Probability that sum of independent uniform variables is less than 1

Question

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.

Answer 1

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 $x>t$.

Answer 2

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.