Could someone please explain what "$J$" consists of in this paper, equation 1.5. $$ J \sim N(\beta, \sigma^2 I/k) $$ What's $\beta$ here? What's $N$?

Also, why are they putting that much effort in deriving $k$ with complicated formulas instead of just saying "use CV to find the best $k$"?


closed as too localized by Nick Cox, gung, whuber Jun 21 '13 at 20:36

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    $\begingroup$ I cannot see any $N$ above! Yes, this seems to be somewhat mystical. To find out what is the idea behind $J$ above I think you must look in the references of the paper you have given. $\beta$ seems to be the true value of the regression parameter, so $J$ seems to be a random variable. Seems to be some sort of randomized estimator? If you are interested in this you must go to the references in your cited paper. $\endgroup$ – kjetil b halvorsen Jan 14 '13 at 16:29

I can answer part of your question, at least. $J$ is a random variable with a normal distribution (that's what $N$ means---normal). $\beta$ is the mean of $J$; $σ^2I/k$ is the standard deviation.

Since $\beta$ is the slope in the original linear regression $Y = X\beta + \epsilon$ (equation 1.1), it's probably appearing in equation 1.5 as a stand-in for $\beta$.

  • $\begingroup$ Welcome to the site, @caedocyon. This looks like a nice contribution--I hope we'll see more. Let me note a couple of (admittedly small) points: I suspect $\sigma^2I/k$ is a diagonal matrix of variances, not SDs; also, since this is a multivariable situation, it might be clearer to state that $\beta$ is the mean vector / the vector of slopes from the original model. (NB, I haven't read the paper, these are guesses, feel free to correct me if I'm wrong.) $\endgroup$ – gung Mar 23 '13 at 16:41

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