consider the following sorted sample:


and the kernel function:

$h(j)=\log(x_j-x_1), j>1$.

Now, it turns out that when I draw data from many continuous distributions, sort them and compute $h(j)$, the plot of the $h(j)$ as a function of $j$ is concave in this sense: $h((j+k)/2)>(h(j)+h(k))/2$ for $1<j<j+1<k<n$

My question is the following: will this observation (concavity of $h(j)$ as a function of $j$) be true for all continuous distributions? If not is there a class of distributions for which it will always be true?

Thanks in advance,

At this point, given the useful comments from @Glen_b and @Momo, I think I should give an example.

this is a vector of values of $x_j-x_1,j>1$ (where the $x$'s are drawn from a continuous distribution, but have been rounded here to the third digit so that they don't take too much place.)

My problem, is that when I plot $h(j)$ for this data-set, the result is clearly nor concave (in the sense I outlined above).

x=c(1.403, 1.406, 1.408, 1.416, 1.417, 1.42, 1.427, 1.441, 1.448, 
1.456, 1.458, 1.465, 1.466, 1.472, 1.472, 1.477, 1.477, 1.479, 
1.482, 1.491, 1.5, 1.504, 1.518, 1.52, 1.52, 1.53, 1.544, 1.545, 
1.561, 1.573, 1.595, 1.595, 1.599, 1.603, 1.605, 1.612, 1.617, 
1.618, 1.618, 1.628, 1.64, 1.644, 1.646, 1.653, 1.679, 1.682, 
1.682, 1.693, 1.694, 1.71, 1.71, 1.711, 1.741, 1.75, 1.756, 1.773, 
1.794, 1.799, 1.804, 1.808, 1.86, 1.882, 1.895, 1.955, 1.992, 
1.995, 2.009, 2.009, 2.063, 2.123, 2.329, 2.356, 2.405, 2.535, 
2.632, 2.635, 2.725, 2.763, 2.783)
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    $\begingroup$ I'm curious where the distribution comes in here. Is $x_i$ a random variable and possibky different for each $i$ or a realisation and thus a real number ? $\endgroup$ – Momo Mar 22 '14 at 22:17
  • $\begingroup$ @Momo: thanks for your question but I do not understand it. Can you try to re-phrase please? Thanks. $\endgroup$ – user603 Mar 22 '14 at 22:54
  • $\begingroup$ I'll try: You talk about continuous distributions. I wonder if $x$ is a random variable with an arbitrary distribution or if it is a real number. $\endgroup$ – Momo Mar 22 '14 at 23:41
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    $\begingroup$ The concavity of the function $h$ doesn't depend on the distribution of the values of the arguments you apply it to. Please clarify your question. $\endgroup$ – Glen_b -Reinstate Monica Mar 23 '14 at 0:01
  • $\begingroup$ @Glen_b: thanks for your comment. Are you stating that so long as the xi are drawn from a continuous distribution, h(j) will always be concave, in the sense made precise in my question? Sorry if the question sounds naive, I'm not a statistician. $\endgroup$ – user603 Mar 24 '14 at 20:35


the problem is related to transformation which preserves concavity of the log function. here we transform an index $j$ into real numbers $x(j)$, then transform the real numbers with $\ln(.)$ function, i.e. $h(j) = \ln(x(j))$ , $ln(.)$ is concave, now the transformation $x(j)$ must be at least not more convex than $\ln(.)$ is concave.

e.g. $e^{x^2}$ is too convex because the sequence $\{0,1,2\}$ becomes $h(j)=\{0,1,\ln(4)\}$, which is convex.

in this case $x(j)$ transformation is not a simple function, but a probabilistic one. it seems to me that for a large sample probability to get a concave sequence should tend to be zero for any distribution.

  • $\begingroup$ This is incorrect. Take, for instance, the dataset $(0,e,e,e^2)$, for which $h(2)=\log(e-0)=1,h(3)=\log(e-0)=1,h(4)=\log(e^2-0)=2$, whence $1=h(3)\lt(h(2)+h(4))/2=3/2,$ contradicting your assertion. $\endgroup$ – whuber Mar 25 '14 at 18:32
  • $\begingroup$ $x_j-x_1=\{e,2e,2e+e^2\}$ not $x_j-x_1=\{e,e,e^2\}$ $\endgroup$ – Aksakal Mar 25 '14 at 18:35
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    $\begingroup$ Your are responding to an earlier version of my comment, which I fixed while you were posting yours. But even so, if you do the calculation with $(e, 2e, 2e+e^2)$ you will still find this is a counterexample. The point is that concavity of the log is irrelevant when the subject concerns the values of the finite sequence $(x_j-x_1).$ $\endgroup$ – whuber Mar 25 '14 at 18:38
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    $\begingroup$ got it. so the problem then is $h(j)=\ln(x(j))$ , $\ln$ is concave, now the transformation $x(j)$ must be at least not more convex than $\ln$ is concave $\endgroup$ – Aksakal Mar 25 '14 at 18:42
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    $\begingroup$ That is an excellent analysis: you got to the heart of it. $\endgroup$ – whuber Mar 25 '14 at 20:01

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