Question

Let \(U_{1}, U_{2}, \ldots\) be independent uniform \((0,1)\) random variables, and define \(N\) by $$ N=\min \left(n: U_{1}+U_{2}+\cdots+U_{n}>1\right\\} $$ What is \(E[N] ?\)

Short answer

The expected value of \(N\), denoted as \(E[N]\), is infinite. It can be found by summing the weighted probabilities for all integer values of \(n \geq 1\). Although this result is counter-intuitive, mathematically the expectation is infinite, demonstrating that we cannot expect a finite number of uniformly distributed random variables to sum to more than 1.

Step-by-step solution

Probability of N

First, we need to find the general expression for the probability that N is equal to a specific value, say \(n\). We look for the probability that the sum exceeds 1 after adding the \(n^{th}\) uniform random variable but not before: \(P(U_1 + U_2 + \cdots + U_{n-1} \leq 1 < U_1 + U_2 + \cdots + U_n)\). We can rewrite this as: $$ P\left( U_1 + U_2 + \cdots + U_{n-1} \leq 1 - U_n < U_1 + U_2 + \cdots + U_{n-1} \right) $$ Since \(U_n\) is uniformly distributed over \((0,1)\), the probability that the sum of the first \(n-1\) variables lies within a certain interval is equal to the size of that interval. #Step 2: Joint probability density function#

Joint PDF

Now, we need to find the joint probability density function (PDF) of the first \(n - 1\) uniform random variables, \(U_1, U_2, \ldots, U_{n-1}\). Since these variables are independent and identically distributed, their joint PDF can be obtained as the product of their individual PDFs: $$ f_{U_1, U_2, \ldots, U_{n-1}}(u_1, u_2, \ldots, u_{n-1}) = \prod_{i=1}^{n-1} f_{U_i}(u_i) $$ Since each \(U_i\) has PDF equal to 1 on the interval \((0,1)\), the joint PDF is also equal to 1 on the \((n-1)\)-dimensional unit cube where \(0 < u_i < 1\) for all \(i\). #Step 3: Probability of the sum being in the given range#

Probability of Sum in Range

Given the joint PDF of the first \(n-1\) random variables, we can find the probability that their sum lies in the given range: $$ P\left( U_1 + U_2 + \cdots + U_{n-1} \leq 1 - U_n < U_1 + U_2 + \cdots + U_{n-1} \right) = \int_{0}^{1} \cdots \int_{0}^{1 - u_n} f_{U_1, U_2, \ldots, U_{n-1}}(u_1, u_2, \ldots, u_{n-1}) du_1du_2 \cdots du_{n-1} $$ Since the joint PDF is equal to 1 on the integration domain, this probability simply becomes: $$ P\left( N=n \right) = \int_{0}^{1} \cdots \int_{0}^{1 - u_n} du_1du_2 \cdots du_{n-1} $$ #Step 4: Finding the expected value of N#

Expected Value of N

Now that we have an expression for the probability of N being equal to a specific value \(n\), we can find the expected value \(E[N]\) by summing the weighted probabilities for all integer values of \(n \geq 1\): $$ E[N] = \sum_{n=1}^{\infty} nP(N=n) = \sum_{n=1}^{\infty} n \int_{0}^{1} \cdots \int_{0}^{1 - u_n} du_1du_2 \cdots du_{n-1} $$ By plugging the expression for \(P(N=n)\) found in Step 3 and carefully evaluating the integral, we can find the expected value: $$ E[N] = \sum_{n=1}^{\infty} n \int_{0}^{1} \cdots \int_{0}^{1 - u_n} du_1du_2 \cdots du_{n-1} = \sum_{n=1}^{\infty} \frac{1}{n} $$ It turns out that this sum diverges, which means \(E[N]\) is infinite. So, the expectation of \(N\) is infinite, demonstrating that we cannot expect a finite number of uniformly distributed random variables to sum to more than 1. It's important to note that this result is quite counter-intuitive, as in practice the sum converges very quickly, but mathematically the expectation is infinite.