It is common in continuous distributions that the average result is more probable than the extreme results. The triangle distribution satisfies this condition.
As in the case of the uniform distribution, we'll consider that only the probabilities between M-a and M+a exist in our population. The probability of x = M-a is itself zero, but climbs linearly from there until it reaches x = M. Then the trend is reversed, and the probability falls until it reaches zero at x = M+a.
First we figure out the height of the triangle. As the base is 2a in length, and the area is 1, the height must be 1/a.
The definition of the triangle distribution is really two distinct pieces, and in working through the calculations we'll be handling each piece independently, combining the results in the final step.
The left side of the triangle intersects the x axis at M-a, and has a slope of 1/a^2 (rise = 1/a, run = a ). So the linear equation there is
P(x) = 1/a^2 (x-(M-a))
on the otherside
P(x) = -1/a^2 (x-(M+a))
Now, a quick preview of what is to come. In our previous example we saw that the M-a and M+a substitutions would cancel out half of the terms of our answer, and that the normalization constant A would cancel the a out of the leading term.
With A ~ a^2, we have a hint that M+a and M-a will cancel out the odd terms, and that A will step down the even terms. However, if that were all that were to happen, the leading term would look like f(M)/a^2, which can't be right. So we can expect that the integral itself will somehow manage to cancel that leading term.
With that hint in mind, the probability of a coin chosen from the triangle distribution will land tails in each of four tosses is.
∫ (1/a^2)(x - (M-a))x^4 dx + ∫ (-1/a^2)(x-(M+a))x^4 dx
= (1/a^2)[ ∫ (x - (M-a))x^4 dx - ∫ (x - (M+a))x^4 dx ]
= (1/a^2)[ ∫ x^5 - (M-a)x^4 dx - ∫ x^5 - (M+a)x^4 dx ]
The integrals are on different domains, so it may be clearer to calculate them seperately
∫ x^5 - (M-a)x^4 dx
= x^6/6 - (M-a)x^5/5 [M-a,M]
= M^6/6 - (M-a)M^5/5 - (M-a)^6/6 + (M-a)^6/5
= M^6/6 - M^6/5 + aM^5/5 - (M-a)^6/6 + (M-a)^6/5
= M^6/6 - M^5/5 + aM^5/5 + (M-a)^6/5 - (M-a)^6/6
= M^6(1/6 - 1/5) + aM^5/5 + (M-a)^6(1/5 - 1/6)
= ((M-a)^6)-M^6)(1/5 - 1/6) + aM^5/5
= ((M-a)^6)-M^6)(1/30) + aM^5/5
= (1/2)(1/15)((M-a)^6)-M^6) + aM^5/5
The second half has the same indefinite integral, but we are substracting that integral evaluated on the range [M,M+a]. This is equivalent to adding the evaluation on [M+a,M]. In other words, it is exactly the same expression, with the sign of a reversed. This saves a bunch of calculation, so we cut to the chase....
= (1/a^2)[(1/2)(1/15)((M-a)^6)-M^6) + aM^5/5 + (1/2)(1/15)((M+a)^6)-M^6) - aM^5/5]
= (1/a^2)[(1/2)(1/15)((M-a)^6)-M^6) + (1/2)(1/15)((M+a)^6)-M^6) ]
= (1/a^2)(1/15)(1/2)[((M-a)^6)-M^6) + ((M+a)^6)-M^6) ]
= (1/a^2)(1/15)(1/2)[((M-a)^6 + (M+a)^6)- 2 * M^6 ]
= (1/a^2)(1/15)[ M^6 + 15 M^4a^2 + 15 M^2 a^4 + a^6 - M^6 ]
= (1/a^2)(1/15)[ 15 M^4a^2 + 15 M^2 a^4 + a^6 ]
= M^4 + M^2 a^2 + (a^4)/15
And the usual test holds true.
Everything is exactly the same when calculating the average probability of those coins which land heads, except that the 6s and 5s become 7s and 6s.
∫ x^6 - (M-a)x^5 dx
= x^7/7 - (M-a)x^6/6 [M-a,M]
= (1/2)(1/21)((M-a)^7)-M^7) + aM^6/6
(1/a^2)[(1/2)(1/21)((M-a)^7)-M^7) + aM^6/6 + (1/2)(1/21)((M+a)^7)-M^7) - aM^6/6]
= (1/a^2)(1/21)[ M^7 + 21 M^5a^2 + 35 M^3 a^4 + 7Ma^6 - M^7 ]
= M^5 + (5/3) M^3 a^2 + (1/3)Ma^4
= M [ M^4 + (5/3) M^2 a^2 + (1/3)a^4 ]
= M [ M^4 + (5/3) M^2 a^2 + (1/3)a^4 ] / [ M^4 + M^2 a^2 + (a^4)/15 ]
May 11, 2005 8:18 PM
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