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  "body": "\n## Definition\n\nLet $X$ be a real $n \\times p$ matrix, let $y \\in \\R^n$, and let $\\lambda \u003e 0$.\nFor $b \\in \\R^p$ write $\\norm{b}_1 = \\sum_{j=1}^p \\abs{b_j}$ and let $\\norm{\\cdot}_2$ be the Euclidean norm.\nA *lasso estimate* with parameter $\\lambda$ is a minimiser over $b \\in \\R^p$ of\n$$\\frac{1}{2} \\norm{y - X b}_2^2 + \\lambda \\norm{b}_1.$$\n\n## Remarks\n\nThe objective is [[convex|07Q7CNM5WF]] and tends to infinity as $\\norm{b}_1 \\to \\infty$, so a minimiser exists.\nIt need not be unique when the columns of $X$ are linearly dependent.\nThere is no closed formula in general; the estimate is computed by coordinate descent or by the LARS algorithm.\nAuthors scale the objective in different ways, for example with $\\frac{1}{2n}$ in front of the squared norm, which only rescales $\\lambda$.\n\nExample: assume $X^T X = I_p$ and put $z = X^T y$, the [[ordinary least squares|YM708KHV2W]] estimate.\nThen $\\norm{y - Xb}_2^2 = \\norm{z - b}_2^2 + \\text{const}$, and the problem splits into the $p$ problems of minimising $\\frac{1}{2}(z_j - b_j)^2 + \\lambda \\abs{b_j}$.\nThe solution is the soft thresholding $\\hat\\beta_j = \\sgn(z_j) \\max(\\abs{z_j} - \\lambda, 0)$.\nSo every $j$ with $\\abs{z_j} \\le \\lambda$ gets $\\hat\\beta_j = 0$.\n\nNon-example: with the squared norm $\\norm{b}_2^2$ in place of $\\norm{b}_1$, the same orthonormal case gives $\\hat\\beta_j = z_j / (1 + 2\\lambda)$, which is never $0$ when $z_j \\ne 0$; that is ridge regression, not the lasso.\n\nEquivalent form: for each $\\lambda \u003e 0$ and each lasso estimate $\\hat\\beta$, the vector $\\hat\\beta$ minimises $\\norm{y - Xb}_2^2$ under the constraint $\\norm{b}_1 \\le \\norm{\\hat\\beta}_1$.\n",
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