How to find the pdf of the minimum of absolute differences of Uniform distributions.












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Let $X_1$,$X_2$ and $X_3$ are independent random variables that are uniformly distributed over $(0;b), b>0$. What is the probability density function of z=min($Y_1$,$Y_2)$, where $Y_1=|X_1-X_2|$ and $Y_2=|X_1-X_3|$.










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    Let $X_1$,$X_2$ and $X_3$ are independent random variables that are uniformly distributed over $(0;b), b>0$. What is the probability density function of z=min($Y_1$,$Y_2)$, where $Y_1=|X_1-X_2|$ and $Y_2=|X_1-X_3|$.










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      $begingroup$


      Let $X_1$,$X_2$ and $X_3$ are independent random variables that are uniformly distributed over $(0;b), b>0$. What is the probability density function of z=min($Y_1$,$Y_2)$, where $Y_1=|X_1-X_2|$ and $Y_2=|X_1-X_3|$.










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      Let $X_1$,$X_2$ and $X_3$ are independent random variables that are uniformly distributed over $(0;b), b>0$. What is the probability density function of z=min($Y_1$,$Y_2)$, where $Y_1=|X_1-X_2|$ and $Y_2=|X_1-X_3|$.







      probability probability-distributions uniform-distribution






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      asked Nov 15 '13 at 1:25









      user108921user108921

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          $begingroup$

          Problems such as these can be both tedious and difficult to do by hand, but can be solved in just one or two lines with the help of automated tools. For this example, the joint pdf of $(X_1,X_2,X_3)$ is say $f(x_1,x_2,x_3)$:



          enter image description here



          Problem: find the pdf of $Z = min[ |X_1-X_2|, |X_1-X_3|]$



          Solution: First, derive the cdf of $Z$, namely $P(Z<z)$:



          enter image description here



          where I am using the Prob function from the mathStatica add-on to Mathematica to do the tedious work for me (I am one of the developers of the former).



          The pdf of $Z$ is, of course, just the derivative of the cdf wrt $z$, i.e.:



          $$ begin{cases}frac{2 (b-z)^2}{b^3} & frac{b}{2} < z < b \ frac{2 left(2 b^2-6 b z+5 z^2right)}{b^3} & 0 < z leq frac{b}{2} \ 0& otherwise end{cases}$$
          All done.



          Here is a plot of the solution (the pdf of $Z$, say when $b = 4$) plotted in red $---$ I have also overlaid a Monte Carlo check (in blue), just to make sure no mistakes have crept in:



          <img src="">






          share|cite|improve this answer











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            1 Answer
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            1 Answer
            1






            active

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            active

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            active

            oldest

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            1












            $begingroup$

            Problems such as these can be both tedious and difficult to do by hand, but can be solved in just one or two lines with the help of automated tools. For this example, the joint pdf of $(X_1,X_2,X_3)$ is say $f(x_1,x_2,x_3)$:



            enter image description here



            Problem: find the pdf of $Z = min[ |X_1-X_2|, |X_1-X_3|]$



            Solution: First, derive the cdf of $Z$, namely $P(Z<z)$:



            enter image description here



            where I am using the Prob function from the mathStatica add-on to Mathematica to do the tedious work for me (I am one of the developers of the former).



            The pdf of $Z$ is, of course, just the derivative of the cdf wrt $z$, i.e.:



            $$ begin{cases}frac{2 (b-z)^2}{b^3} & frac{b}{2} < z < b \ frac{2 left(2 b^2-6 b z+5 z^2right)}{b^3} & 0 < z leq frac{b}{2} \ 0& otherwise end{cases}$$
            All done.



            Here is a plot of the solution (the pdf of $Z$, say when $b = 4$) plotted in red $---$ I have also overlaid a Monte Carlo check (in blue), just to make sure no mistakes have crept in:



            <img src="">






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Problems such as these can be both tedious and difficult to do by hand, but can be solved in just one or two lines with the help of automated tools. For this example, the joint pdf of $(X_1,X_2,X_3)$ is say $f(x_1,x_2,x_3)$:



              enter image description here



              Problem: find the pdf of $Z = min[ |X_1-X_2|, |X_1-X_3|]$



              Solution: First, derive the cdf of $Z$, namely $P(Z<z)$:



              enter image description here



              where I am using the Prob function from the mathStatica add-on to Mathematica to do the tedious work for me (I am one of the developers of the former).



              The pdf of $Z$ is, of course, just the derivative of the cdf wrt $z$, i.e.:



              $$ begin{cases}frac{2 (b-z)^2}{b^3} & frac{b}{2} < z < b \ frac{2 left(2 b^2-6 b z+5 z^2right)}{b^3} & 0 < z leq frac{b}{2} \ 0& otherwise end{cases}$$
              All done.



              Here is a plot of the solution (the pdf of $Z$, say when $b = 4$) plotted in red $---$ I have also overlaid a Monte Carlo check (in blue), just to make sure no mistakes have crept in:



              <img src="">






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Problems such as these can be both tedious and difficult to do by hand, but can be solved in just one or two lines with the help of automated tools. For this example, the joint pdf of $(X_1,X_2,X_3)$ is say $f(x_1,x_2,x_3)$:



                enter image description here



                Problem: find the pdf of $Z = min[ |X_1-X_2|, |X_1-X_3|]$



                Solution: First, derive the cdf of $Z$, namely $P(Z<z)$:



                enter image description here



                where I am using the Prob function from the mathStatica add-on to Mathematica to do the tedious work for me (I am one of the developers of the former).



                The pdf of $Z$ is, of course, just the derivative of the cdf wrt $z$, i.e.:



                $$ begin{cases}frac{2 (b-z)^2}{b^3} & frac{b}{2} < z < b \ frac{2 left(2 b^2-6 b z+5 z^2right)}{b^3} & 0 < z leq frac{b}{2} \ 0& otherwise end{cases}$$
                All done.



                Here is a plot of the solution (the pdf of $Z$, say when $b = 4$) plotted in red $---$ I have also overlaid a Monte Carlo check (in blue), just to make sure no mistakes have crept in:



                <img src="">






                share|cite|improve this answer











                $endgroup$



                Problems such as these can be both tedious and difficult to do by hand, but can be solved in just one or two lines with the help of automated tools. For this example, the joint pdf of $(X_1,X_2,X_3)$ is say $f(x_1,x_2,x_3)$:



                enter image description here



                Problem: find the pdf of $Z = min[ |X_1-X_2|, |X_1-X_3|]$



                Solution: First, derive the cdf of $Z$, namely $P(Z<z)$:



                enter image description here



                where I am using the Prob function from the mathStatica add-on to Mathematica to do the tedious work for me (I am one of the developers of the former).



                The pdf of $Z$ is, of course, just the derivative of the cdf wrt $z$, i.e.:



                $$ begin{cases}frac{2 (b-z)^2}{b^3} & frac{b}{2} < z < b \ frac{2 left(2 b^2-6 b z+5 z^2right)}{b^3} & 0 < z leq frac{b}{2} \ 0& otherwise end{cases}$$
                All done.



                Here is a plot of the solution (the pdf of $Z$, say when $b = 4$) plotted in red $---$ I have also overlaid a Monte Carlo check (in blue), just to make sure no mistakes have crept in:



                <img src="">







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 4 '18 at 13:46

























                answered Nov 15 '13 at 8:27









                wolfieswolfies

                4,1662923




                4,1662923






























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