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In an interview question I was asked what would be the effect of adding a Gaussian noise to the features $\mathbf X\in\mathbb R^{n\times k}$ over the confidence interval of the parameters $\mathbf\beta\in\mathbb R^k$.

I could not answer, so afterwards I went checking around how these confidence intervals look like, and found that the parameters have associated a $t$-statistics $(\beta_j-\beta_{0,j})/\sigma_j$, where $\sigma_j$ is the standard error of the $j$-th parameter:

$$ \sigma_j=\sqrt{\frac{\hat\sigma^2}{\Sigma_{jj}}},$$

where $\hat\sigma^2$ is the estimator of the variance of the residuals and $\mathbf\Sigma=\mathbf X^T\mathbf X$ is the sum-of-squares matrix of the features.

This suggests me that the confidence interval improves increasing the variance of the features, that is honesty counterintuitive, and that therefore adding Gaussian noise should also improve it. This is even more counterintuitive: shuffling the deck improves predictions about it? How would you explain this?

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  • $\begingroup$ Could you clarify what you mean by "improving" a confidence interval? $\endgroup$
    – whuber
    Commented Nov 9, 2020 at 14:13
  • $\begingroup$ the confidence interval gets shrinked, I interpret it as an improvement. $\endgroup$
    – marco
    Commented Nov 9, 2020 at 14:14
  • $\begingroup$ Thank you; that helps us understand your question. Please note, though, that there are two problems with your formulation: the equation for the standard error is incorrect and you haven't considered the equation for the confidence interval itself. $\endgroup$
    – whuber
    Commented Nov 9, 2020 at 14:24
  • $\begingroup$ I see, I took the standard error by couple of different sources, and this question: stats.stackexchange.com/questions/288774/…, where it looks like I am considering the same expression about the standard error. Would you please point me to the mistake? $\endgroup$
    – marco
    Commented Nov 9, 2020 at 14:52
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    $\begingroup$ You are missing a square root. Note, too, that $\Sigma$ is not a covariance matrix (it is a "sum of squares and products" matrix). $\endgroup$
    – whuber
    Commented Nov 9, 2020 at 15:23

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