Question 1166278: 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! Yes, the random process $Y(t) = X(at)$ is wide-sense stationary (WSS) if $X(t)$ is WSS, provided $a$ is a non-zero constant.
To prove that $Y(t)$ is WSS, we must verify the two conditions for wide-sense stationarity:
1. The **mean** of $Y(t)$, $E[Y(t)]$, is a constant (independent of $t$).
2. The **autocorrelation function** of $Y(t)$, $R_Y(t_1, t_2)$, depends only on the time difference $\tau = t_2 - t_1$.
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## 1. Mean of $Y(t)$ is Constant
Since $X(t)$ is WSS, its mean, $E[X(t)]$, is a constant, which we'll call $\mu_X$.
$$E[X(t)] = \mu_X$$
Now we find the mean of $Y(t)$:
$$E[Y(t)] = E[X(at)]$$
Since $X(t)$ is WSS, the mean of $X$ evaluated at *any* time (including the time $at$) is the constant $\mu_X$.
$$E[Y(t)] = \mu_X$$
Since $\mu_X$ is a constant (independent of $t$), the first condition is satisfied.
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## 2. Autocorrelation of $Y(t)$ Depends Only on Time Difference ($\tau$)
Since $X(t)$ is WSS, its autocorrelation function, $R_X(t_1, t_2)$, depends only on the time difference $\tau = t_2 - t_1$. We denote this as $R_X(\tau)$.
$$R_X(t_1, t_2) = E[X(t_1)X(t_2)] = R_X(t_2 - t_1)$$
Now we find the autocorrelation function of $Y(t)$:
$$R_Y(t_1, t_2) = E[Y(t_1)Y(t_2)]$$
Substitute the definition $Y(t) = X(at)$:
$$R_Y(t_1, t_2) = E[X(at_1)X(at_2)]$$
The term inside the expectation is the autocorrelation of $X(t)$ evaluated at times $t' = at_1$ and $t'' = at_2$. Since $X(t)$ is WSS, this expectation depends only on the difference between these times, $t'' - t'$:
$$R_Y(t_1, t_2) = R_X(t'' - t') = R_X(at_2 - at_1)$$
Factor out the constant $a$:
$$R_Y(t_1, t_2) = R_X(a(t_2 - t_1))$$
Let $\tau = t_2 - t_1$.
$$R_Y(t_1, t_2) = R_X(a\tau)$$
Since $R_X(a\tau)$ is a function that depends only on the time difference $\tau$ (and the constant $a$), the second condition is satisfied.
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Since both the mean of $Y(t)$ is constant and the autocorrelation of $Y(t)$ depends only on the time difference, the process $Y(t) = X(at)$ is **wide-sense stationary**.
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