Yesterday, I was asked the following question: the sum of two uniformly distributed random variables with the same support has a triangular distribution. Is there a random variable
X such that
X + Y has a distribution which is uniform, where
X and
Y are independent and identically distributed?
I don't know the answer, but I started thinking as follows. First, it's enough to show that there aren't independent identically distributed
X, Y, such that
X+
Y has a distribution uniform on [-1, 1]; linearly scaling gets the general result. Now, the
characteristic function of a uniform distribution on [-1, 1] is
φ(
t) = (sin
t)/
t. The characteristic function of
X+
Y is the product of the characteristic functions of
X and
Y. (If you're more familiar with analysis than probability, note that characteristic functions are basically Fourier transforms, and the probability density function of
X+
Y is the convolution of the probability density functions of
X and
Y.) Thus, if
X exists it has characteristic function
ψ(
t) = [(sin
t)/
t]
1/2 -- this is already a bit problematic, because we want ψ to be continuous, but even with that restriction we still have to specify which square root is being taken on each of the intervals ... [-3π, -2π], [-2π, -π], [-π, π], [π, 2π], [2π, 3π] ... (Informally, we have to make a new choice every time (sin
t)/
t goes through 0.
At this point I think one wants to use
Bochner's theorem, which says that the functions which are characteristic functions of measures on the real line are exactly the
positive definite functions -- but how does one show that this function is positive definite?
The other thing to do is to look at the discrete analogue; consider the
probability generating function of a random variable which is uniformly distributed on the set {0, 1, ...,
n-1}. This is χ(
x) = (1+x+x
2+...+x
n-1)/n. Now, if this random variable were the sum of two independent identically distributed random variables, its p.g.f. would be the square of a polynomial with positive real coefficients. It's not.
But what about the continuous case?