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For simplicity, imagine that the function only consists of 11 measurements made at $x=0, 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000$ For each of these $x$ values, you have a $y$ value. Let the lagged $y$ value be $y_l$. For lag = 0, $y_l=y$: x y y_l 0 0 0 100 0 0 200 0 0 300 0 0 400 ...

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Since all you have are the means and the covariances, the most natural thing to do would seem to be to assume a multivariate Gaussian with those parameters, plug in $x_2,\ldots,x_n$ and normalise to get the conditional distribution for $x_1$; the estimate would then be the mean of that distribution (which is the same as its mode and median, since it's normal)...

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... if $R_1$ and $R_2$ both denote a matrix we get $R_1$ and $R_2$ should be vectors. Then $Var(R_1+R_2)=\Sigma_1+\Sigma_2+2\rho_{12}\sqrt{\Sigma_1} \sqrt{ \Sigma_2}$ where $\Sigma_i$ denotes the variance matrix of $R_1$ and $R_2$ respectively.

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When working with multivariate variances and covariances, it's good to keep this notational advice in mind. I'll stick with your notation and use $\operatorname{Var}(R)$ to denote the (co)variance matrix of the random vector $R$, i.e. $\operatorname{Var}(R)=\operatorname{cov}(R,R)$. Then \begin{align} \operatorname{Var}(R_1+R_2)&=\operatorname{cov}(...

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You probably meant that a $\textit{pair}$ of independent variables are highly correlated ($0.9$). Basically, although technically it's fine, this may indicate that both variables measure (almost) the same factor. Hence, generally speaking, bring the same information to the model, as such may cause redundancy. The fact that both variables are significant may ...

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Three facts: Symmetric semi positive definite matrix has all eigenvalues real and greater or equal zero. Symmetric positive definite matrix has all eigenvalues real and greater than zero. Determinant of any matrix is equal to product of all eigenvalues. Hence, if determinant of symmetric semi positive definite matrix $A$ is nonzero, then $A$ is positive ...

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(I've revised my answer to account for the authors' use of the Dirac delta function, which I had misinterpreted as a Kronecker delta function. This is sometimes called a delta-correlated random process.) The authors appear to apply the following simulation procedure (for heuristics, see here): If $\{X(z)\}_{z\in\mathbb{R}}$ is a continuous Gaussian ...

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As already said in comments, it could be good to have more data points. It is not sure that a linear model is the most appropriate. You effectively obtained $$y=1.374 -0.000697369 x$$ to which correspond a sum of squares equal to $0.0381$ and $R^2=0.9951$. Using this model, what would happen when $\text{Hz} > 2000$ ? Is a negative duration of the ease-...

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I was able to plot the data set and find the linear regression. Intercept (a): 1.3739956457219 Slope (b): -0.0006973690931099 So, Y = a + bX This works with a pretty decent level of accuracy for my purposes – accurate to .02 seconds.

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I'd recommend getting a few more frequencies and measuring the ease-in duration for those frequencies. With just two points it's complete guesswork. All we can say now is that it appears to be some kind of inverse relationship: higher frequency implies shorter ease-in duration.

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