Find $P(x_2/x_3 leq a)$ where $x_1,x_2,dots,x_n $ are iid $operatorname{Uniform}(0,1)$












-2












$begingroup$


Let $x_1,x_2,...,x_n $ be independent uniformly distributed on $(0,1)$.



How to find $P(x_2/x_3 leq a)$?




Can I consider $$ x_2, x_3 $$ as two random variables or do I need to do something special? my attempt is below



$$f(x_2, x_3) = 1$$



$$P(frac {x_2}{x_3} leq a) = int _0^{1} int _0^{a*x_3} dx_2 dx_3 = frac{a}{2} $$




Edit, I need to find the probability by conditioning. S my approach is not right.




Update: after reviewing the comments below, I summarized them into:



$$Pleft(frac {x_2}{x_3} leq aright) = begin{cases}
int_0^1 int_0^{a*x_3} dx_2 dx_3 = a/2 && a le 1\
int_0^1 int_1^{x_2/a} dx_3 dx_2 = 1 - frac{1}{2*a} && a > 1
end{cases}
$$



these results are somewhat different from some of the comments below and they are not obtained by conditioning, however, they seem to be aligned with Mathematica results



I'm still thinking of the best way to get these results using conditional probability






probability-theory uniform-distribution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
    $endgroup$
    – wolfies
    Dec 10 '13 at 7:11










  • $begingroup$
    $color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
    $endgroup$
    – achille hui
    Dec 10 '13 at 7:24










  • $begingroup$
    Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
    $endgroup$
    – Macavity
    Dec 10 '13 at 7:37










  • $begingroup$
    @Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
    $endgroup$
    – dark blue
    Dec 10 '13 at 13:02












  • $begingroup$
    @wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
    $endgroup$
    – dark blue
    Dec 10 '13 at 13:03


















-2












$begingroup$


Let $x_1,x_2,...,x_n $ be independent uniformly distributed on $(0,1)$.



How to find $P(x_2/x_3 leq a)$?




Can I consider $$ x_2, x_3 $$ as two random variables or do I need to do something special? my attempt is below



$$f(x_2, x_3) = 1$$



$$P(frac {x_2}{x_3} leq a) = int _0^{1} int _0^{a*x_3} dx_2 dx_3 = frac{a}{2} $$




Edit, I need to find the probability by conditioning. S my approach is not right.




Update: after reviewing the comments below, I summarized them into:



$$Pleft(frac {x_2}{x_3} leq aright) = begin{cases}
int_0^1 int_0^{a*x_3} dx_2 dx_3 = a/2 && a le 1\
int_0^1 int_1^{x_2/a} dx_3 dx_2 = 1 - frac{1}{2*a} && a > 1
end{cases}
$$



these results are somewhat different from some of the comments below and they are not obtained by conditioning, however, they seem to be aligned with Mathematica results



I'm still thinking of the best way to get these results using conditional probability






probability-theory uniform-distribution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
    $endgroup$
    – wolfies
    Dec 10 '13 at 7:11










  • $begingroup$
    $color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
    $endgroup$
    – achille hui
    Dec 10 '13 at 7:24










  • $begingroup$
    Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
    $endgroup$
    – Macavity
    Dec 10 '13 at 7:37










  • $begingroup$
    @Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
    $endgroup$
    – dark blue
    Dec 10 '13 at 13:02












  • $begingroup$
    @wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
    $endgroup$
    – dark blue
    Dec 10 '13 at 13:03
















-2












-2








-2





$begingroup$


Let $x_1,x_2,...,x_n $ be independent uniformly distributed on $(0,1)$.



How to find $P(x_2/x_3 leq a)$?




Can I consider $$ x_2, x_3 $$ as two random variables or do I need to do something special? my attempt is below



$$f(x_2, x_3) = 1$$



$$P(frac {x_2}{x_3} leq a) = int _0^{1} int _0^{a*x_3} dx_2 dx_3 = frac{a}{2} $$




Edit, I need to find the probability by conditioning. S my approach is not right.




