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Tim De Meyer
Posted: Mon Mar 12, 2007 12:01 am
Guest
Hi all,

I could use some help on the following problem:
Suppose I draw one random sample x out of a normal distribution(pdf N(x,mu=0,stdev=1)). I calculate the corresponding P-value as the integral of N between -inf and x. I repeat this for inf times so I get a distribution of P-values. Then I want to calculate the integral of this new distribution between -inf and a certain P-value p.
In my problem p is given and I want to calculate this last integral. Mathematically, suppose norminv is the inverse of the normal cdf, then I want to calculate:

integral between -inf and p of N(norminv(p),0,1)

Does anyone knows if there exists a way to calculate this? If somebody can confirm it doesn't exist I can use a sampling approach...

Thanks a lot!
Tim
Jack Tomsky
Posted: Mon Mar 12, 2007 4:31 am
Guest
Quote:
Hi all,

I could use some help on the following problem:
Suppose I draw one random sample x out of a normal
distribution(pdf N(x,mu=0,stdev=1)). I calculate the
corresponding P-value as the integral of N between
-inf and x. I repeat this for inf times so I get a
distribution of P-values. Then I want to calculate
the integral of this new distribution between -inf
and a certain P-value p.
In my problem p is given and I want to calculate this
last integral. Mathematically, suppose norminv is the
inverse of the normal cdf, then I want to calculate:

integral between -inf and p of N(norminv(p),0,1)

Does anyone knows if there exists a way to calculate
this? If somebody can confirm it doesn't exist I can
use a sampling approach...

Thanks a lot!
Tim


THe integral is equal to p. That's because p is uniformly distributed between zero and one.

Jack
 
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