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  "body": "\n## Definition\n\nLet $\\tau \\in (0, 1)$.\nThe *check function* is $\\rho_\\tau \\colon \\R \\to \\R$, $\\rho_\\tau(u) = u\\big(\\tau - \\mathbf{1}\\set{u \u003c 0}\\big)$, so $\\rho_\\tau(u) = \\tau u$ for $u \\ge 0$ and $\\rho_\\tau(u) = (\\tau - 1)u$ for $u \u003c 0$.\nLet $(x_1, y_1), \\dots, (x_n, y_n)$ be data with $x_i \\in \\R^p$ and $y_i \\in \\R$.\nThe *linear quantile regression* estimate at level $\\tau$ is any minimiser\n$$\\hat\\beta(\\tau) \\in \\arg\\min_{\\beta \\in \\R^p} \\sum_{i=1}^{n} \\rho_\\tau\\big(y_i - x_i^{T} \\beta\\big).$$\nIt estimates $\\beta(\\tau)$ in the model in which the conditional [[quantile function|0B3Q3XDM5X]] of $Y$ given $X = x$ at level $\\tau$ is $x^{T}\\beta(\\tau)$.\n\n## Remarks\n\nFor a real random variable $Y$, the minimisers of $b \\mapsto E[\\rho_\\tau(Y - b)]$ are exactly the $\\tau$-quantiles of $Y$ (when $E\\abs{Y} \u003c \\infty$).\nThis is why the check [[loss|3BTG6VVCGZ]] targets quantiles, just as squared loss targets the mean.\nThe minimisation is a linear program.\nWhen $x_1, \\dots, x_n$ span $\\R^p$, some minimiser fits at least $p$ data points exactly.\nFor $\\tau = 1/2$, $\\rho_{1/2}(u) = \\abs{u}/2$, and the method is least absolute deviations regression.\n\nExample: the intercept-only model, $p = 1$ and $x_i = 1$, with $y = (1, 2, 3, 4, 5)$ and $\\tau = 1/2$.\nThe function $b \\mapsto \\sum_i \\abs{y_i - b}/2$ is minimised at $b = 3$, the sample median.\n\nNon-example: minimising $\\sum_i (y_i - x_i^{T}\\beta)^2$ is [[linear regression|41FVC4P5JB]] by least squares.\nIts loss is quadratic, not a check function, and it estimates the conditional mean.\nFor skewed data the conditional mean and the conditional median differ, so the two fits differ.\n",
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