# LASSO relationship between $\lambda$ and $t$

My understanding of LASSO regression is that the regression coefficients are selected to solve the minimisation problem:

$$\min_\beta \|y - X \beta\|_2^2 \ \\s.t. \|\beta\|_1 \leq t$$

In practice this is done using a Lagrange multiplier, making the problem to solve

$$\min_\beta \|y - X \beta\|_2^2 + \lambda \|\beta\|_1$$

What is the relationship between $\lambda$ and $t$? Wikipedia unhelpfully simply states that is "data dependent".

Why do I care? Firstly for intellectual curiosity. But I am also concerned about the consequences for selecting $\lambda$ by cross-validation.

Specifically, if I'm doing n-fold cross validation, I fit n different models to n different partitions of my training data. I then compare the accuracy of each of the models on the unused data for a given $\lambda$. But the same $\lambda$ implies a different constraint ($t$) for different subsets of the data (i.e., $t=f(\lambda)$ is "data dependent").

Isn't the cross validation problem I really want to solve to find the $t$ that gives the best bias-accuracy trade-off?

I can get a rough idea of the size of this effect in practice by calculating $\|\beta\|_1$ for each cross-validation split and $\lambda$ and looking at the resulting distribution. In some cases the implied constraint ($t$) can vary quiet substantially across my cross-validation subsets. Where by substantially I mean the coefficient of variation in $t>>0$.

• Upvoting to cancel out the unexplained downvote. The question is well outside my expertise but it seems reasonably formulated.
– mkt
Dec 22, 2017 at 10:20

$$\beta = \left( X'X + \lambda I \right) ^{-1} X'y$$

We also know that $$\| \beta \| = t$$, so it must be true that

$$\| \left( X'X + \lambda I \right) ^{-1} X'y \| = t$$.

which is possible, but not easy to solve for $$\lambda$$.

Your best bet is to just keep doing what you're doing: compute $$t$$ on the same sub-sample of the data across multiple $$\lambda$$ values.

This question relates to Is the magnitude coefficient vector in Ridge regression monotonic in lambda? which sketches a situation for ridge regression, but it is similar for Lasso. Consider the relationship of the optimal RSS as a function of the value of $$t = \vert \beta \vert$$. Say that this function is $$RSS = f(t)$$.

The goal of lasso is to find the $$\beta$$ which minimizes $$\text{Cost}(\beta) = RSS(\beta) + \lambda \vert\beta\vert$$

We could describe the cost as well as a function of the magnitude of the coefficients $$t$$

$$\text{Cost}(t) = f(t) + \lambda t$$

this is minimized when

$$\frac\partial{\partial t} \text{Cost}(t) = \frac\partial{\partial t} f(t) + \lambda = 0$$

And the relationship between $$\lambda$$ and $$t$$ is

$$\lambda = - \frac\partial{\partial t} f(t)$$

This function $$f(t)$$, the size of the RSS for a given size of the estimates of the coefficients, is dependent on the data.