It's a useful exercise, I believe, to consider the pythagorean approximation in its abstract form, to contemplate the features which must be present in the correct expression.
"In order to make it easier to think", I modify the scale of the previous solutions.
R = RS + RA
r = RS - RA
x = r/R
As before, we consider the consequences of fixing R. x is allowed to range from [-1,1]. We thus have the following boundary cases
W(0) = 0
W(1) = .5
W(-x) = -W(x)
The last of these conditions is most important (though it makes the first redundant); as before it allows us to restrict the Taylor expansion of W to the odd terms alone.
W(x) = A(R)x + B(R)x^3 + C(R)x^5 +...
There are a few additional conditions that we can work out, from the nature of the problem itself. For instance, it should never make a teams winning percentage worse to score a larger percentage of the runs. So the first derivative my be positive everywhere - not just for all x, but also for all R. It is likely that the correct solution also exhibits the law of diminishing returns - so while the first derivative is positive everywhere, it should be maximum as 0. So the second derivative (which is an odd function) should be negative whenever x in (0,1), for all R.
Working from here, A(R) > 0, for all R. B(R) < 0 for all R.
Actually, we can do a bit better than that in our expression for A(R). The derivative must be maximum at x = 0, where only the A(R) term is in play. Furthermore, by the fundamental theorem of calculus, the derivative of W(r) has to be 1/2 somewhere: [ W(1) - W(-1) ] / 1 - (-1) = 1/2. Therefore, A(R) has to be greater than 1/2 for all R.
Now, if the derivative is very close to 1/2 near the origin, it is going to be very close to 1/2 everywhere (it has to be, to meet the terms of the boundary conditions), in which case we're dealing with a line: all of the higher order coefficients are zero.
This is exactly what has to happen in very low scoring conditions. When the run environment is low enough, neither team is capable of scoring two runs, and in effect the team that wins will be that which scores the run. So winning percentage is directly proportional to x.
Of course, A = 1/2 requires that the higher order coefficients cancel out (to achieve the boundary condition W(1) = 1/2. But this naturally follows from the fact that all of the higher order coefficients are zero.
Also note that we never actually achieve this condition - A(0) = 1/2, but W itself makes no sense when R=0, because x = r/R is undefined. The reflects the fact that our definition of the boundary of W (W=.5 if RA=0; W=-.5 if RS=0, W=0 if RS=RA) has an unresolvable contradiction when RS=RA=0.
If A(0) = 1/2, then the previous guesses at A(R) are wrong. We need 1/2 plus some function of R with a root at zero, which is monotonically increasing and tending toward infinity.
One solution form is A(R) = 1/2 + b asinh( cR ), which in the region of interest produces a pretty good fit with b = .330227 and c = .215276, and you can probably find a simpler fit A(R) = 1/2 + b ln ( 1 + cR ).
Of course, you can also simply choose to fit the data as best you can with a polynomial: A(R) = .5 + .0535R + .0088R^2 - .0024R^3 + .0003R^4 + ...
July 26, 2004 12:24 AM
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