In my lecture note, it states that $\hat{y} \sim\mathcal{N}(X\beta, \sigma^{2}X(X^{T}X)^{-1}X^{T}$)

but isn't $y \sim\mathcal{N}(X\beta, \sigma^{2}I_n$) ?

Which one is accurate or are they the same thing but with different representation?


1 Answer 1


$\mathbf{y \neq \hat y}$

  • The $\sigma^2(X^TX)^{-1}$ is the covariance table $\text{Cov}(\hat\beta)$ for the estimated coefficients
  • Then $X Cov(\hat\beta) X^T$ is the covariance $\Sigma$ for the error of the estimated values $\hat y = X \hat \beta$

    (that is different from the true sample values $y$).


Imagine a linear regression line which is always less accurate at the ends due to the uncertainty in the slope.

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Different notation

Possibly it may become more clear when we use different notation (use the $\hat \mu$ instead of $\hat y$)

  • The true mean (conditional on $X$) is $$\mu_X = X\beta$$ the $y$ and $\hat y$ are different derivatives of this.
    • The estimated mean $\hat y$, or the regression line. You could better interpret $\hat y$ as the estimated mean, ie the regression line. Then $$\hat \mu_X \sim \mathcal{N}(\mu, \sigma^2 X (X^TX)^{-1} X^T )$$ is expressing the sample variation of this regression line around $X\beta$ (how your estimate of the conditional mean $\hat \mu_X$ will vary from experiment to experiment, or sample to sample)
    • The sampled data $y$. The interpretation of $y$ and $\hat{y}$ is a bit different. The one is a data point the other is the mean. The sampled data points $y$ are distributed around the true mean $X\beta$ and, if the variance is homoscedastic and normal distributed then it is $$Y \sim \mathcal{N}(\mu_X,\sigma^2 I_n)$$

In the case of prediction, if you would like to estimate an error for the estimate of a new value, then you would actually use the sum of the two variances expressed above (the estimate of the mean plus the 'error' of a sampled value with respect to the mean).


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