Random sums of iid Uniform random variables [duplicate]

Let $\{X_r : r\ge 1\}$ be independently and uniformly distributed on $[0,1]$. Let $0<x<1$ and define

$$N=\min\{n\ge 1 : X_1 + X_2 +\ldots+X_n> x\}$$

Show that $$P(N>n) = \frac{x^n}{n!}$$ and hence find the mean and variance of $N$.

I cant see how $N$ can be greater than $n$, as each $X_r$ is less than or equal $1$, so if each $X_r= 1$, then the sum would equal $n$, but not be greater. I am obviously interpreting this wrong but I cant see where.

Let $X_1, X_2, \dots$ denote successive random draws from a Uniform(0,1) parent, where the sample is terminated as soon as the Irwin-Hall sum $X_1 + X_2 + \dots + X_n > a$ is attained. [To avoid notational confusion, I am using $a$ (rather than $x$) for the target sum.] Importantly, it is given that $0< a< 1$. Then:

• $P(N>1) = P(X_1 < a) = a$

• $P(N > 2) = P(X_1 + X_2 < a) = \frac{a^2}{2}$ i.e. the area defined by the isosceles right triangle:

• $P(N > 3) = P(X_1 + X_2 + X_3 < a) = \frac{a^3}{6}$ which is the volume defined by the trirectangular tetrahedron:

In each case, to advance to a higher dimension, we multiply the preceding probability by $P(X<a)=a$ and divide by the number of 'sides' ($n$) i.e. $$P(N>n) = \frac{a}{n} P(N>n-1) \quad \text{ and so } \quad P(N>n) = \frac{a^n}{n!}$$

The cdf of $N$ is $P(N \leq n) = 1 - P(N>n)$. It is then straightforwards to derive the pmf of $N$ and derive any desired moments.

You are mixing up indices. $N$ is defined to be the least number such that $X_1+...+X_N>x$, (notice it's $N$ not $n$). The problem is just recycling $n$. In other words the $n$ in the definition for $N$ has nothing to do with the $n$ in $P(N>n)$.