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In this link it says that $Y$ variables has zero covariance (because covariance matrix has only diagonal terms) which implies they are independent.

Actually in linear regression $Y$ takes its expectation values from linear function of $X$ variables whereas takes its variance from error terms. Also $Y$ is normally distributed because of the fact that errors are normally distributed.

So, if $Y$ variables has zero covariance then errors must be zero covariance which means they are independent. Then in parallel with last answer sum of $(Y_i-\hat{Y}_i)^2$ divided by $\sigma^2$ which means sum of error squared divided by $\sigma^2$ must be chi-square with $n$ degrees of freedom. Why do we have $n-p-1$ dgf afterwards?

Could you help me overcome that contradiction?

kjetil b halvorsen
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mertcan
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1 Answers1

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The beginning of your question is a bit confusing, but the issue on why the degrees of freedom can be directly addressed. The proof is already in the question you point to, so I'll try a quick intuiton.

You see, $\frac{\sum_{i=1}(Y-\hat{Y_i})^2}{\sigma^2}$ is a function of $\hat{Y}_i$, a value that's obtained from a model with $p+1$ parameters (you have a constant plus a $p$ $x_i$ variables in $X$).

From a statistical intuition point of view, it's natural to expect you'd subtract these number of parameters from the total number of observations $n$, hence giving you $n-(p+1)$ degress of freedom.

Lucas Farias
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  • My point is: in regression we accept that error terms are normally distributed and independent. Then if we we divide square of errors by $\sigma^2$, those are converted to standard normal distributed and still independent. In short each $\frac{(Y-\hat{Y_i})^2}{\sigma^2}$ must be independent and chi square must have n degree of freedom. This is my confusion. Could you help me? Also if we have n-p-1 degree of freedom then it means our error terms actually does not fully independent right? – mertcan Mar 30 '19 at 14:48
  • By the way if error terms are not independent, covariance matrix of $Y$ should not have only diagonal terms as it set in the given link before right? – mertcan Mar 30 '19 at 14:59
  • For instance in [this link](http://users.stat.umn.edu/~helwig/notes/mvlr-Notes.pdf) "slide10" it is expressed that error vector is multivariate normal and if we assume that errors are uncorrelated then errors must have independent normal distribution because of multivariate normal distribution assumption. As a result of that each $\frac{(Y-\hat{Y_i})^2}{\sigma^2}$ must be independent standard normal. Therefore sum of them must equal to chi square distribution with n degree of freedom. What do you think about that? – mertcan Mar 31 '19 at 11:06
  • You are mixing the observed and predicted values. The expression for the error is a function of $Y$ *and* $\hat{Y}$. – Lucas Farias Apr 02 '19 at 20:21