I did some numerical computations and for non-uniform priors the LOS depends on the number of draws:

E.g. for the (non-normalized prior)

Code: Select all

`(4*score*(1-score))**20`

(this a prior that indicates that two engines are close in strength, a typical situation)

I get

Code: Select all

```
W=0 D=0 L=2 LOS=0.333
W=0 D=1 L=2 LOS=0.296
W=0 D=2 L=2 LOS=0.267
W=0 D=3 L=2 LOS=0.245
W=0 D=4 L=2 LOS=0.227
W=0 D=5 L=2 LOS=0.213
W=0 D=6 L=2 LOS=0.202
W=0 D=7 L=2 LOS=0.192
W=0 D=8 L=2 LOS=0.185
W=0 D=9 L=2 LOS=0.178
W=0 D=10 L=2 LOS=0.173
W=0 D=11 L=2 LOS=0.168
W=0 D=12 L=2 LOS=0.164
W=0 D=13 L=2 LOS=0.160
W=0 D=14 L=2 LOS=0.157
```

So stating that draws convey no information about LOS is a bit misleading.

Below is de code

Code: Select all

```
from __future__ import division
import scipy.integrate
import math
def uniform(w,l):
return 1
def standard_prior(w,l):
return w**10*l**10
def close_prior(w,l):
d=1-w-l
s=w+(1/2)*d
return (4*s*(1-s))**20
def LOS_integrand(prior,w,l,W,D,L):
return prior(w,l)*(w**W)*((1.0-w-l)**D)*(l**L)
def LOS_denominator(prior,W,D,L):
return scipy.integrate.dblquad(lambda l,w: LOS_integrand(prior,w,l,W,D,L),0.0,1.0,lambda w:0.0,lambda w:1.0-w)
def LOS_numerator(prior,W,D,L):
return scipy.integrate.dblquad(lambda l,w: LOS_integrand(prior,w,l,W,D,L),0.0,1.0,lambda w:0.0,lambda w:min(w,1.0-w))
if __name__=='__main__':
W=0
L=2
prior=close_prior
for D in xrange(0,15):
print "W=%d D=%d L=%d LOS=%.3f" % (W,D,L,LOS_numerator(prior,W,D,L)[0]/LOS_denominator(prior,W,D,L)[0])
```

Ideas=science. Simplification=engineering.

Without ideas there is nothing to simplify.