Find $E[N]$ for $N =\min\{n:U_1+U_2\cdots+U_n>1\}$ where $U_1,U_2, . . .$ are independent uniform (0,1) random variables.

Question

Find $E[N]$ for $N = \min\{n:U_1+U_2\cdots+U_n>1\}$ where $U_1,U_2, . . .$ are independent uniform (0,1) random variables.

I may be way off base with my thinking here, but is $\min\{n:U_1+U_2\cdots+U_n>1\}$ the same as $\max\{n:U_1+U_2+\cdots+U_n\leq1\}$? If so, I'm thinking of this as a renewal process where n is the maximum number of "arrivals" by time = 1. Then I'm just finding E[N(t)] which is the renewal function m(t) that satisfies the integral equation $m(t)=F(t)+\int_0^tm(t-x)f(x)\cdot dx$ and I know where to go from there.

Answer 1

$P_K\equiv P(N>K)=P(\sum_1^K{U_i<1})=\int_0^1{P(\sum_1^{K-1}{U_i<1-U_K})d{U_K}}$

$P_1=1$

Let $U_i'=\frac{U_i}{1-U_K}$, we have

$P_K=\int_0^1{P(\sum_1^{K-1}{U_i<1-U_K,U_1<1-U_K,...,U_{K-1}<1-U_K})d{U_K}}$

$=\int_0^1{P(\sum_1^{K-1}{U_i'<1})\prod_1^{K-1}P(U_i<1-U_K)d{U_K}}$

$=\int_0^1{P_{K-1}(1-U_K)^{K-1}d{U_K}}$

$=P_{K-1}\frac{1}{K}$

$=\frac{1}{K!}$

Finally ($P(N=1)=0$):

$E(N)=\sum_{K=2}^\infty{P(N=K)\cdot K}=\sum_{K=2}^\infty{(P_{K-1}-P_K)\cdot K}$

$=\sum_{K=2}^\infty{(\frac{1}{(K-1)!}-\frac{1}{K!})\cdot K}$

$=\sum_{K=2}^\infty{(\frac{1}{(K-2)!})}$

$=e$

Answer 2

Defining $N(t) = \max\{n:\sum_{i=1}^{n}U_i < t\}$, note that $(N(t)+1)$th renewal is the first renewal after time $t$. So, what you are looking for is,

$$\mathbb{E}[N(1)+1] = m(1)+1 = e$$

See this answer for the proof of last equality.