# Inequality involving quantiles

Suppose that $X$ and $Y$ are r.v.s such that $F_X$ (the cdf of $X$) is continuous and $$\sup_{r\in\mathbb{R}}|F_X(r)-F_Y(r)|\le \epsilon.$$ Is it true that $\mathsf{P}(X\le q_Y(\alpha))\le \alpha+\epsilon$, where $q_Y(\alpha)$ is the $\alpha$-quantile of $Y$?

If $q_Y(\alpha)$ is a continuity point of $F_Y$, then it is true. Also by the uniform bound the jumps of $F_Y$ cannot exceed $2\epsilon$ so that the bound holds when $F_Y$ is discontinuous at $q_Y(\alpha)$ ($\because$ $F_Y$ is cadlag).

## 1 Answer

You have for all $r\in\Bbb R$ $$F_X(r) - F_Y(r) \le \sup_{r\in\mathbb{R}}|F_X(r)-F_Y(r)|\le \epsilon$$

Hence for all $r\in\Bbb R$ it holds $$F_X(r) \le F_Y(r) + \varepsilon$$ $q_Y(\alpha)$ is the $\alpha$-quantile of Y hence we have for each $n\in\Bbb N$ $$F_Y\left(q_Y(\alpha) - \frac{1}{n}\right) = P\left( Y \le q_Y(\alpha) - \frac{1}{n}\right) \le P\left( Y < q_Y(\alpha) \right) \le \alpha$$

And because $F_X$ is continuous we get: \begin{align*}P\left(X \le q_Y(\alpha)\right) &= F_X\left(q_Y(\alpha)\right) \\ &= \lim_{n\to\infty} F_X\left(q_Y(\alpha) - \frac{1}{n}\right) \\&\le \lim_{n\to\infty} F_Y\left(q_Y(\alpha) - \frac{1}{n}\right) + \varepsilon \\ &\le \alpha + \varepsilon\end{align*}

• There is nothing to be known… it's the definition of the cdf. $F_Y$ is cdf of $Y$ iff $F_Y(y) = P(Y \le y)$. Independently of $F_Y$ being continuous or not. – Gono Jun 20 '18 at 19:17