What you want to do is reward for the prediction being within the interval and penalize for distance from the interval as well as for the width of the interval. For example:
W could even be 1, IDK. Lowest score wins.
Code:
I= upper_limit -lower_limit
W=Weight of Interval width penalty
P= Actual
if( P>lower_limit AND PD=0
}else{
D=min [ abs ( P - upper_limit); abs ( P - lower_limit) ]
}
Score= D+W*I
W=Weight of Interval width penalty
P= Actual
if( P>lower_limit AND PD=0
}else{
D=min [ abs ( P - upper_limit); abs ( P - lower_limit) ]
}
Score= D+W*I
W could even be 1, IDK. Lowest score wins.
Sorry for coming off as very negative today, I really don't mean to smack down everything, but this wouldn't work either. My counterexample: I would choose upper_limit = lower_limit = 0 and win every time (assuming P is not zero)! I would get D=0 and I=0, which would give me Score=0, regardless of W. Furthermore 0 is the lowest possible score since D>=0 and I>=0, so I would win every time!
