A random variable $N$ whose support is natural numbers is such that $$\sum_{i=1}^{N} X_i \geq 1$$ and $$\sum_{i=1}^{N-1} X_i < 1.$$
$X_i$'s are independent and identically distributed as $U(0,1)$.
What is $\mathbb{E}[N]$?
I want an approach using the tools of renewal theory, stochastic processes.
My try:
Using the classical approach one can find that $\Pr \{N=n\} = \frac{n-1}{n!}$. And we thus get $\mathbb{E}[N] = e.$
But how do we approach via stochastic processes?
Thanks in advance.
To compute the expectation of $N=\inf\{n\mid X_1+\cdots+X_n>1\}$ without computing its distribution, the most direct approach might be to enlarge the setting, considering, for every nonnegative $x$, the mean number $n(x)$ of random variables uniform on $(0,1)$ that are necessary to get a sum larger than $1$, assuming that one starts at level $x$.
Thus, $n(x)=E(N_x)$ where $N_x=\inf\{n\mid x+X_1+\cdots+X_n>1\}$ and one is after $n(0)$.
First, $n(x)=0$ for every $x>1$. Second, for every $x$ in $[0,1]$, conditioning on the uniform random variable $X_1$, one gets the relation $$n(x)=1+E(n(x+X_1))$$ that is, $$n(x)=1+\int_0^{1-x}n(x+u)du=1+\int_x^1n(u)du$$ Differentiating, one gets $$n'(x)=-n(x)$$ for every $x$ in $[0,1]$, with $n(1)=1$, hence $$n(x)=e^{1-x}$$ for every $x$ in $[0,1]$, in particular,
Exercise: Using the same approach, compute $E(N^{(2)})$, where $N^{(2)}=\inf\{n\mid X_1+\cdots+X_n>2\}$. You should find $E(N^{(2)})=e^2-e$. Give a simple argument to explain why $E(N^{(2)})<2E(N)$.