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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.2012-04-25
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    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.2012-04-25
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    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 :-)2012-04-25
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    Silly me! I thought it was $1+t$2012-04-25
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    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?2012-04-26
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    I probably should use that as $t\rightarrow 1$ the probability, that $X_k \leq t$ tends to $1$. Hence $F_n (t)-t$ becomes smaller (I'm looking at the negative part).2012-04-26

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