Question 1166371
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**.