# How to find the probability $P\{X_1 + X_2 \leq X_3\}$?

Suppose $X_1,X_2,X_3$ be three independent and mutually identically distributed random variable with uniform distribution on closed interval [0,1]. What is the probability $P\{X_1 + X_2 \leq X_3\}$?

We may also do it by convolution of pdfs.

Suppose the random variable $Z=X_1+X_2$, then by convolution, the pdf of $Z$ is:

\begin{align*} f(z) &= \left\{\begin{matrix} z & 0\le z < 1\\ 2-z & 1\le z < 2 \\ 0 & \text{otherwise} \end{matrix}\right. \end{align*}

Hence

\begin{align*} \mathbb{P}\left(Z<X_3\right) &= \int_0^1\, \int_0^{x_3} z\, dz\, dx_3 \\ &= \frac{1}{6} \end{align*}

• I know this question was answered months ago, but I just ran across it searching for something similar. Could you please explain what convolution is? – pocketlizard Dec 3 '14 at 20:30
• Hi, read about convolution in this book, has several good problems. – gar Dec 8 '14 at 11:10

Hint: What is the volume of the region in $\mathbb R^3$ satisfying $0 \le x \le 1$, $0 \le y \le 1$, $x+y \le z \le 1$?

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align}&\color{#66f}{\large% \left.\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\dd z\,\dd y\,\dd x\, \right\vert_{z\ >\ x\ +\ y}} =\left.\int_{0}^{1}\int_{0}^{1}\int_{x + y}^{1}\dd z\,\dd y\,\dd x\, \right\vert_{x\ +\ y\ <\ 1} \\[3mm]&=\left.\int_{0}^{1}\int_{0}^{1}\pars{1 - x - y}\,\dd y\,\dd x\, \right\vert_{y\ <\ 1\ -\ x} =\int_{0}^{1}\int_{0}^{1 - x}\pars{1 - x - y}\,\dd y\,\dd x \\[3mm]&=\int_{0}^{1}\bracks{\pars{1 - x}^{2} - {\pars{1 - x}^{2} \over 2}}\,\dd x =\half\int_{0}^{1}x^{2}\,\dd x = \color{#66f}{\large{1 \over 6}} \end{align}