Let $U_1, U_2, \ldots$ be a sequence of independent random variables with PDF $f(u) = 1, \;\; 0 < u < 1$. Let $U_1, U_2, \ldots$ be a sequence of independent random variables with PDF
$$f(u) = 1, \;\; 0 < u < 1$$
Find $E[N]$ when $N = \min\left\{n\mid \sum\limits_{i = 1}^n U_i > 1\right\}$.
Hint: For $x\in (0,1)$, define $N(x) = \min\Big\{n\mid \sum\limits_{i = 1}^n U_i > x\Big\}$.
I am absolutely lost on this question. Any help would be greatly appreciated.
 A: For $x\ge 0$, let $N(x)=\min\left\{n:\sum_{i=1}^n U_i>x\right\}$ as in the hint, so that $N=N(1)$.  We want to calculate $$p(x,n)=\Bbb P\big[N(x)=n\big]$$ when $n$ is a positive integer and $0\le x\le 1$.  
For a positive integer $k$, let $S_k=U_1+U_2+\ldots+U_k$ with probability density $f_k$.    For $0\le x\le 1$, observe that
$$f_k(x)=\frac{d}{dx}\int_0^x\int_0^{x-u_1}\ldots \int_0^{x-\sum_{j=1}^{k-2}u_j}\int_0^{x-\sum_{j=1}^{k-1}u_j}du_k\ du_{k-1}\ \ldots\ du_{2}\ du_1.$$
So $$f_k(x)=\frac{d}{dx}\left(\frac{x^k}{k!}\right)=\frac{x^{k-1}}{(k-1)!}$$ for $0\le x\le 1$.
If $N(x)=n$, then $S_{n-1}\le x<S_n$.  We claim that $p(n,x)=\frac{x^{n-1}}{(n-1)!}-\frac{x^n}{n!}$ for every positive integer $n$.  If $n=1$, then clearly $S_1=U_1$, so $$p(x,1)=\Bbb{P}[S_1>x]=\Bbb{P}[U_1>x]=1-x=\frac{1}{(1-1)!}-\frac{x^1}{1!}.$$  We now assume that $n>1$.  Therefore
$$p(x,n)=\int_0^x  \Bbb P\big[U_n>x-s\big]f_{n-1}(s)ds=\int_0^x\big(1-(x-s)\big)f_{n-1}(s)ds.$$
So
\begin{align}p(x,n)&=\int_0^x(1-x+s)\frac{s^{n-2}}{(n-2)!}ds=(1-x)\frac{x^{n-1}}{(n-1)!}+\frac{x^{n}}{n\cdot (n-2)!}\\&=\frac{(n-x)x^{n-1}}{n!}=\frac{x^{n-1}}{(n-1)!}-\frac{x^n}{n!}.\end{align}
(It should be verified that $\sum_{n=1}^\infty p(x,n)=1$.)  
Specifically
$$\Bbb{P}[N=n]=p(1,n)=\frac1{(n-1)!}-\frac1{n!}=\frac{n-1}{n!}.$$
It follows that
$$\Bbb{E}\big[N(x)\big]=\sum_{n=1}^\infty n\Bbb{P}\big[N(x)=n\big]=\sum_{n=1}^\infty n\left(\frac{x^{n-1}}{(n-1)!}-\frac{x^n}{n!}\right)=e^x,$$
so $\Bbb{E}[N]=e$.
Write $p'(x,k)$ for $\frac{d}{dx}p(x,k)$.  For $x>1$, we can show that
$$p(x,n)=-\int_{x-1}^x (1-x+s)\sum_{r=1}^{n-1}p'(s,r) ds.$$
That is, we have a recursion
$$p(x,n)=\sum_{r=1}^{n-1}\left(- p(x,r)+\int_{x-1}^x p(s,r)ds\right).$$
If $P(x,n)=\sum_{j=1}^np(x,n)=\Bbb{P}\big[N(x)\le n\big]$, then we get
$$P(x,n)=\int_{x-1}^x P(s,n-1) ds.$$
Since
$$P(x,1)=p(x,1)=\left\{\begin{array}{ll}1-x&\text{if }0\le x< 1\\
0&\text{if }x\ge1,
\end{array}\right.$$
and $$P(x,n)=1-\frac{x^n}{n!}$$ for $0\le x\le 1$, we can in principle find all $P(x,n)$ and $p(x,n)$ for every positive integer $n$ at any $x\ge 0$.  For example,
$$P(x,2)=\left\{\begin{array}{ll}1-\frac{x^2}{2}&\text{if }0\le x<1\\
2-2x+\frac{x^2}{2}&\text{if }1\le x<2\\
0&\text{if }x\ge 2,
\end{array}\right.$$
$$p(x,2)=\left\{\begin{array}{ll}x-\frac{x^2}{2}&\text{if }0\le x<1\\
2-2x+\frac{x^2}{2}&\text{if }1\le x<2\\
0&\text{if }x\ge 2,
\end{array}\right.$$
$$P(x,3)=\left\{\begin{array}{ll}1-\frac{x^3}{6}&\text{if }0\le x<1\\
\frac{1}{2}+\frac{3x}{2}-\frac{3x^2}{2}+\frac{x^3}{3}&\text{if }1\le x<2\\
\frac{9}{2}-\frac{9x}{2}+\frac{3x^2}{2}-\frac{x^3}{6}&\text{if }2\le x<3\\
0&\text{if }x\ge 3,
\end{array}\right.$$
$$p(x,3)=\left\{\begin{array}{ll}\frac{x^2}{2}-\frac{x^3}{6}&\text{if }0\le x<1\\
-\frac{3}{2}+\frac{7x}{2}-2x^2+\frac{x^3}{3}&\text{if }1\le x<2\\
\frac{9}{2}-\frac{9x}{2}+\frac{3x^2}{2}-\frac{x^3}{6}&2\le x<3\\
0&\text{if }x\ge 3.
\end{array}\right.$$  I am not seeing a pattern to be able to find $P(x,n)$ or $p(x,n)$ for arbitrary $x$ and $n$, except: $$P(x,n)=p(x,n)=\frac{(n-x)^n}{n!}$$ if $n-1\le x\le n$, and trivially $P(x,n)=p(x,n)=0$ when $x>n$.  Also, we have $$P(x,n)+P(n-x,n)=1$$ when $0\le x\le n$.
