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This question already has an answer here:

Why increasing lambda parameter in L2-regularization makes the co-efficient values converge to zero?

I have just tried to do the math, but it's a little bit rusted.

Lets say that we have a simple linear model as follows: $y=w_1\cdot x$

we could write the cost function for ridge regression is to be minimized:

$cost(\hat{w_1}, \lambda)= (y - \hat{w_1} \cdot x)^2 + \lambda \cdot \hat{w_1}^2$

it means that if we consider the problem as min-max:

$\frac{\hat{dw_1}}{dc} = -2 \cdot x \cdot (y - \hat{w_1}) + 2\cdot \lambda \cdot \hat{w_1} = 0$ so,

$y = (1 + \frac{\lambda}{x}) \cdot \hat{w_1}$

Since the y and x are invariants, it is to be expected increasing $\lambda$ make the co-efficient decrease as the equation holds.

Is that the right way to reason?

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marked as duplicate by Michael Chernick, kjetil b halvorsen, mkt, jpmuc, Ferdi Jan 28 at 7:47

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    $\begingroup$ Apply the formula given in the question (and answers) at stats.stackexchange.com/questions/69205/…. Find out more by searching our site for "ridge regression." $\endgroup$ – whuber Jan 22 at 21:23
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    $\begingroup$ Yes, you are reasoning is fine. $\endgroup$ – usεr11852 Jan 22 at 21:28
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Yep, that is one way to think about it, although it seems a tad obscure to me.

I think it's simpler to just look at your $\text{cost}$ equation:

$\text{cost}(\hat{w_1}, \lambda) = (y - \hat{w_1} \cdot x)^2 + \lambda \cdot \hat{w_1}^2$

We can see from this that, for large $\lambda$, our cost increases quadratically with the absolute size of $\hat{w_1}$. That is, we are penalising our model for having a large weight: thus to reduce the cost, our $\hat{w_1}$ coefficient is shrunk towards zero.

If $\lambda$ is small, or zero, this second term doesn't really affect the cost, so $\hat{w_1}$ is free to grow as large as it needs to, to minimise the other component of the cost function.

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