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Nov 4, 2020 at 15:09 comment added whuber @Henry Then please take a look at my link or at the answer posted here by Firebug.
Nov 4, 2020 at 15:06 comment added Henry @whuber - I can accept ridge regression is a least squares method, but I would not see it as ordinary: you are minimising the sum of the squares of the residuals plus the square of something else
Nov 4, 2020 at 14:56 comment added whuber @Henry and when $\lambda$ otherwise is positive, it is still OLS regression.. In short, this objective function is a squared Euclidean distance to a point, whence it is (obviously) strictly convex.
Nov 4, 2020 at 0:00 history tweeted twitter.com/StackStats/status/1323777001421942784
Nov 3, 2020 at 16:18 history edited Firebug
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Nov 3, 2020 at 16:17 answer added Firebug timeline score: 2
Nov 3, 2020 at 15:25 comment added Henry If $\lambda$ is $0$ then this is ordinary least squares rather than ridge regression
Nov 3, 2020 at 12:35 history became hot network question
Nov 3, 2020 at 5:19 vote accept user8714896
Nov 3, 2020 at 5:18 answer added Thomas Lumley timeline score: 21
Nov 3, 2020 at 5:00 comment added Sycorax Since the second partial derivative of $f$ is a matrix, you'll need to adapt your definition to the case of a matrix. You can show that $\frac{\partial ^2 f}{\partial \beta^2}$ is positive semi-definite. What remarks about convexity can you make about p.s.d. matrices?
Nov 3, 2020 at 4:47 comment added user8714896 @Sycorax I assumed that no entries in the resulting matrix can be negative in order for the function to be strictly convex. Or is that not the case? Is convexity different if it's a matrix?
Nov 3, 2020 at 4:44 comment added Sycorax $X^\top X$ is a matrix, not scalar. In what sense do you mean "negative"?
Nov 3, 2020 at 4:32 history asked user8714896 CC BY-SA 4.0