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We have $y_1,\cdots,y_M \in \mathbb{R}^r$ vectors. Suppose that the last entry of each vector $y_i$ is $i$, so that the vectors are sorted by the last entry. I solved the quadratic programming problem:

minimize $|y_M-y_1-w|^2$ subject to $w^t y_{i+1} > w^t y_i$

This is always possible, since the vectors are sorted. Let $w$ be the solution to this problem and $w_r$ be the last entry of $w$. Then it empirically occurred to me that (for large $M$):

1) $w^t y_{i+1} - w^t y_i$ is normally distributed $N(w_r,\sigma^2)$

2) $w^t y_i$ is uniform distributed.

3) $\sigma < w_r$

4) $w_r = M/2$

Now I am asking myself is there is a chance to prove 2) under the condition that 1) & 3) are true. Hence I am considering the following problem:

Let $a < a_1 < a_2 \cdots < a_n < b$ ($a,b$ constant) , $a_i$ be independent random variables such that $a_{i+1}-a_i \sim N(n/2, \sigma^2)$ is normally distributed. Assume that $\sigma << n/2$.

Can one prove or is it true or does somebody know any heuristic, that $a_i$ is uniform distributed over the interval $(a,b)$ ?

One can do an experiment in R for example:

> rn <- rnorm(1000,1000/2,60)
> a <- cumsum(rn)
> b <- runif(length(a),min(a),max(a))
> ks.test(a,b)

    Two-sample Kolmogorov-Smirnov test

    data:  a and b
    D = 0.024, p-value = 0.9356
    alternative hypothesis: two-sided

This phenomenon seems to happen only if the expected value is much greater than the standard deviation. For example:

> rn <- rnorm(1000,0,1)
> a <- cumsum(rn)
> b <- runif(length(a),min(a),max(a))
> ks.test(a,b)

    Two-sample Kolmogorov-Smirnov test

 data:  a and b
 D = 0.142, p-value = 3.499e-09
 alternative hypothesis: two-sided

Does anybody have an explanation for this?

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1 Answer 1

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Unfortunately, the result you are trying to prove is false from beginning to end.

  • The $a_i$ cannot be independent random variables since they are constrained not just by the upper and lower limits but because we know that $a_{i+1} > a_i$.
  • Even if we did not have the restriction on the range, or the ordering of the random variables, and choose to have the $a_i$ be independent, a standard result (see this answer for details) is that if the sum of two independent random variables is normal, then the two random variables are necessarily normal too. If $X$ and $Y$ are independent, so are $X$ and $-Y$ independent random variables, showing that the normality of the difference $X-Y$ of independent $X$ and $Y$ also implies the normality of $X$ and $Y$.
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  • $\begingroup$ Ok, but how do you explain the numeric result? $\endgroup$
    – user16491
    Jul 8, 2017 at 13:57
  • $\begingroup$ "There are lies, damned lies, and statistics"? :-) $\endgroup$ Jul 8, 2017 at 14:01
  • $\begingroup$ I think the question remains open, as the main point is the numerical example. Thanks for your answer! $\endgroup$
    – user16491
    Jul 8, 2017 at 14:05

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