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 $t>>0$.
 A: This is the standard solution for ridge regression:
$$
\beta = \left( X'X + \lambda I \right) ^{-1} X'y 
$$
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.
A: 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.