Update: after reviewing the comments below, I summarized them into:



$$Pleft(frac {x_2}{x_3} leq aright) = begin{cases}
int_0^1 int_0^{a*x_3} dx_2 dx_3 = a/2 && a le 1\
int_0^1 int_1^{x_2/a} dx_3 dx_2 = 1 - frac{1}{2*a} && a > 1
end{cases}
$$



these results are somewhat different from some of the comments below and they are not obtained by conditioning, however, they seem to be aligned with Mathematica results



I'm still thinking of the best way to get these results using conditional probability






probability-theory uniform-distribution







share|cite|improve this question











$endgroup$




Let $x_1,x_2,...,x_n $ be independent uniformly distributed on $(0,1)$.



How to find $P(x_2/x_3 leq a)$?




Can I consider $$ x_2, x_3 $$ as two random variables or do I need to do something special? my attempt is below



$$f(x_2, x_3) = 1$$



$$P(frac {x_2}{x_3} leq a) = int _0^{1} int _0^{a*x_3} dx_2 dx_3 = frac{a}{2} $$




Edit, I need to find the probability by conditioning. S my approach is not right.




Update: after reviewing the comments below, I summarized them into:



$$Pleft(frac {x_2}{x_3} leq aright) = begin{cases}
int_0^1 int_0^{a*x_3} dx_2 dx_3 = a/2 && a le 1\
int_0^1 int_1^{x_2/a} dx_3 dx_2 = 1 - frac{1}{2*a} && a > 1
end{cases}
$$



these results are somewhat different from some of the comments below and they are not obtained by conditioning, however, they seem to be aligned with Mathematica results



I'm still thinking of the best way to get these results using conditional probability






probability-theory uniform-distribution




probability-theory uniform-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 9:03









user10354138

7,4422925




7,4422925










asked Dec 10 '13 at 6:15









dark bluedark blue

130110




130110








  • 1




    $begingroup$
    Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
    $endgroup$
    – wolfies
    Dec 10 '13 at 7:11










  • $begingroup$
    $color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
    $endgroup$
    – achille hui
    Dec 10 '13 at 7:24










  • $begingroup$
    Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
    $endgroup$
    – Macavity
    Dec 10 '13 at 7:37










  • $begingroup$
    @Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
    $endgroup$
    – dark blue
    Dec 10 '13 at 13:02












  • $begingroup$
    @wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
    $endgroup$
    – dark blue
    Dec 10 '13 at 13:03
















  • 1




    $begingroup$
    Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
    $endgroup$
    – wolfies
    Dec 10 '13 at 7:11










  • $begingroup$
    $color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
    $endgroup$
    – achille hui
    Dec 10 '13 at 7:24










  • $begingroup$
    Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
    $endgroup$
    – Macavity
    Dec 10 '13 at 7:37










  • $begingroup$
    @Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
    $endgroup$
    – dark blue
    Dec 10 '13 at 13:02












  • $begingroup$
    @wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
    $endgroup$
    – dark blue
    Dec 10 '13 at 13:03










1




1




$begingroup$
Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
$endgroup$
– wolfies
Dec 10 '13 at 7:11




$begingroup$
Looks like you are just seeking the distribution of the ratio of any 2 independent standard Uniform random variables.
$endgroup$
– wolfies
Dec 10 '13 at 7:11












$begingroup$
$color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
$endgroup$
– achille hui
Dec 10 '13 at 7:24




$begingroup$
$color{red}{text{i}}$id = $color{red}{text{independent}}$ and identically distributed. Being independent, if an iid variable doesn't appear in the expression for determining a probability, you can simply forget it.
$endgroup$
– achille hui
Dec 10 '13 at 7:24












$begingroup$
Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
$endgroup$
– Macavity
Dec 10 '13 at 7:37




$begingroup$
Your approach seems good for $a le 1$. You need $dx_3$ In there instead of $dy$.
$endgroup$
– Macavity
Dec 10 '13 at 7:37












$begingroup$
@Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
$endgroup$
– dark blue
Dec 10 '13 at 13:02






$begingroup$
@Macavity, I don't have any limits on a. Do you just mean that for a>0 it's 0 (which is a standard art)? but how do i do it by conditioning?
$endgroup$
– dark blue
Dec 10 '13 at 13:02














