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We generate $n$ observations from 1D normal distribution. The observations are denoted as $x_1, x_2, \cdots, x_n$.

Given these observations, consider sequence $\mathbf{x} \in \mathbb{R}^n$:

$$\mathbf{x} = \begin{bmatrix} x_1 \\ x_1+x_2 \\ x_1+x_2+x_3 \\ \vdots \\ x_1+x_2+\cdots + x_n \end{bmatrix} $$

What will be the autocorrelation (and power density function) of this generated sequence $\mathbf{x}$?

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  • $\begingroup$ You should write "a sample of size $n$" or "a sample consisting of $n$ observations", etc., instead of "$n$ samples". $\qquad$ $\endgroup$ – Michael Hardy Apr 23 at 0:44
  • $\begingroup$ @MichaelHardy thanks for suggestion. Sorry for the confusion. Forgive me, I meant each $x_i$ is a sample observation, after these observations we consider the sequence $\mathbf{x}$ of the accumulation of these observations. $\endgroup$ – WDC Apr 23 at 0:48
  • $\begingroup$ Hi: The resulting covariance matrix is n×n. The diagonal elements are the variances. These are equal to the largest subscript of the corresponding summed element times σ2. This is because the variance of an iid sum of n normals has variance nσ2. For the off diagonal elements which are covariances, do the same thing but take the smaller of ( column subcript, row subscript ). For the spectrum, you could transform the covariance matrix to the corresponding correlation matrix and then compute the z transform of the autocorrelations but I'm not sure if it would simplify nicely. $\endgroup$ – mlofton Apr 23 at 2:50
  • $\begingroup$ Hi: Note that, for the spectrum, in order to go from the covariances to the correlations, you will need to divide the covariance elements by the corresponding standard deviations because $\rho_{x,y}$=covariance(x,y)/(std(x)std(y)). $\endgroup$ – mlofton Apr 23 at 2:51
  • $\begingroup$ @mlofton thanks for the hints really appreciate it $\endgroup$ – WDC Apr 23 at 10:03

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