# Probability that sum of independent uniform variables is less than 1

I would like to determine the probability $\mathbb{P}(X_1+\dots+X_n\leq 1)$, where $X=(X_i)_{1\leq i\leq n}$ is a family of independent uniform random variables on $[0,1]$. My first idea is to do this by induction. The first three base cases are straightforward to determine and give us $\mathbb{P}(X_1\leq 1)=1$, $\mathbb{P}(X_1+X_2\leq 1)=\frac{1}{2}$ and $\mathbb{P}(X_1+X_2+X_3\leq 1)=\frac{1}{6}$, which suggests that $\mathbb{P}(X_1+\dots+X_n\leq 1)=\frac{1}{n!}$. Supposing this is true for a certain arbitrary integer $n$, I am having difficulties establishing the result for $n+1$, i.e. $\mathbb{P}(X_1+\dots+X_n+X_{n+1}\leq 1)=\frac{1}{(n+1)!}$. I believe the starting point should be: $$\mathbb{P}(X_1+\dots+X_n+X_{n+1}\leq 1)=\mathbb{P}(X_1+\dots+X_n\leq 1-X_{n+1}),$$ and then somehow condition on $X_{n+1}$, but I am stuck at this point of the calculation. Any ideas of references to literature or even an alternative direct proof would be greatly appreciated.

• en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution
– user140541
Mar 4, 2016 at 23:21
• @d.k.o. Yes I am aware of the Irwin Hall distribution, however I would still like to establish the result as per above. Mar 4, 2016 at 23:24
• Oh. Maybe geometric approach will suffice. This probability is represented by a volume of a part of a (hyper)cube (in 2 dim it's the lower triangle)
– user140541
Mar 4, 2016 at 23:26
• @d.k.o. I used the geometric approach for the base cases $n=2$ and $n=3$, but it is harder to prove for an arbitrary integer Mar 4, 2016 at 23:35
• Jun 29, 2020 at 7:57

Prove by induction the more general result: If $0\le t\le 1$, then $$P(S_n\le t)=\frac{t^n}{n!},$$ where $S_n$ denotes the sum $X_1+\cdots+X_n$. The base case $n=1$ is clear. If holds for $n$, then calculate for $0\le t\le 1$: $$P(S_{n+1}\le t)=\int_0^1P(S_n+x\le t)f(x)dx\stackrel{(1)}=\int_0^t\frac{(t-x)^n}{n!}\,dx=\frac{t^{n+1}}{(n+1)!}$$ Note that in (1) the quantity $P(S_n\le t-x)$ is zero when $x>t$.

• thanks for your answer, does this still hold for $t=1$? Mar 4, 2016 at 23:43
• @user223935: it does and you then get $\mathbb{P}(S_n\le 1)=\frac{1^n}{n!}$ but the more general hypothesis is easier to prove by induction Mar 4, 2016 at 23:46
• @user223935 Sure! And even if it $t=1$ was not included in the proof, you can take the limit as $t\uparrow 1$ (since $S_n$ has a continuous CDF). Mar 4, 2016 at 23:47
• Where did the expression $P(S_{n+1} \leq t) = \int_{0}^1 P(S_n+x\leq t) f(x) dx$ come from?
– 24n8
May 4, 2020 at 0:50
• @Iamanon I am using $P(S_{n+1}\le t)=P(S_n+X_{n+1}\le t)=\int P(S_n+X_{n+1}\le t\mid X_{n+1}=x)f(x)\,dx$. Notice now that $P(S_n+X_{n+1}\le t\mid X_{n+1}=x)=P(S_n+x \le t\mid X_{n+1}=x)=P(S_n+x\le t)$ since $S_n$ and $X_{n+1}$ are independent. May 4, 2020 at 1:26

A geometric argument should suffice.   Given that $\{X_k\}_\infty$ are all iid Uniform$(0;1)$ random variables, then:

$\mathsf P(X_1+X_2\leq 1)$ is the probability that points distributed uniformly over the unit square lie in the lower left triangle; which is $1/2$ the area of the unit square.

$\mathsf P(X_1+X_2+X_3\leq 1)$ is the probability that points distributed uniformly over the unit cube lie in the $(0,0,0)$-corner pyramid; which is $1/6$ the volume of the unit cube.

$\mathsf P(X_1+X_2+X_3+X_4\leq 1)$ is the probability that points distributed uniformly over the unit tesseract lie in $(0,0,0,0)$-corner pentachron; which is $1/24$ of the hypervolume of the unit tesseract.

And so forth.

$\mathsf P(\sum\limits_{k=1}^n X_k\leq 1)$ is the probability that points distributed uniformly over a unit $n$-hypercube lie in a corner $n$-hyperpyramid; which is $1/n!$ of the $n$-hypervolume of the unit $n$-hypercube.

• This is true, but it rather presupposes that you know the hypervolume of a corner hyperpyramid Mar 4, 2016 at 23:48