Probability of a sum of uniform random variables

Question

Let $X_1,X_2,X_3,X_4$ and $X_5$ be uniformly distributed random variables over $(0,1)$.

What is the probability that $X_1+X_2+X_3+X_4+X_5 < 1$? I would like to know the shortest way to do this.

Answer 1

Hint: Can you describe the region $\{(x_1,x_2,x_3,x_4,x_5) \in [0,1]^5 : x_1+x_2+x_3+x_4+x_5<1\}$ as a geometric object? What is this object's volume? Try with $5$ replaced by $2$ or $3$ first if you don't see it.

Hint 2: it looks like you don't yet have intuition about the relationship of expectations and integrals (that okay! you have to start somewhere). Consider that if a random variable $X$ taking values in $\mathbb{R}^d$ has density $f$, then $$ P( X \in A ) = \int_A f(x)dx $$ for nice sets $A$. Note that the set $\{X_1+\cdots + X_5 < 1\}$ is the set where the random variable $X:=(X_1,\ldots,X_5)$ is in the set $A:=\{(x_1,x_2,x_3,x_4,x_5) \in [0,1]^5 : x_1+x_2+x_3+x_4+x_5<1\}$. Take for granted that this set is nice enough. Assuming the $X_i$ are independent, we know the density of $X$ is $f(x) = 1_{x_1,\ldots,x_5 \in [0,1]}$. Check that integrating this over $\mathbb{R}^n$ gives $1$. In particular $$ P(X_1+\cdots+X_5 < 1) = P(X \in A) = \int_{A}1_{x_1,\ldots,x_5 \in [0,1]} 1dx = \int_{A} 1 dx. $$ Now, we know that the integral of $1$ over a region is the volume of that region. Thus to calculate the integral, we can try to geometrically recognize what kind of shape $A$ is and just calculate it's volume.

Lets looks at the shape in a few different dimensions, replacing $5$ here with $n$.

$n=1$: $[0,1)$ thats an interval with length $1$.

$n=2$: $\{(x,y) \in [0,1]^2: x+y<1\}$ is a triangle with area $1/2$.

$n=3$: $\{(x,y,z) \in [0,1]^2: x+y+z<1\}$ is a tetrahedron with volume $1/(2\cdot 3)$.

In general the shape is a simplex with volume $1/n!$. To calculate this we can just do the integral $$ \int_0^1\int_0^{1-x_1}\int_0^{1-x_1-x_2}\cdots \int_0^{1-x_1-x_2\ldots-x_{n-1}} 1 dx_n\cdots dx_3 dx_2 dx_1 = \frac{1}{n!}. $$