$begingroup$
@wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
$endgroup$
– dark blue
Dec 10 '13 at 13:03






$begingroup$
@wolfies, wolfies, thank you for putting it in words. so does my approach make sense? I don't understand how to do it by conditioning
$endgroup$
– dark blue
Dec 10 '13 at 13:03












2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint: You should reason that (for positive a)
$$Pleft(frac {x_2}{x_3} leq aright) = int _0^{1} min(a ,x_3, 1) , dx_3 = begin{cases}
int_0^1 a , x_3 , dx_3 && a le 1\
int _0^{frac1a} a , x_3 , dx_3 + int _{frac1a}^{1} 1 , dx_3 && a > 1
end{cases}
$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    For $xinleft(0,1right)$ we have by independency:



    $Pleft{ X_{1}/X_{2}leq amid X_{2}=xright} =Pleft{ X_{1}leq axright} $



    So if function $f$ is defined by $fleft(xright)=begin{cases}
    1 & text{if }axgeq1\
    0 & text{if }axleq0\
    ax & text{otherwise}end{cases}$



    Then:



    $Pleft{ X_{1}/X_{2}leq amid X_{2}right} =fleft(X_{2}right)$
    and $Pleft{ X_{1}/X_{2}leq aright} =mathbb{E}fleft(X_{2}right)=int_{0}^{1}fleft(xright)dx$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thank you, but why is the answer different from the above?
      $endgroup$
      – dark blue
      Dec 10 '13 at 18:53










    • $begingroup$
      It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
      $endgroup$
      – drhab
      Dec 11 '13 at 10:14













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint: You should reason that (for positive a)
    $$Pleft(frac {x_2}{x_3} leq aright) = int _0^{1} min(a ,x_3, 1) , dx_3 = begin{cases}
    int_0^1 a , x_3 , dx_3 && a le 1\
    int _0^{frac1a} a , x_3 , dx_3 + int _{frac1a}^{1} 1 , dx_3 && a > 1
    end{cases}
    $$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Hint: You should reason that (for positive a)
      $$Pleft(frac {x_2}{x_3} leq aright) = int _0^{1} min(a ,x_3, 1) , dx_3 = begin{cases}
      int_0^1 a , x_3 , dx_3 && a le 1\
      int _0^{frac1a} a , x_3 , dx_3 + int _{frac1a}^{1} 1 , dx_3 && a > 1
      end{cases}
      $$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint: You should reason that (for positive a)
        $$Pleft(frac {x_2}{x_3} leq aright) = int _0^{1} min(a ,x_3, 1) , dx_3 = begin{cases}
        int_0^1 a , x_3 , dx_3 && a le 1\
        int _0^{frac1a} a , x_3 , dx_3 + int _{frac1a}^{1} 1 , dx_3 && a > 1
        end{cases}
        $$






        share|cite|improve this answer











        $endgroup$



        Hint: You should reason that (for positive a)
        $$Pleft(frac {x_2}{x_3} leq aright) = int _0^{1} min(a ,x_3, 1) , dx_3 = begin{cases}
        int_0^1 a , x_3 , dx_3 && a le 1\
        int _0^{frac1a} a , x_3 , dx_3 + int _{frac1a}^{1} 1 , dx_3 && a > 1
        end{cases}
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 10 '13 at 14:42

























        answered Dec 10 '13 at 14:30









        MacavityMacavity

        35.3k52554




        35.3k52554























            1












            $begingroup$

            For $xinleft(0,1right)$ we have by independency:



            $Pleft{ X_{1}/X_{2}leq amid X_{2}=xright} =Pleft{ X_{1}leq axright} $



            So if function $f$ is defined by $fleft(xright)=begin{cases}
            1 & text{if }axgeq1\
            0 & text{if }axleq0\
            ax & text{otherwise}end{cases}$



            Then:



            $Pleft{ X_{1}/X_{2}leq amid X_{2}right} =fleft(X_{2}right)$
            and $Pleft{ X_{1}/X_{2}leq aright} =mathbb{E}fleft(X_{2}right)=int_{0}^{1}fleft(xright)dx$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              thank you, but why is the answer different from the above?
              $endgroup$
              – dark blue
              Dec 10 '13 at 18:53










