Bomb Problem (part II)

Assume that point A is at location -1 and point B is at 1. If one source produced both A and B and the next shot is expected to be produced by the same source, then the best fit is the normal distribution with mean 0 and dispersion 1: N(0,1,x) - gives the maximum probabilities for points A and B. On the other hand, if different sources produced A and B, the best guess would be that the third shot is normally distributed with one of the sources F(x). Now since no information present about whether one or two sources produced A and B, give the same chances to either way:

f(x)=(N(0,1,x)+F(x))/2

For F(x), give the same chances to points A and B:

F(x)=(F(-1,x)+F(1,x))/2

And the best guess for F(y,x) would be N(y,(x-y)/2,x), because the only information about the dispersion for F(y,x) is the distance between x and y, and the dispersion (x-y)/2 is the best fit. Note, that the function F(y,x) is not proper distribution for variable x, because the probability is given for fixed dispersion and it depends on x, i.e. to be normalized the dispersion has to be fixed. The overall distribution then is:

f(x)=(N(0,1,x)+(N(-1,(x+1)/2,x)+N(1,(x-1)/2,x))/2)/2

where

N(y,s,x)=1/sqrt(2Pi*s²)*Exp(-0.5(x-y)²/s²)

The graph for function f(x) is presented above. The minimum values are at points: -0.47 and +0.47. This means that the safest place is about a quarter distance from either end.