Question 1166371: Suppose the random variable X(t)is a wide-sense stationary random process.
Prove that the random process Y(t)=X(at) is wide-sense stationary, where is a constant.
Answer by CPhill(2030)   (Show Source):
You can put this solution on YOUR website! This is a proof in the theory of random processes. To prove that $Y(t) = X(at)$ is a **Wide-Sense Stationary (WSS)** random process, we must show that it satisfies the two conditions for WSS:
1. Its mean is constant (independent of time $t$).
2. Its autocorrelation function depends only on the time difference, $\tau = t_2 - t_1$.
We are given that $X(t)$ is WSS. Therefore, $X(t)$ satisfies:
$$E[X(t)] = \mu_X \quad (\text{a constant})$$
$$R_X(t_1, t_2) = E[X(t_1)X(t_2)] = R_X(t_2 - t_1) = R_X(\tau)$$
Let's check the two WSS conditions for $Y(t) = X(at)$.
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## 1. Mean of $Y(t)$
The mean of $Y(t)$ is $E[Y(t)]$.
$$E[Y(t)] = E[X(at)]$$
Since $X(t)$ is WSS, its mean is constant, $\mu_X$, regardless of the time index used (whether it is $t$ or $at$).
$$E[X(at)] = \mu_X$$
Therefore, the mean of $Y(t)$ is:
$$\mu_Y = E[Y(t)] = \mu_X$$
Since $\mu_X$ is a **constant**, $\mu_Y$ is **independent of time $t$**. The first condition for WSS is met.
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## 2. Autocorrelation of $Y(t)$
The autocorrelation function of $Y(t)$ is $R_Y(t_1, t_2) = E[Y(t_1)Y(t_2)]$.
$$R_Y(t_1, t_2) = E[X(at_1)X(at_2)]$$
Since $X(t)$ is WSS, its autocorrelation $R_X$ depends only on the time difference, $\tau_X = at_2 - at_1$.
$$R_Y(t_1, t_2) = R_X(at_2 - at_1)$$
$$R_Y(t_1, t_2) = R_X(a(t_2 - t_1))$$
Let $\tau = t_2 - t_1$ be the time difference for $Y(t)$.
$$R_Y(t_1, t_2) = R_X(a\tau)$$
Let $g(\tau) = R_X(a\tau)$. Since $\mu_X$ and $a$ are constants, $g(\tau)$ is a function that depends **only on the time difference $\tau$**, and not on $t_1$ or $t_2$ individually.
Therefore, the autocorrelation of $Y(t)$ is:
$$R_Y(t_1, t_2) = R_Y(\tau)$$
The second condition for WSS is met.
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## Conclusion
Since $Y(t)$ has a constant mean and its autocorrelation function depends only on the time difference $\tau$, the random process $Y(t)=X(at)$ is **Wide-Sense Stationary**.
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