What the penalty buys and costs

Write $\hat\theta_r$ for the retrieved maps (or time courses) in realisation $r$, $\bar\theta$ for their average over $R$ realisations and $\theta^\ast$ for the target. Summed over every entry, the mean squared error splits exactly into two parts:

The target is what the same pipeline returns without noise. Standardise $X_0 = \mathrm{TC}\cdot\mathrm{SM}$ and regress it on the time courses, giving $A^\ast = (D^\top D)^{-1}D^\top X_0$, which works out as the true maps scaled by $\sqrt{N/(N-1)}$, and $D^\ast = X_0 A^{\ast\top}$. Errors are divided by $\lVert\theta^\ast\rVert^2$, so 1.0 means an error as large as the target itself, which is exactly what you get by returning all zeros. The MSE carries a 95% interval over realisations. Bias² and variance are shown as point estimates of its two parts. Averaging only $R$ realisations leaves noise in $\bar\theta$, which inflates the plug-in bias² by about $\operatorname{tr}\Sigma / R$, where $\Sigma$ is the covariance of $\hat\theta_r$. The table also gives a debiased bias², $\lVert \bar\theta - \theta^\ast\rVert^2 - \sum_e s_e^2 / R$, with $s_e^2$ the sample variance of entry $e$ across realisations.

Every estimator and penalty is fitted to the same datasets (common random numbers), so the curves differ only because the estimators differ. Realisation 1 is the dataset from step 1. The full design, and why the lasso here comes in two versions, are on the methods page.