases do not appear in the calculation of ensemble autocorrelation function [itex]R_X(t_1,t_2)[/itex]. Why do I get to take the two cases into account when I use the formulation in terms of [itex]A[/itex] and [itex]B[/itex], and these two cases do not appear when I calculated [itex]R_X(t_1, t_2)[/itex] using the formulation with [itex]\alpha[/itex] as discussed in the start of the problem?
     
  2. jcsd
  3. Jun 4, 2015 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

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

    [itex]Cov(X(t_1)X(t_2))=E(X(t_1)X(t_2))-E(X(t_1))E(X(t_2))[/itex]
     
  4. Jun 5, 2015 #3

    Stephen Tashi

    User Avatar
    Science Advisor

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

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