On Fri, 29 Apr 2011, David Winsemius wrote:
On Apr 29, 2011, at 1:29 PM, Mike Miller wrote:
On Fri, 29 Apr 2011, Giovanni Petris wrote:
Well, but the original poster also refers to 0.2 and 0.8 as "expected min
and max", in which case we are back to a joke...
Well, he is a lot better with English than I am with Mandarin. He seemed
to like the truncated normal answers, so we'll let those be his answers.
It is possible to choose parameters for a normal distribution with 500
observations such that the expected value of the maximum is .8 and the
expected value of the minimum is .2. Obviously, the mean would be .5,
not 1, but what would the variance then have to be to provide the
correct expected max and min? That's another legitimate question.
You would need to specify an N since the expected first and last order
statistic would decrease/increase with increasing N.
Right -- I chose N=500, as did the OP. I think the order statistics for
the normal are pretty complex, but it wouldn't be hard to use the density
for order statistics for the uniform to compute the appropriate values for
a standard normal, then rescale.
http://en.wikipedia.org/wiki/Order_statistic#The_order_statistics_of_the_uniform_distribution
You'd have to multiply the beta density times the inverse normal cdf and
get the weighted average for a set of points. It doesn't sound terribly
difficult but I don't want to do it! ;-)
Mike
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