Distribution of a smallest index

Question

We have the i.i.d. sequence $\{X\}$ where $X_i \sim U[0,1]$. Define the number $N = \text{min}\{n\geq1: X_1 + \ldots +X_n >x\} $ for $x \in (0,1)$. What is Pr$\{N > n\}$?

My attempt at a solution: Pr$\{N > n\} =$ Pr$\{X_1 + \ldots + X_n \leq x\} = \int_{A_x}\mathrm{d}x_1 \ldots \mathrm{d}x_n$, where $A_x = \{(x_1,\ldots,x_n):0\leq \sum^n_{i=1} x_i \leq x\}$

Correct answer is: Pr$\{N > n\} = \frac{x^n}{n!}$

Need help with:

• compute integral
• Is Pr$\{N > n\} =$ Pr$\{X_1 + \ldots + X_n \leq x\}$ even correct?

Thanks

Answer 1

Here's a hint for a solution: Look at the case that $N > 2$. Then:

$$\begin{align} \mathbf{P}(N > 2) & = \mathbf{P}(X_1 \leq x, X_2 < x - X_1) \\ & = \int_0^{x-x_1}\int_0^xdx_1dx_2 \\ & = \int_0^x(x-x_1)dx_1 \\ & = \frac{x^2}{2}. \\ \end{align}$$

You can then show the general case by induction. And yes, your equality is correct.

Edit: I removed an introductory remark which made no sense for only positive random variables. Also corrected some obvious typos.