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  "body": "\n## Definition\n\nLet $X$ and $Y$ be real [[random variables|0PNRKMRC55]], not necessarily on the same probability space.\n$X$ is *smaller than $Y$ in the usual stochastic order*, written $X \\le_{\\mathrm{st}} Y$, if\n$$P(X \u003e t) \\le P(Y \u003e t) \\qquad \\text{for all } t \\in \\R.$$\n\n## Remarks\n\nIn terms of [[distribution functions|0H8H7ZTHW2]], $X \\le_{\\mathrm{st}} Y$ means $F_X(t) \\ge F_Y(t)$ for all $t$.\nIt is equivalent to $E[u(X)] \\le E[u(Y)]$ for every nondecreasing function $u \\colon \\R \\to \\R$ for which both [[expectations|0WQNRCNY3M]] exist.\nIt is also equivalent to the existence of a [[coupling|HTR7J0W66V]] $(X', Y')$ with $X'$ distributed as $X$, $Y'$ distributed as $Y$ and $X' \\le Y'$ almost surely.\nOne can take $X' = F_X^{-1}(U)$ and $Y' = F_Y^{-1}(U)$ for $U$ uniform on $(0, 1)$, where $F^{-1}$ is the [[quantile function|0X60VE98HN]].\nThe relation depends only on the laws, and it is a partial order on the probability measures on $\\R$.\nIf $X \\le_{\\mathrm{st}} Y$ and $E[X] = E[Y]$ is finite, then $X$ and $Y$ have the same law.\nExample: if $X$ is exponential with rate $\\lambda$ and $Y$ is exponential with rate $\\lambda' \\le \\lambda$, then $P(X \u003e t) = e^{-\\lambda t} \\le e^{-\\lambda' t} = P(Y \u003e t)$ for $t \\ge 0$, and both sides are $1$ for $t \u003c 0$; so $X \\le_{\\mathrm{st}} Y$.\nNon-example: if $X$ is normal with mean $0$ and variance $1$ and $Y$ is normal with mean $0$ and variance $4$, then $P(X \u003e 1) \u003c P(Y \u003e 1)$ but $P(X \u003e -1) \u003e P(Y \u003e -1)$, so neither $X \\le_{\\mathrm{st}} Y$ nor $Y \\le_{\\mathrm{st}} X$.\nThe name stochastic ordering also covers other orders, such as the convex order ($E[u(X)] \\le E[u(Y)]$ for convex $u$) and the hazard rate and likelihood ratio orders.\n",
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