# How does $\mathbb{E}\left[||X_\tau - X_{\tau+t}||^2\right] = 2\cdot\sigma^2(X_\tau)(1-A(X_\tau, X_{\tau+t}))$ generalize to vectors?

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 ?

## 1 Answer

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.

• Thanks a lot for answer! Jul 19 '21 at 6:37