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Suppose $X_1,\cdots,X_n$ are i.i.d. continuous random variables from distribution with cdf $F(x)$.Let $F_n(x)$ be a random variable defined by $$F_n(x)=\frac{1}n\sum_{i=1}^nI\{X_i\le x\}.$$ Define the $p$th quantile for cdf $F(x)$ as $$\xi_p\equiv F^{-1}(p)=\inf\{x:F(x)\ge p\},$$ and the $p$th quantile for empirical cdf $F_n(x)$ as $$\hat\xi_p\equiv F_n^{-1}(p)=\inf\{x:F_n(x)\ge p\}.$$ Show that $$\Big|F^{-1}\big(F_n(\xi_p)\big)-\hat\xi_p\Big|\overset{a.s}\to0$$


How can we prove the result? By LLN, we have $F_n(x)\overset{a.s}\to F(x)$. Can we write the equation as $$\Big|F^{-1}\big(F_n(\xi_p)\big)-\hat\xi_p\Big|=\Big|F^{-1}\big(F_n(\xi_p)\big)-F_n^{-1}\big(F(\xi_p)\big)\Big|,$$ and get the result directly?

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    $\begingroup$ For a fixed $x$, $F_n(x) = \frac1n \sum_{k=1}^n 1_{X_k \le x}$ is a random variable with binomial distribution $Pr(F_n(x) \le l/n)= \sum_{m=0}^{\lfloor l \rfloor} {n \choose m} F(x)^m (1-F(x))^m$. So $\mathbb{E}[F_n(x)]= F(x)$ and $\mathbb{E}[(F_n(x)-F(x))^2]= \frac1n F(x)(1-F(x))\to 0$. Whence $F_n(x) \to F(x)$ almost surely and since $F^{-1}$ is continuous then $F^{-1}(F_n(x)) \to F^{-1}(F(x))$ a.s. $\endgroup$ – reuns Dec 29 '18 at 7:38
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    $\begingroup$ @reuns The continuity of $F(x)$ does not make $F^{-1}$ continuous. Consider uniform distribution on the union of intervals $(0,1)$ and $(2,3)$ and you will get $F^{-1}(0.5)=1$, $F^{-1}(0.5+)=2$ (right limit). $\endgroup$ – NCh Dec 29 '18 at 15:15
  • $\begingroup$ Not so easy. If $F^{-1}$ is discontinuous only at finitely many point it is easier. $\endgroup$ – reuns Jan 8 at 11:27
  • $\begingroup$ @reuns Thank you for your kind comments! $\endgroup$ – J.Mike Jan 8 at 17:59

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