Expected number of (0,1) distributed continuous random variables required to sum upto 1

Solution 1:

I asked a related question in MathOverflow a while back. Here is the link. Let $N$ be the number of $U(0,1)$ random variables needed for the sum to cross $1$. It is in fact possible to derive the pmf of $N$ and compute $E(N)$ as David Bar Moshe has indicated. But there is a nifty derivation if we are only interested in $E(N)$ that involves conditioning on the very first random variable $U_1$ that we pick. I have sketched an outline below. Let me know if any part of the derivation isn't clear.

Let $N(x)$ be expected number of $U(0,1)$ random variables needed for the sum to cross $x$ where $0 \leq x \leq 1$. Then, a recursion can be derived for $f(x) = E(N(x))$ as

$f(x) = \int_{0}^{1} E(N(x) \mid U_1 = y) dy$.

This integral naturally splits into two - that between $0$ and $x$ and that between $x$ and $1$. If $U_1 < x$, $E(N(x))$ is simply $1 + f(x-U_1)$. If $U_1 \> x$, $E(N(x))$ is just $1$. This will give us an integral equation for $f(x)$ that can be solved to get the solution (with suitable initial condition) as $f(x) = e^x$.

Solution 2:

This problem is well-known. For an answer, see, for example, the first part of Section 2 in http://myweb.facstaff.wwu.edu/curgus/Papers/27Unexpected.pdf, or Equations (7)-(10) in http://mathworld.wolfram.com/UniformSumDistribution.html