Sum of Two IID Uniform Variables¶

Proposition¶

If $X,Y$ are independent $\operatorname{Unif}(0,1)$ variables, then $S=X+Y$ has density

f_S(s)=\begin{cases}s,&0<s\le1,\\2-s,&1<s<2,\\0,&\text{otherwise}.\end{cases}

The density peaks at $s=1$ $1$ .

Each Uniform Distribution on $(0,1)$ has density one there. The Convolution of Independent Random Variables integral reduces to the length of the overlap between $(0,1)$ and $(s-1,s)$ $1, s)$ , producing $s$ for $0<s\le1$ and $2-s$ for $1<s<2$ $2$ .

Interpretation¶

Two independent uniform values have more ways to add up near the centre than near either endpoint. Their sum therefore has a triangular density on $(0,2)$ with its peak at $1$ .

The triangular density of the sum of two independent uniforms on 0 to 1, peaking at 1, divided into regions with probabilities 1/8, 3/4 and 1/8.

The red triangle on $(0,1/2]$ has base $1/2$ and height $1/2$ , so $\mathbb{P}(S\le1/2)=1/8$ $1/8$ ; by symmetry the right tail has the same probability, and the middle part from $1/2$ to $3/2$ carries the remaining $3/4$ , as in the Examples.

Move $s$ : $f_S(s)$ is the length of the overlap of $(0,1)$ and $(s-1,s)$ $1, s)$ , which grows to $1$ at $s=1$ and then shrinks.

The result can also be read in the unit square, where the pair $(X,Y)$ is uniform. The event $\{S\le s\}$ is the part of the square below the line $x+y=s$ $s$ ; for $0<s\le1$ it is a triangle of area $s^2/2$ , which is the CDF, and its derivative $s$ is the density. The density is symmetric about $1$ because $(1-X)+(1-Y)=2-S$ has the same distribution as $S$ . Mean and variance follow without the density: $\mathbb{E}S=1/2+1/2=1$ $1$ , and by independence $\operatorname{Var}(S)=1/12+1/12=1/6$ $1/6$ .

This is the continuous counterpart of the triangular distribution of the sum of two dice, which peaks at $7$ . Adding more independent uniforms makes the shape smoother and more bell-shaped, an effect that returns later as the central limit theorem.

The sum of two uniforms is not uniform on $(0,2)$ : that would give $\mathbb{P}(S\le1/2)=1/4$ $1/4$ , twice the correct value $1/8$ .

Key Results¶

$\mathbb{E}(X+Y)=1$ and $\operatorname{Var}(X+Y)=1/6$ $1/6$ .
The CDF is $s^2/2$ on $0<s\le1$ and $1-(2-s)^2/2$ on $1<s<2$ $2$ .

Examples¶

$\mathbb{P}(X+Y\le1/2)=(1/2)^2/2=1/8$ $1/8$ .
$\mathbb{P}(1/2<X+Y<3/2)=1-1/8-1/8=3/4$ $3/4$ .

Prerequisites¶

Variance and Standard Deviation

Sources¶

Nicky van Foreest · Probability Distributions · Week 3

Keywords: triangular distribution

Last updated: 20 October 2026 03:02