🚀 go-pugleaf

Harald Giese wrote:

> For a normal distribution A and B are independent random variables.
> Therefore, the likelihood for P(A>B) = 0.5 (i.e. 50%) and
> P(B>A) = 1 - 0.5 = 0.5 (P(u) + P(v) = 1, if v is the complementary event
> of u  - think of a coin: P(head) + P(tail) = 1).
> You are just dividing a set of N random numbers into two sets of N/2
> elements by random! (However, if you had specified something like A > m
> + sd this would make a different case.) Of course both sets are random
> representations of the same distribution with the same mean and standard
> deviation - but slightly different sample means and sample standard
> deviations as those are only estimators of the "true" parameters of the
> distribution. For N -> Infinity: mean(x) = mean(y) =
> mean_of_normal_distribution.

Dear Harald,

Thankyou for your reply. I don't think you understood my problem correctly or
perhaps I misinterpreted your reply.

I have 1 normal distribution (in fact I have 2 identical normal distributions
but I posed the problem this way to make it simpler). I take random sample A
and then random sample B. if A>B then the value of A goes to group X. if A<B
then value A (value A again not B) goes to Y. Thus it is intuitive that X
will have a greater mean than Y. They will differ from the original by the
same amounts and be mirror images of each other about the mean.

I have done some more thinking about this and the distributions of X and Y
obviously must sum to the original distribution. If you draw this you get 2
non normal distributions with differences from the mean of about 0.5 sd. My
question again though is how do you go through a mathematical proof showing
the exact values of the means and the common new sd? Is there an equation
giving the shape of the new distributions? Is it just multiplying a
cumulative gaussian (prob that A>B) by the original distribution - if so how
do you do this?

many thanks

Jon