1
$\begingroup$

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 $P\{X_1 + X_2 \leq X_3\}$?

$\endgroup$
3
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – pocketlizard Dec 3 '14 at 20:30
  • $\begingroup$ Hi, read about convolution in this book, has several good problems. $\endgroup$ – gar Dec 8 '14 at 11:10
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.