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Let $F_n(t)=\frac{1}{n} \sum_{k=1}^n 1_{X_k \leq t}$ be the emprical distribution function of i.i.d. random variables $X_k\sim U[0,1]$. Define for $t\in [0,1)$ $$M_n(t)=\frac{F_n(t)-t}{1-t}.$$ I have shown that this is a martingale. If anybody is interested, I'm posting it. Now I want to show that if we put $M_n(t)=1$ for $t\geq 1$, then $M_n$ is not UI, but bounded in $L^1$. But then, I am not sure what $\lim_{t\rightarrow 1} M_n(t)$ is, can you help me with that? I guess, I need that to show that it's not UI, right? Furthermore, I'm really stuck with the boundedness, none of the inequalities I found, lead anywhere. Any ideas?

Thank you very much!

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I think the $X_i$ are uniform. $\sqrt(n)M_n$ converges to a brownian bridge, $W_0$, and $\frac {W_0}{1-t}$ is not $\mathbb L^1$ bounded. – mike Apr 25 '12 at 10:55
I've got an extra information: My $X_i$ are actually uniformly distributed on [0,1]. I didn't see that earlier, sorry. Does that help anyhow? That $M_n$ is not UI I figured out, I think, but I used a different approach. I showed that $$\lim_{t\rightarrow 1} E(M_n (t))\neq E(\lim_{t\rightarrow 1} M_n (t))$$ and then used a theorem, that states, that this would hold if $M_n$ was UI. – kelu Apr 25 '12 at 15:31
So my main question is now how to show boundedness in $L_1$. Since I now have the distribution of the $X_i$ I'm trying to just compute the expectation. Maybe I can regard the positive and the negative part seperately. Only, I'm not sure at all about how to do it, so if you think, this is a stupid approach and can think of something better, please let me know :-) – kelu Apr 25 '12 at 16:04
Silly me! I thought it was $1+t$ – math Apr 25 '12 at 19:24
I think a good approach will be to use that $F_n (t)$ is bounded by 1. Then I want to look at $$E|M_n (t)|=E\left(\frac{F_n (t)-t}{1-t}\right)^{+} + E\left(\frac{F_n (t)-t}{1-t}\right)^{-},$$ so the expectation of the positive and the negative part seperately. Then for the positive part I have $$E\left(\frac{F_n (t)-t}{1-t}\right)^{+}\leq E\left(\frac{1-t}{1-t}\right)=1,$$ is that correct? – kelu Apr 26 '12 at 11:20

About the boundedness in $\mathcal{L}_{1}$: First we note that $F_{n}(t)\in[0,1]$ . We consider $$E\left|M_{n}(t)\right|=E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}+E\left(\frac{F_{n}(t)-t}{1-t}\right)^{-},$$ where $\left(x\right)^{+}=\max\left\{ 0,x\right\}$ and $\left(x\right)^{-}=\max\left\{ 0,-x\right\}$. Of course, for $t\geq 1$ the boundedness is trivial since then $M_{n}(t)=1$, so we focus on $t\in[0,1)$. On this interval I have shown, that $M_{n}$ is a martingale, which gives us $E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}-E\left(\frac{F_{n}(t)-t}{1-t}\right)^{-}=E\left[M_{n}(t)\right]=E\left[M_{0}(t)\right]=0$. Thus we have $E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}=E\left(\frac{F_{n}(t)-t}{1-t}\right)^{-}$, so we can focus on $E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}$. We exclude the special case, that the fraction is zero, because then the boundedness is trivially true as well. We notice that for $\frac{F_{n}(t)-t}{1-t}$ to belong to the positive part, we need to have that either either both, nominator and denominator, are positive or both of them are negative. Since we are considering $t\in[0,1)$, we have that both have to be positive. Since $F_{n}(t)\in[0,1]$ we obtain $$E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}\leq E\left(\frac{1-t}{1-t}\right)=1.$$ Since $E\left(\frac{F_{n}(t)-t}{1-t}\right)^{+}=E\left(\frac{F_{n}(t)-t}{1-t}\right)^{-}$ then also the negative part is bounded by 1 and we conclude, that for all $t\geq 0$ it holds that $E\left|M_{n}(t)\right|\leq 2.$

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Note that $M_n(t)=1-\frac{1-F_n(t)}{1-t}$ and that $1-F_n(t)\geqslant0$ with $\mathbb E(F_n(t))=t$ hence $$ \mathbb E(|M_n(t)|)\leqslant1+\frac{1-\mathbb E(F_n(t))}{1-t}=2. $$ On the other hand, $\mathbb E(M_n(t))=0$ for every $t\lt1$ and $M_n(t)=1$ for every $t\geqslant T_n$ with $T_n=\max\{X_k\mid 1\leqslant k\leqslant n\}$. Since $T_n\lt1$ almost surely, $M_n(t)\to M^*_n=1$ when $t\to1$, $t\lt1$.

If $(M_n(t))_{0\leqslant t\lt1}$ was UI, $\mathbb E(M^*_n)=1$ would be the limit of $\mathbb E(M_n(t))$ when $t\to1$, $t\lt1$, that is, $0$. Since $1\ne0$, $(M_n(t))_{0\leqslant t\lt1}$ is not UI.

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