Expected number and variance of number when sum exceeds 1 (Irwin-Hall distribution)

Question

I am given a random variable (uniformly distributed) between 0 and 1. To this, I add a second such random variable. I keep on adding these variables until sum exceeds 1, and then stop. Let us call number of such random variables needed is N. What is expectation of N ? What is the variance of N ?

To this my approach was following:

The sum of n independent and identically distributed U(0,1) random variables follow the Irwin-Hall distribution. Let $X_i = \sum_{k=0}^i{U_k}$ where $U_k$s are uniform on [0,1] and i.i.d random variables.

1. Is it true that $\mathbb{E}[N]=\sum_{i=1}^{\infty} i \mathbb{P}(\{X_i \geq 1\})$ ?

I am not sure it is because in that case, by the p.d.f of the Irwin-Hall distribution (see the Wikipedia link above), we see \begin{align} \mathbb{P}(X_i \geq 1) &=1-\mathbb{P}(X_i < 1) \\ &=1-\frac{1}{(i-1)!}\sum_{k=0}^i(-1)^k \binom{i}{k} \int_{0}^1 (x-k)_+^{i-1} dx\\ &=1-\frac{1}{(i-1)!}\int_{0}^1 x^{i-1} dx = 1 - \frac{1}{i!}. \end{align} But then we note, $\mathbb{E}[N]=\sum_{i=1}^{\infty} i \mathbb{P}(\{X_i \geq 1\}) = \mathbb{E}[N]=\sum_{i=1}^{\infty} i (1 - \frac{1}{i!})$. This last sum seems to be diverging ????

2. Same question applies when I try to calculate the Variance

Answer 1

The probability distribution for the number $N$ of trials needed for the sum to first exceed $1$ is given by $$\Pr[N = n] = \frac{1}{(n-1)! + (n-2)!}, \quad n \in \{2, 3, \ldots \},$$ so in particular, $$\begin{array}{c|c} n & \Pr[n = n] \\ \hline 2 & \frac{1}{2} \\ 3 & \frac{1}{3} \\ 4 & \frac{1}{8} \\ 5 & \frac{1}{30} \\ 6 & \frac{1}{144} \\ 7 & \frac{1}{840} \\ \end{array}$$

I state this result without proof. Thus $\operatorname{E}[N] = e$ and $\operatorname{Var}[N] = (3-e)e.$

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Addendum. The MathWorld entry on the uniform sum distribution, formulas $(7)$ through $(10)$, give the relevant details for a proof of the above.