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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.

Marios Gretsas
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  • The sum of the $n$ first $X_i$ follows an Irwin-Halle distribution of density $$ f_S(x,n)=\frac{1}{2(n-1)!}\sum_{k=0}^n (-1)^k{n \choose k} (x-k)^{n-1}\sign(x-k)$$ – EDX Jan 10 '21 at 00:10
  • Also try to make first case for the min = 1 et the min=2 etc... then using conditional probability should help. – EDX Jan 10 '21 at 00:35
  • Surely $P(N(x)\geq x)=1$ trivially as $X_i<1$ with probability $1$? – jlammy Jan 10 '21 at 00:41
  • Look at https://math.stackexchange.com/questions/572273/what-is-the-expectation-of-the-following-random-variable – EDX Jan 10 '21 at 00:46
  • https://math.stackexchange.com/q/214399/321264 – StubbornAtom Jan 10 '21 at 06:27
  • https://math.stackexchange.com/questions/3022309/expected-number-of-terms-needed-to-get-a-sum-greater-than-t-for-i-i-d-random –  Jan 21 '21 at 21:41

1 Answers1

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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 $t\leq1$.

jlammy
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