Cosine of a Uniform Random VariableDate: 05/31/2005 at 10:08:47 From: Maurice Subject: Cosine of a uniform random variable Given a uniformly distributed random variable between [0,pi/2], I want to know the cosine of this variable. I already know the answer, and it is 2/pi. The problem is, I don't know WHY this is true. I can easily simulate the problem to verify the answer, but I have found no clues on how to solve it. I find it very strange that pi occurs in the result of a cosine. I'm quite good at math, but not at statistics. Perhaps I should take some integral somewhere, I just don't know. I searched half the Internet but could not even find a lead that got me started. The problem originates from computing the average lighting contribution. Given the fact that gradients are uniformly distributed, the dot product of this random vector and another, constant vector can be modeled as the cosine of a uniform random variable between [0,pi/2]. Date: 05/31/2005 at 12:37:59 From: Doctor Douglas Subject: Re: Cosine of a uniform random variable Hi, Maurice. Yes, you should evaluate two integrals to find the expected value of a function f(x) over a distribution p(x): Integral{a,b} f(x) p(x) dx <f(x)> = -------------------------- Integral{a,b} p(x) dx In your case f(x) = cos(x), p(x) = 1 [p is uniform, independent of x], and the limits of integration are a=0 and b=pi/2. The top integral leads to sin(x) evaluated between 0 and pi/2, which reduces to 1-0 = 1. The bottom integral is just b-a, which is pi/2. And finally the average value of f(x) is just the ratio of these two quantities, which is 1/(pi/2) = 2/pi, as you already know. This equation above is essentially the definition of an expectation taken over a distribution, and you can see that it is kind of like a weighted average (each value of f is weighted by the strength of the distribution p at that particular value of x). You must divide, or "normalize" the result by the total amount of p, which explains why we have the integral in the denominator. If we had set things up such that Integral{a,b}p(x)dx = 1, then p would already have been normalized. To accomplish this we would have had to set p(x)=2/pi for 0 < x < pi/2 (this uniform distribution is now normalized). Note that you can use the above equation for whatever f and p you like [for example, try f(x) = cos^2(x) and a uniform distribution p on the interval between 0 and 2*pi, or 0 and pi]. Here are some other answers in our archives that deal with closely related topics: Average Radial Distance of Points within a Circle http://mathforum.org/library/drmath/view/62529.html Monte Carlo Approximation of Pi http://mathforum.org/library/drmath/view/51909.html - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/