Andrew M. Ross wrote in news:[EMAIL PROTECTED]:
> Li Qiang wrote:
>> Hi, I have a problem that I'm not able to solve because of my
>> non-statistics background.
>>
>> Assume we do a Monte Carlo experiment with M=1000 samples, each of
>> which has N=100 data drawing from a p.d.f. p(x) of normal
>> distribution. For each sample, we first sort the 100 data in an
>> increasing order X1 <= X2 <= ... <= X99 <= X100, and calculate the
>> probability that a random datum obeying p(x) is located between minus
>> infinity and X1, that is, the integral INT_{-infinity}^{X1} p(x) dx.
>> If we calculate the mean of this probability for M=1000 samples, then
>> the mean will approach 1/(N+1)=1/101. That is to say, the
>> expectation of this probability is 1/(N+1).
>> My Monte Carlo experiment showes that the above is very likely
>> correct. Could anyone mathematically prove whether the above is true?
>> Furthermore, if we replace the 1st minimum value X1 with nth minimum
>> value Xn, then the expectation of the probability will be n/(N+1).
>> If it is true for a normal distribution function, is it also true for
>> any other p.d.f. p(x). It seems to me that the guess is also true
>> for a uniform distribution.
>
> It's easier to analyze for XN (the maximum value) rather than X1.
> I am more accustomed to using f(x) for the p.d.f. and F(x) for the
> c.d.f., so I will stick to those.
> First, the c.d.f. of XN is Pr{max<x}=Pr{sample 1<x}*Pr{sampe 2<x}*...
> which means that the c.d.f. of XN is (F(x))^N
> and hence its p.d.f. is N * f(x) * (F(x))^(N-1)
> Let Y = the variable you describe, the integral of f(x) from -infinity
> to XN. We can write
> Y = F( XN )
> The expected value of Y is
> (using the Law of the Unconscious Statistician):
> E[Y] = integral from -infinity to infinity of
> F(y) * pdf of XN at y * dy
> The integrand is then
> F(y) * N * f(y) * (F(y))^(N-1) dy =
> N * f(y) * (F(y))^N dy
> and, since f(y) = the derivative of F(y),
> we get
> integral N * f(y) * (F(y))^N = N (F(y))^(N+1) / (N+1)
> At -infinity, F(y)=0; at +infinity, F(y)=1
> so we get N/(N+1) as expected, without even using
> the fact that this is a Normal random variable.
>
> For the minimum, we have to throw in a lot of 1- terms.
> I haven't tried to work it all out, but it seems reasonable
> that it will work. For something other than the maximum or
> minimum, the algebra/integration will get a bit hairier. By
> the way, the general topic we're dealing with here is
> "Order Statistics"; there are several books on it that may help
> if you have deeper questions.
>
Agreed. Isn't the original questioner is also correct in thinking that
the mapping from the ordered set of X's to the normal pdf is not
relevant? Any random sampling from a continuous distribution would give
the same result, since it is only the ordering (Y < X1) of a random
variate that is asked for, not anything that depends on the distribution.
--
David Winsemius
If the statistics are boring, then you've got the wrong numbers.
-Edward Tufte
.
.
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