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We have n IID random variables $X_1, X_2, \ldots, X_n$. Let $R_i$ be $X_i$'s rank in the set $\{X_1, X_2, \ldots, X_3 \}$ when we order from large to small. How to prove $R_i, \forall i \in \{1, 2, \ldots, n\}$, is uniformly distributed on $\{1, 2, \ldots, n\}$?

My first guess is that, for any position $j$ in ordered sequence of $X$s, as $X_1, X_2, \ldots, X_n$ are equally likely to be the $j$th largest,

$$\Pr \{ R_i = j \} = \frac 1 n.$$

So $R_i$ is uniformly distributed on $\{1, 2, \ldots, n\}$.

Another way to think about it, is to count how many possible cases are there when $R_i = j$. As $X_i$'s position in ordered sequence is fixed at $j$, then we can just permute the rest of $X$s to get all possible ordered sequence. Since there are $n-1$ variables left, there are $(n-1)!$ situations. As for any $R_i = j, 1 \le j \le n$, there are always $(n-1)!$ possible ordered sequences, we can say $R_i$ is uniformly distributed on $\{1, 2, \ldots, n\}$.

Are these 2 proof rigorous?


As cardinal pointed out, an additional condition is needed for the proof, and a sufficient such condition is that $X_i$ to be a continuous random variable.

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You need either an additional hypothesis (e.g., $X_i$ are continuous random variables) or you need to specify how ties are broken for this to hold. For example, consider $X_i \sim \mathrm{Ber}(1/2)$. – cardinal Sep 14 '11 at 14:25
Why $X_i$ are continuous? Because there might be duplications? – ablmf Sep 14 '11 at 15:07
Yes, because of duplications. – Robert Israel Sep 14 '11 at 15:29
I posted an answer and deleted it. I'll be back..... – Michael Hardy Sep 14 '11 at 15:43
Now I've done several edits within a few minutes and if you looked during that time, you might have been confused, but now I think I have it about where I want it. – Michael Hardy Sep 14 '11 at 19:33
up vote 0 down vote accepted

Suppose $Y_1 = X_2$, $Y_2=X_3$, and $Y_3 = X_1$. If you can show that the joint distribution of the vector $(Y_1,Y_2,Y_3)$ is the same as the joint distribution of the vector $(X_1,X_2,X_3)$, then it follows that the probability that $X_1$ has a certain rank is the same as the probability that $Y_1$ has that rank. But the latter is the probability that $X_2$ has that rank. Thus the probability that $X_1$ has a certain rank is the same as the probability that $X_2$ has that rank. As with $1$ and $2$, so also with the others. Thus $$ \Pr(R_1 = j) = \Pr(R_2=j)=\cdots=\Pr(R_n=j). $$ Since these are mutually exclusive and exhaustive events, each must have probability $1/n$. Notice that we didn't need to know the value of $j$ for this. Therefore $$ \Pr(R_1 = 1) = \Pr (R_1=2) = \Pr(R_1=3) = \cdots = \Pr(R_1=n) = \frac1n, $$ and similarly with any of the other indices besides $1$.

So your first guess was right, but one can say a few things to demonstrate that it's right, and call that a proof.

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Of course, more pedestrian combinatorial arguments also work. – Michael Hardy Sep 14 '11 at 19:35
I think this is rigorous proof. But to be sure, I will leave it for 1 or 2 days to see if others agrees. – ablmf Sep 14 '11 at 20:57
As far as I can tell, this argument suffers from the same pitfall as the OP's attempts. I believe my example still shows why. – cardinal Sep 14 '11 at 23:39
I will put that into know condition. – ablmf Sep 15 '11 at 1:55
@Michael: Precisely. Since others will (hopefully) be reading this in the future, my thought was that it might be important to actually state this, lest one draw the wrong conclusion otherwise. A version of Lebesgue's decomposition theorem tells us exactly what the measures with the required hypothesis must look like. – cardinal Sep 15 '11 at 11:23

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