The first distribution to be considered is the Uniform distribution. In a uniform distribution, there is a continuous set of possible values, each of which are equally likely.
In the general case, we suppose that all values in the range [M-a,M+A] are equally likely, and that outside that range the probability is zero. P(x) is going to be some constant value A, which must satisfy the equation
∫ A dx = 1
The calculation is pretty trivial to work through, but not so trivial as recognizing that we have a rectangle, A units high and 2a units wide, with area 1. Therefore
A = 1/2a.
Now the probability of the coin landing tails four times in a row is
∫ A x^4 dx
= A [ (M+a)^5 - (M-a)^5 ] / 5
= [ (M+a)^5 - (M-a)^5 ] / (2 * 5 * a)
Because we are subtracting, the even powers of a cancel out, and the odd powers remain, cancelling out the 2 in the denominator.
= [ 5M^4a + 10 M^2a^3 + a^5 ]/ (5 * a)
= M^4 + 2 M^2a^2 + (a^4)/5
Quick inspection shows that in the limit a->0, this looks exactly like our discrete case when all coins have probability M of landing tails, so the general form is certainly correct.
To determine the average probability of the coins that landed tails in all four trials, we do a similar integral
∫ A x^5 dx
= A [ (M+a)^6 - (M-a)^6 ] / 6
= [ (M+A)^6 - (M-a)^6 ] / ( 2 * 6 * a )
= [ 6 M^5a + 20 M^3a^3 + 6 Ma^5 ] / ( 6 * a )
= M [ M^4 + (10/3)M^2a^2 + a^4 ]
but that needs to be normalized in turn...
= M [ M^4 + (10/3)M^2a^2 + a^4 ] / [ M^4 + 2 M^2a^2 + (a^4)/5 ]
And this also checks in the case that a = 0.
May 11, 2005 8:12 PM
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