The checks on the equations of the uniform distributions aren't particularly satisfactory, insofar as they really only confirm that the leading term is correct. For a more detailed test, we calculate a continuous form of our earlier discrete two class problem, to see if we get the same answer.
What we do here is consider a uniform distribution that appears in two distinct pieces. The probability is zero everywhere except when M-a Now, we can go through the bother of calculating two integrals (one from M-a to M-b, and the other from M+b to M+a), but there's a shortcut available. These two integrals together are equivalent to the integral on [M-a,M+a] minus the integral on [M-b,M+b], and we've already calculated that, which allows us to jump ahead several steps. 1/(a-b) [ M^4a + 2M^2a^3 + (a^5)/5 ] - [ M^4b + 2M^2b^3 + (b^5)/5 ] Our discrete case is really just the limit of this problem, in the limit b->a. = M^4 + 2M^2(a^2+aa+a^2) + (1/5)(a^4+a^3a+a^2a^2+aa^3+a^4) which checks with our previous result. We find the same is true when we look at the calculation of the average probability as well... 1/(a-b) [ M(M^4a+(10/3)M^2a^3+a^5) - M(M^4b+(10/3)M^2b^3+b^5) ] once again taking the limit b->a. With consistent results, we can feel confident that the analysis is sound. May 11, 2005 8:15 PM
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= 1/(a-b) [ M^4(a-b) + 2M^2(a^3-b^3) + (1/5)(a^5-b^5) ]
= 1/(a-b) [ M^4(a-b) + 2M^2(a-b)(a^2+ab+b^2) + (1/5)(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4) ]
= M^4 + 2M^2(a^2+ab+b^2) + (1/5)(a^4+a^3b+a^2b^2+ab^3+b^4)
= M^4 + 2M^2(3 a^2) + (1/5)(5 a^4)
= M^4 + 6M^2a^2 + a^4
= 1/(a-b) M [ M^4(a-b)+(10/3)M^2(a^3-b^3)+(a^5-b^5) ]
= 1/(a-b) M [ M^4(a-b)+(10/3)M^2(a-b)(a^2+ab+b^2)+(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4) ]
= M [ M^4+(10/3)M^2(a^2+ab+b^2)+(a^4+a^3b+a^2b^2+ab^3+b^4) ]
= M [ M^4 + 10M^2a^2 + 5a^4 ]