The Irwin-Hall distribution: A Problem to do with a Sum of Uniform Distributions

The following question appears at http://www.spellscroll.com/questionfull/184/ where it appears to trace back to http://www.wilmott.com/

X_1, X_2,..., X_n are independent random variables, uniformly distributed on [0,1].

What is the probability that X_1 +X_2 +... +X_n < 1 ?

An attempt:

A general sum of independent identically distributed U[0,1] random variables seems to be called the Irwin-Hall distribution. It is clearly continuous and of interest at least in the general problem of finding the distribution of a sum of random variables.

We can see that a general X_k, k= 1,...,n has characteristic function (given as a Fourier transform of its probability density function),

ϕ_k (t) = E[exp (t.X_k)] = i.(exp(it) - 1) / t

We have that,

E[exp (t.(X_1 +X_2 +... +X_n))] = E[(exp (t.X_1)) ... (exp (t.X_n))] = E[exp (t.X_1)] ... E[exp (t.X_n)]

(this last pair of equalities hold for any independent X_1, ..., X_n in fact)

Whence the characteristic function of our sum,

ϕ(t) = E[exp (t.(X_1 +X_2 +... +X_n))] = [i.(exp(it) - 1) / t]^n (= ϕ_1 (t) ... ϕ_n (t))

The approach thus far is clear, invoking the characteristic function was due to it's sum to product property.

(continued...)

Solving cubics - Cardano's Method

With apologies for the lack of updates, now for something completely different!

Below is a general method of tackling cubics. I have seen the method explained in many an account, but conciseness and clarity are often amiss (if ironically, in the mistaken hope that the needless additions and illustrations help further understanding!).

This is interesting not for actual computational use (though perhaps some will like that), but for the idea behind it. Simply put we reduce what we need to a solving a quadratic and then extract cube roots at the end.

So suppose our cubic is of the form,
ax^3 + bx^2 + cx + d = 0
First off we use the substitution y = x - b/(3a) to reduce this to what's known as a 'depressed cubic' of the form,
y^3 = 3Gy + H

Now recall the expansion,
(p+q)^3 = 3pq(p+q) + p^3 + q^3

We note on comparison then that y=p+q is a solution to our cubic if we take,
pq = G and p^3 + q^3 = H

Now form the quadratic with p^3 and q^3 as roots, that is,

t^2 - Ht + G^3

Solving this we get two values which we can arbitrarily call p^3 and q^3 respectively.
Now we take cube roots for each, and in each case we get three cube roots (call the cube roots of unity 1, w and w^2 say, where w = (-1+i.sqrt(3))/2 or (-1-i.sqrt(3))/2, which you can verify for yourself).

Since y=p+q we need be careful which pairs to match up, but on recalling that pq=G we get,
p+q, wp+(w^2)q and (w^2)p+wq

as the three desired cube roots!