Re: Confidence interval on mean for a set of numbers

From: George Kahrimanis (anakreon_at_hol.gr)
Date: 09/18/04 

Date: 18 Sep 2004 04:03:05 -0700

Peter Michaux (petermichaux@yahoo.com) wrote on 2004-09-17:

>I have a list of numbers all of which are nonnegative. I have no idea
>from what distribution this list of numbers comes. I also have no time
>to check which distribution because I have many of these lists.
>
>I can calculate and plot the mean. I also want to have a confidence
>interval on the mean.

and on 2004-09-17:
>[...] I realize that I don't really want just a
>confidence interval on the mean. [...] What I also want to do is
>characterize the width of the underlying distribution.

I agree with "andre" (andrevh@sci.kun.nl), that a quick (and not
too-dirty) way is to assume normality in the distribution of the
mean (provided that the size of the sample is large enough). Here
I post my suggestion on how we can "do it right". Like Peter
Michaux prefers, we avoid estimating the PDF of each outcome;
we work instead with the corresponding predictive CDF
("cumulative distribution function").

E.g., say that we have recorded one thousand outcomes, lumping
all old samples together. Let us rank these outcomes. Then the
probability of the next outcome being, say, between the
outcomes ranked #393 and #394 is 1/1001, provided that we admit
no prior information of any kind regarding the true underlying
distribution (in other words, we apply only
"low structure assumptions"). This "posterior distribution of
percentiles" is still debatable wrt the foundation. (Imho, it
*is* possible to supply a classical proof (i.e. based on
properties of conventional, orthodox probability) but no such
proof is in print yet, afaik.)
For references, if you need them, see Section 3a of my
news:<3ce8f26b.0409070825.7c799b23@posting.google.com>
"prediction versus parameter estimation (was: literature: ...)"
7 Sep 2004 09:25:57 -0700).

If we accept (OK, just provisionally) the above posterior
distribution of percentiles, we obtain an inexact CDF for the
next outcome. E.g., the CDF at x=#393 is 393/1001, the CDF at
x=#394 is 394/1001, and the CDF at any intermediate point is
undefined as an exact value but defined as an "interval value":
[393/1001, 394/1001].

Now it is elementary to derive a CDF for the mean of the next
sample of size n, given the CDF of single iid outcomes. The
upshot is, in this problem we obtain an interval-valued CDF
for the mean.
(If we wanted the CDF of the median rather than the CDF of the
mean, it would be an easier calculation.)
Moreover, if we have recorded a large number of outcomes, this
"second-order uncertainty" will be negligible.

Not only this method is trivial to implement, but also it is
*the* correct way, imo. (Of course I will welcome any comment
or disagreement.)

Good Day, ~ George Kahrimanis

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