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### Cosine of a Uniform Random Variable

```Date: 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/
```
Associated Topics:
College Statistics

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