            • $begingroup$
              It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
              $endgroup$
              – drhab
              Dec 11 '13 at 10:14


















            1












            $begingroup$

            For $xinleft(0,1right)$ we have by independency:



            $Pleft{ X_{1}/X_{2}leq amid X_{2}=xright} =Pleft{ X_{1}leq axright} $



            So if function $f$ is defined by $fleft(xright)=begin{cases}
            1 & text{if }axgeq1\
            0 & text{if }axleq0\
            ax & text{otherwise}end{cases}$



            Then:



            $Pleft{ X_{1}/X_{2}leq amid X_{2}right} =fleft(X_{2}right)$
            and $Pleft{ X_{1}/X_{2}leq aright} =mathbb{E}fleft(X_{2}right)=int_{0}^{1}fleft(xright)dx$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              thank you, but why is the answer different from the above?
              $endgroup$
              – dark blue
              Dec 10 '13 at 18:53










            • $begingroup$
              It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
              $endgroup$
              – drhab
              Dec 11 '13 at 10:14
















            1












            1








            1





            $begingroup$

            For $xinleft(0,1right)$ we have by independency:



            $Pleft{ X_{1}/X_{2}leq amid X_{2}=xright} =Pleft{ X_{1}leq axright} $



            So if function $f$ is defined by $fleft(xright)=begin{cases}
            1 & text{if }axgeq1\
            0 & text{if }axleq0\
            ax & text{otherwise}end{cases}$



            Then:



            $Pleft{ X_{1}/X_{2}leq amid X_{2}right} =fleft(X_{2}right)$
            and $Pleft{ X_{1}/X_{2}leq aright} =mathbb{E}fleft(X_{2}right)=int_{0}^{1}fleft(xright)dx$






            share|cite|improve this answer









            $endgroup$



            For $xinleft(0,1right)$ we have by independency:



            $Pleft{ X_{1}/X_{2}leq amid X_{2}=xright} =Pleft{ X_{1}leq axright} $



            So if function $f$ is defined by $fleft(xright)=begin{cases}
            1 & text{if }axgeq1\
            0 & text{if }axleq0\
            ax & text{otherwise}end{cases}$



            Then:



            $Pleft{ X_{1}/X_{2}leq amid X_{2}right} =fleft(X_{2}right)$
            and $Pleft{ X_{1}/X_{2}leq aright} =mathbb{E}fleft(X_{2}right)=int_{0}^{1}fleft(xright)dx$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '13 at 14:50









            drhabdrhab

            101k545136




            101k545136












            • $begingroup$
              thank you, but why is the answer different from the above?
              $endgroup$
              – dark blue
              Dec 10 '13 at 18:53










            • $begingroup$
              It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
              $endgroup$
              – drhab
              Dec 11 '13 at 10:14




















            • $begingroup$
              thank you, but why is the answer different from the above?
              $endgroup$
              – dark blue
              Dec 10 '13 at 18:53










            • $begingroup$
              It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
              $endgroup$
              – drhab
              Dec 11 '13 at 10:14


















            $begingroup$
            thank you, but why is the answer different from the above?
            $endgroup$
            – dark blue
            Dec 10 '13 at 18:53




            $begingroup$
            thank you, but why is the answer different from the above?
            $endgroup$
            – dark blue
            Dec 10 '13 at 18:53












            $begingroup$
            It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
            $endgroup$
            – drhab
            Dec 11 '13 at 10:14






            $begingroup$
            It is not different. Under the extra conditions $a>0$ the function $f$ defined in my answer coincides on $left(0,1right)$ with $minleft(ax,1right)$ mentioned in the answer of Macavity. My answer is a bit more general (it leaves open the possibility that $a<0$) and secondly it leaves a bit more to you (I didn't work out the integration that far as Macavity did). Macavity is right in excluding aleq0 though. I am just a bit lazier.
            $endgroup$
            – drhab
            Dec 11 '13 at 10:14




















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