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If $X(t)$ is a WSS process with autocorrelation function $R_{xx}(T)$ and $Y(t) = X(t+a) - X(t)$, Show that $$R_{yy}(T) = 2R_{xx}(T) - R_{xx}(T + 2a) - R_{xx}(T + 2a).$$

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    It doesn't seem right. Where did you get that from? – 2017-01-02
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    I'm getting $R_{yy}(T) = 2R_{xx}(T)-R_{xx}(T+a)-R_{xx}(T-a)$. @msm does that look better? – 2017-01-02
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    Yes, I think that is the correct equation @JimmyK4542. – 2017-01-02

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$$\begin{align} R_{yy}(\tau)&=\mathsf{E}\{y(t)y(t+\tau)\}\\ &=\mathsf{E}\{[x(t+a)-x(t)][x(t+\tau+a)-x(t+\tau)]\}\\ &=\mathsf{E}\{x(t+a)x(t+\tau+a)-x(t+a)x(t+\tau)-x(t)x(t+\tau+a)+x(t)x(t+\tau)]\}\\ &=R_{xx}(\tau)-R_{xx}(\tau-a)-R_{xx}(\tau+a)+R_{xx}(\tau)\\ &=2R_{xx}(\tau)-R_{xx}(\tau-a)-R_{xx}(\tau+a) \end{align}$$

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    The $+$ sign in the last line should be a $-$ sign. Otherwise, it looks fine. – 2017-01-02
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    Thanks @JimmyK4542 fixed it. – 2017-01-02
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    How is E{x(t+a)x(t+τ+a)} = Rxx(τ) ? – 2017-01-02
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    Assume $t_1=t+a$. Then it is equal to $\mathsf{E}\{x(t_1)x(t_1+\tau)\}$, which is equal to $R_{xx}(\tau)$ given that $x(t)$ is WSS. – 2017-01-02