John Cook, at his Probability Fact twitter feed (@ProbFact), asked (I’ve cleaned up the notation):
What is the expected amplitude for the sum of 
sines with random phase? i.e. sum of 
where ![\phi_i ~ uniform[0, 2\pi]](https://s0.wp.com/latex.php?latex=%5Cphi_i+%7E+uniform%5B0%2C+2%5Cpi%5D&bg=ffffff&fg=000000&s=0&c=20201002)
Intuitively one expects something on the order of 
, since we’re adding together what are essentially independent random variables. It’s not too hard to throw together a quick simulation, without even bothering with any trigonometry, and this was my first impulse. This code just picks the 
uniformly at random, and takes the maximum of 
for values of 
which are multiples of 
.
x = (0:200)/(2*Pi)
n = 1:100
num.samples = 100
max.of.sines = function(phi){
max(rowSums(outer(x, phi, function(x,y){sin(x+y)})))
}
mean.of.max = function(n, k){mean(replicate(k, max.of.sines(runif(n, 0, 2*pi))))}
averages = sapply(n, function(n){mean.of.max(n, num.samples)})
This is a bit tricky: in the matrix in max.of.sines, output by outer, each column gives the values of a single sine function , and rowSums adds them together.
We can then plot the resulting averages and fit a model 
. I get 
from my simulation, which is close enough to 
to ring a bell:
C = lm(averages^2~n+0)$coefficients
qplot(n, averages, xlab="n", ylab="mean", main="means for 100 samples") +
stat_function(fun = function(x){sqrt(C*x)})
At this point we start thinking theory. If you’re me and you haven’t looked at a trig function in a while, you start at the wikipedia page, and discover that it actually does all the work for you:
where
.
That is, the sum of a bunch of sinusoids with a period is a single sinusoid with the same period, and an amplitude easily calculated from the amplitudes and phases of the original sinusoids. There’s a formula for 
as well, but it’s not relevant here.
In our case all the 
are 1 and so we get
If you take the expectation of both sides, and recognize that 
is 1 if 
(it’s 
) and 0 if 
(just the average of the cosine function), then you learn 
where is the number of summands. That agrees with our original guess, and is enough to prove that 
by Jensen’s inequality.
To get the exact value of 
we can expand on David Radcliffe’s comment: “Same as mean dist from origin after N unit steps in random directions. Agree with sqrt(N*pi/4)”. In particular, consider a random walk in the complex plane, where the steps are given by 
where 
is uniform on the interval 
. We can work out that its sum after steps is
and so, breaking up into the real and imaginary components,
.
Rewriting the squared sums as double sums gives
and combining the double sums gives
and by the formula for the cosine of a difference we get
which is exactly the 
given above. So the amplitude of our sum of cosines is just the distance from the origin in a two-dimensional random walk!
It just remains to show that the expected distance from the origin of the random walk with unit steps in random directions after steps is 
. A good heuristic demonstration is as follows: clearly the distribution of the position 
is rotationally invariant, i. e. symmetric around the origin. The position 
is the sum of independent variables each of which is distributed like the cosine of a uniformly chosen angle; that is, it has mean 
and variance 
. So the -coordinate after steps is approximately normally distributed with variance 
. The overall distribution, being rotationally symmetric with normal marginals, ought to be approximately jointly normal with and 
both having mean 0, variance , and uncorrelated; then 
is known to be Rayleigh-distributed, which finishes the proof modulo that one nasty fact.
God plays dice 