Calculate expectation and variance

Question

Let $(X_n)$ be a sequence of independent RVs which are uniformly distributed on $[0,1]$ interval. For $0<x\le 1$ we define $$N(x):=\inf\{n:X_1+\dots+X_n\ge x\}.$$ Show that $$\mathbb{P}(N(x)\ge n)=\frac{x^{n-1}}{(n-1)!}.$$ Calculate $\mathbb{E}N(x)$ and $D^2 N(x)$.

I have absolutely no idea how to start. Thanks for any help.

Answer 1

Conditioning on $X_1$, one sees that, for every $x$ in $(0,1)$, $N(x)=1+N'(x-X_1)$ if $X_1\lt x$ and $N(x)=1$ if $X_1\geqslant x$, where $N'$ is distributed like $N$ and independent of $X_1$. Thus, the idea is to work with every $N(x)$ simultaneously.

First, $P(N(x)\geqslant1)=1$ for every $x$. Second, for every $n\geqslant1$, $$ P(N(x)\geqslant n+1)=P(N(x)\geqslant n+1;X_1\lt x)=\int_0^xP(N(x-t)\geqslant n)\,\mathrm dt, $$ from which one sees readily that the functions $u_n:x\mapsto P(N(x)\geqslant n)$ are uniquely determined by the initial condition that $u_1(x)=1$ and by the inductive relations that, for every $n\geqslant1$, $u_{n+1}(0)=0$ and $u_{n+1}'(x)=u_n(x)$. Surely you can deduce from this the distribution of $N(x)$.

To compute the expectation and the variance of every $N(x)$, use the fact that, for every integer nonnegative random variable $M$, $$ E(M)=\sum_{n\geqslant1}P(M\geqslant n),\qquad E(M(M+1))=\sum_{n\geqslant1}2nP(M\geqslant n). $$

Answer 2

We're calculating

$$P(N(x)\ge n| N(x) \ge n-1)=\frac{P(N(x)\ge n)}{P(N(x)\ge n-1)}$$ Obviously $$P(N(x)\ge 1)=1=\frac{x^{1-1}}{(1-1)!}$$ Assume that $$P(N(x)\ge n-1)=\frac{x^{n-2}}{(n-2)!}$$ So we have $$P(N(x)\ge n)=P(N(x)\ge n| N(x) \ge n-1) \cdot P(N(x)\ge n-1)=\\P(N(x)\ge n| N(x) \ge n-1) \cdot\frac{x^{n-2}}{(n-2)!}$$ Now we only need to show that $$P(N(x)\ge n| N(x) \ge n-1)=\frac{x}{n-1}$$

How can I show this equality? Any hint?