Assume that $X_1,X_2....$ are random variables uniformly distributed in $(0,1)$ and we write $N(x)=\min\{k \in \Bbb{N}:X_1+...+X_k \geq x\}$ and $G_n(x)=P(N(x) \geq n)$.
Compute $P(N(x) \geq x),E[N(x)],Var[N(x)]$ and $P(N(x)=n)$
I managed to prove the relation $G_n(x)=\int_0^x G_{n-1}(x-y)dy$ using the convolution of distribution functions and i believe that this relation will help me find $P(N(x) \geq x)$ but i cannot continue any further!
How can i compute these quantities?
Can someone help me with this?
Thank you in advance.
For $t\leq1$, note that $N(t)>n$ iff $X_1+\dots+X_n<t$. So $\mathbb P(N(t)\geq n)$ is the volume of an $n$-dimensional simplex with side length $t$, which is $\frac{t^n}{n!}$. So $$\mathbb E[N(t)]=\sum_{n=0}^\infty\mathbb P(N(t)>n)=\sum_{n=0}^\infty\frac{t^n}{n!}=e^t.$$ Also $\mathbb P(N(t)=n)=\mathbb P(N(t)>n-1)-\mathbb P(N(t)>n)=\frac{t^{n-1}(n-t)}{n!}$.
This is all for $t\leq1$. For $t>1$, things get messy fast.
An alternative way to find $\mathbb E[N(t)]$. Assume $X_1=x$. If $x>t$, then $N(t)=1$. If $x<t$, then $N(t)=1+N'$, where $N'$ has the distribution of $N(t-x)$. So $$m(t)=\mathbb E[N(t)]=1+\int_0^tm(t-x)\mathop{}\!\mathrm{d}x=1+\int_0^tm(x)\mathop{}\!\mathrm{d}x.$$ So $m'=m$, and as $m(0)=1$, we again get $m(t)=e^t$, for $t\leq1$.