What is the intuitive meaning of autocorrelation - when is $r_xx[0] = 1$ Consider a linear model $y_n = \beta x_n + v_n$
where $n$ represents the time index, $x$ is the input data source
$\hat{\beta}_{OLS} = (X^TX)^{-1}X^Ty$ is the ordinary least square estimate of the parameter $\beta$. This parameter is obtained assuming the distribution of the observation $y$ is $i.i.d$ Gaussian.  
Consider, the covariance of the estimate $\hat{\beta} = \sigma^2{(X^TX)}^{-1}$ 
The answers to the following questions are extremely hard for me to obtain from books, therefore I have posted the question. Shall appreciate help. Also, please correct me if any information is wrong. 
1) What is the meaning of autocorrelation and how it will change in the following scenario --  If the measurement noise $v$ is White Gaussian then it means that the noise samples are $i.i.d$. If the measurement noise is colored Gaussian, then   the noise samples are not $i.id$ but correlated which means that the property of independence and uncorrelated samples cannot be applied. When the noise is white Gaussian then expectation, $E[vv^T] = 
\sigma^2$ I think. What does the autocorrelation function look like in white and colored Gaussian case and what would expectation of colored noise samples give?
2)  UPDATE: Consider, a new data source denoted by $x_{new}$ which has some different properties than $x$.  
If the covariance $\sigma^2{(X^TX)}^{-1}$ is higher than another estimator of $\hat{\beta}$ given as $\hat{\beta_{new}} = \sigma^2{(X_{new}^TX_{new})}^{-1}$ obtained using the new data input $x_{new}$, then what does having a lower variance say about the nature of the data $x_{new}$? Can I say that $x_{new}$ input transfers more energy, has higher power than the data $x$?
3) Does $X^TX$ term have a name?  
 A: Disclaimer: I'm not entirely clear on everything myself, this is my best-effort, open for improvements.


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*If $$y_{n + 1} = \beta\cdot y_n + \epsilon_n$$ and $$\epsilon_n \sim \text{norm}(0, 1)$$, then autocorrelation can be simplified by linearity and because the errors are independent, $$\text{cov}(y_n, y_{n - 1}) = \beta\cdot\text{cov}(y_{n-1}, y_{n-1}) = \beta\cdot\text{var}(y_{n-1})$$. If $\beta$ is zero then this is "white noise" and the autocorrelation is zero. For larger gaps the same simplifications can be used to show $\text{cov}(y_n, y_{n-k}) = \beta^k\cdot\text{var}(y_k)$. 

*OLS is the maximum likelihood estimate of $\beta$ if the error is Gaussian. But if noise is non-Gaussian, then this may not work that well. Let's say the noise is skewed "upwards", so errors are a bit bigger upwards, let's say lognormal, then the maximum likelihood estimate for the mean is different. So if you use it, it's not optimal. 

*If you are estimating the mean of a lognormal distribution, see the link above for the nonlinear estimator of that. 

EDIT, some topics below 


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*I think the formula in the original question changed, now it's $y_n = \beta\cdot x_n + \epsilon_n$. Unless there is information on how $x_n$ or $y_n$ depend on variables at time $n-1$, there is not much to say about autocorrelation. 

*Autocorrelation is correlation with previous values, which is covariance with previous values divided by standard deviations. To simplify, I've only shown a derivation with covariance. 

*In the comments below, I think you mean variance of $\beta$, not it's covariance (otherwise, covariance with what?).

*The assumptions of the question regarding $X_{\text{new}}$ are unclear to me, and I suspect it's a question I don't necessarily know the answer to, or maybe there is not a single answer. Basically, if you assume Gaussian error, then the OLS estimate is the maximum likelihood estimate and also the unbiased estimator of $\beta$ with the lowest possible variance. If you change the distribution of the noise, then the bets are off. Maybe you can start with assuming different noise model and the see what happens, before trying to get the general answer. To be precise, the OLS estimate is the best unbiased estimator under broader assumptions, see https://en.wikipedia.org/wiki/Gauss%E2%80%93Markov_theorem.  

