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I have following equation:

$\mathbb{E}\left[||X_\tau - X_{\tau+t}||^2\right] = 2\cdot\sigma^2(X_\tau)(1-A(X_\tau, X_{\tau+t}))$

Where $X_\tau$ and $X_{\tau+t}$ come from the same distribution and thus have the same variance, they are just shifted by $t$ in time. Furthermore, $A(X_\tau)$ is autocorrelation for $X_\tau$, for specific lag $t$ (so basically correlation between the two).

My question is, what if $X$ is a column vector $\vec{X}$ ?

$\mathbb{E}\left[||\vec{X}_\tau - \vec{X}_{\tau+t}||^2\right] = ?$

Autocorrelation becomes a autocorrelation matrix ? What with variance ? It's going to be a column vector or a matrix with diagonal as variances of the vector elements ?

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To simplify this, you can use the fact that the norm is an inner product $$ \| \vec{X}_{\tau} - \vec{X}_{\tau + t}\|^2 = \langle \vec{X}_{\tau} - \vec{X}_{\tau + t},\ \vec{X}_{\tau} - \vec{X}_{\tau + t} \rangle\, . $$ If $\vec{X}_{t} = (X_t^{(1)},\ldots,X_t^{(d)})$, then $$ \mathbb{E}\left[ \| \vec{X}_{\tau} - \vec{X}_{\tau + t}\|^2 \right] = \sum_{i=1}^d \mathbb{E} \left[ \left( X_{\tau}^{(i)} - X_{\tau + t}^{(i)} \right)^2 \right] \, , $$ which is now amenable to your original 1D equation. Therefore, $$ \mathbb{E}\left[ \| \vec{X}_{\tau} - \vec{X}_{\tau + t}\|^2 \right] = 2 \sum_{i=1}^d \sigma_i(X_{\tau}^{(i)})^2 \left(1 - A_i(X_{\tau}^{(i)}, X_{\tau + t}^{(i)}) \right) \, , $$ where $\sigma_i^2$ and $A_i$ are the individual variance and autocorrelation functions for each component in the vector. Because the norm is still a 1D quantity you do not need to worry about finding a covariance or cross-covariance matrix.

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  • $\begingroup$ Thanks a lot for answer! $\endgroup$ Jul 19 at 6:37

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