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Considering a box containing balls labeled 1 through $\theta$ from which we sample $n$ balls (hence, a discrete uniform distribution on the interval of $(0,\theta),$ so the probability of picking $x_i$ is $\frac{1}{\theta}$). We randomly sample $n$ elements from the sample space of $(x_1,x_2,...,x_m)$ (without replacement, so $x_1\ne x_2$ etc). Using the method of moments to approximate $\theta$, I get $E(X)=\bar{x}$, so $\frac{\theta}{2}=\bar{x}$, and $\theta=2\bar{x}$, where $\bar{x}=\frac{x_1+x_2+...+x_n}{n}$.

I was asked to find $Pr(2\bar{x}<\max(x_1,x_2,...,x_n))$ for n=2 (since when this is true our estimate for $\theta$ is clearly incorrect). Of course, this is trivial, since we get

$Pr(\frac{2(x_1+x_2)}{2}<\max(x_1,x_2))=Pr(x_1+x_2<x_1)\cup Pr(x_1+x_2<x_2)=0$.

It's fairly obvious, however, that for values of $n$ greater than 2, this is not the case. In class my teacher mentioned solving for this probability by hand gets increasingly difficult as $n$ gets larger, but it should be fairly doable for $n=3$. I'm just curious as to how one would approach this. I'm not really sure what the joint pmf for $x_1+x_2+x_3$ is (they're not independent, I believe), and also a little confused as to where I would set the bounds on the triple integral that follows. (If its not clear, I'm looking for $Pr(x_1+x_2+x_3<\max(x_1,x_2,x_3)))$

I realize that this is not a practical approach really (the book suggests using simulations for higher values of n to approximate the probability), and that the MOM is not a great way to do this in general, but I'm curious nonetheless!

Any thoughts?

Edit: While unable to actually come up with the proper integral, I used some simulations in R to determine that the proper answer is $\frac{1}{8}$, I'm not sure if that helps anything but I figured I'd put it up here just in case.

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What is $\theta$ again? –  jay-sun Jan 26 '13 at 22:54
    
Apologies, $\theta$ is m (the number of balls in box). I'll edit that for clarity. Thanks! –  user1257768 Jan 26 '13 at 22:57
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