Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If I average two independent realizations of $N(0,1)$, I get a random variable with distribution $N(0,1/2)$.

Now suppose $X_1,\ldots,X_n$ are independent realizations of $N(0,1)$. Sort them in ascending order, and of the $n-1$ pairs of adjacent realizations, randomly select $k\leq n-1$ of them without replacement, and average each pair to form the random variables $Y_1,\ldots,Y_k$.

Are these new random variables independent realizations of $N(0,\sigma^2)$ for some $\sigma$?

share|improve this question

2 Answers 2

No they're certainly not independent. Suppose you have $k=n-1$ and you happen to know that the first $Y$ that you sample takes the value 100000. This can happen in many ways (all of them extremely unlikely), but the overwhelmingly most likely way for it to happen is for two of the $X$'s to be both very close to 100000 (this is a simple computation with the convexity of $x^2$). This means that given the information that you have so far, another of the $Y$'s will be (with very high probability) very close to 50000. Of course this doesn't happen for independent variables.

share|improve this answer

When you average a chain of ascending numbers, for two of the averages to be close to each other, three of the original numbers must be close to each other. This means that the probability of two of the $Y_k$ being within $\epsilon$ of each other declines like $\epsilon^2$, not like $\epsilon$ as it would if they were independent.

share|improve this answer
    
Good point, but order statistics are more likely to be close together (they are not independent). Or am I missing something? –  Dustin G. Mixon Sep 1 '13 at 14:32
1  
I'm thinking about the probability that two among all of the variables are close together, so the order I put them in doesn't matter. The probability is a constant times the probability that two independent Gaussians are close. –  Will Sawin Sep 1 '13 at 15:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.