Question on probability of two random variables for example $p(X < a, X < Y-b)$












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Suppose we have two random variables $X$ and $Y$, that are independent. Suppose they have the associated distribution functions $F(x)$ and $G(y)$. Now, assume that $X$ and $Y$ are uniformly distributed. More specifically, let $X$ be uniformly distributed in $[-phi,phi]$ $Y$ be uniformly distributed in interval $[0,bar{y}]$.



Given some constants, $a,b$, how would we compute this probability?



$p(X < a, X < Y-b)$



First, I can see that this is the same as $p(X leq min{a, Y-b})$



So we would have:



$int_x int_y mathbb{1}[x leq min{a, y-b}] dG(y) dF(x)$



If we assume that $bar{y}$ is sufficiently large, how would we compute this integral?






probability integration probability-theory







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    Suppose we have two random variables $X$ and $Y$, that are independent. Suppose they have the associated distribution functions $F(x)$ and $G(y)$. Now, assume that $X$ and $Y$ are uniformly distributed. More specifically, let $X$ be uniformly distributed in $[-phi,phi]$ $Y$ be uniformly distributed in interval $[0,bar{y}]$.



    Given some constants, $a,b$, how would we compute this probability?



    $p(X < a, X < Y-b)$



    First, I can see that this is the same as $p(X leq min{a, Y-b})$



    So we would have:



    $int_x int_y mathbb{1}[x leq min{a, y-b}] dG(y) dF(x)$



    If we assume that $bar{y}$ is sufficiently large, how would we compute this integral?






    probability integration probability-theory







    share|cite|improve this question

























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      0







      Suppose we have two random variables $X$ and $Y$, that are independent. Suppose they have the associated distribution functions $F(x)$ and $G(y)$. Now, assume that $X$ and $Y$ are uniformly distributed. More specifically, let $X$ be uniformly distributed in $[-phi,phi]$ $Y$ be uniformly distributed in interval $[0,bar{y}]$.



      Given some constants, $a,b$, how would we compute this probability?



      $p(X < a, X < Y-b)$



      First, I can see that this is the same as $p(X leq min{a, Y-b})$



      So we would have:



      $int_x int_y mathbb{1}[x leq min{a, y-b}] dG(y) dF(x)$



      If we assume that $bar{y}$ is sufficiently large, how would we compute this integral?






      probability integration probability-theory







      share|cite|improve this question













      Suppose we have two random variables $X$ and $Y$, that are independent. Suppose they have the associated distribution functions $F(x)$ and $G(y)$. Now, assume that $X$ and $Y$ are uniformly distributed. More specifically, let $X$ be uniformly distributed in $[-phi,phi]$ $Y$ be uniformly distributed in interval $[0,bar{y}]$.



      Given some constants, $a,b$, how would we compute this probability?



      $p(X < a, X < Y-b)$



      First, I can see that this is the same as $p(X leq min{a, Y-b})$



      So we would have:



      $int_x int_y mathbb{1}[x leq min{a, y-b}] dG(y) dF(x)$



      If we assume that $bar{y}$ is sufficiently large, how would we compute this integral?






      probability integration probability-theory




      probability integration probability-theory






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 20 at 10:22









      Steve

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          Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.



          Then:



          $$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
          & =intint[x<min(a,y-b)]dF(x)dG(y)\
          & =int F(min(a,y-b))dG(y)\
          & =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
          & =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
          end{aligned}
          $$



          In second equality independence is used.



          In third equality it is used that $F$ is a continous CDF.




          Apply this to your mentioned special case.






          share|cite|improve this answer





















          • Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
            – Steve
            Nov 21 at 1:23












          • Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
            – drhab
            Nov 21 at 7:18











          Your Answer





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          Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.



          Then:



          $$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
          & =intint[x<min(a,y-b)]dF(x)dG(y)\
          & =int F(min(a,y-b))dG(y)\
          & =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
          & =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
          end{aligned}
          $$



          In second equality independence is used.



          In third equality it is used that $F$ is a continous CDF.




          Apply this to your mentioned special case.






          share|cite|improve this answer





















          • Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
            – Steve
            Nov 21 at 1:23












          • Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
            – drhab
            Nov 21 at 7:18
















          0














          Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.



          Then:



          $$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
          & =intint[x<min(a,y-b)]dF(x)dG(y)\
          & =int F(min(a,y-b))dG(y)\
          & =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
          & =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
          end{aligned}
          $$



          In second equality independence is used.



          In third equality it is used that $F$ is a continous CDF.




          Apply this to your mentioned special case.






          share|cite|improve this answer





















          • Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
            – Steve
            Nov 21 at 1:23












          • Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
            – drhab
            Nov 21 at 7:18














          0












          0








          0






          Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.



          Then:



          $$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
          & =intint[x<min(a,y-b)]dF(x)dG(y)\
          & =int F(min(a,y-b))dG(y)\
          & =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
          & =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
          end{aligned}
          $$



          In second equality independence is used.



          In third equality it is used that $F$ is a continous CDF.




          Apply this to your mentioned special case.






          share|cite|improve this answer












          Let $[x<min(a,y-b)]$ denote the function $mathbb R^2tomathbb R$ that takes value $1$ if $x<min(a,y-b)$ and takes value $0$ otherwise.



          Then:



          $$begin{aligned}P(X<a,X<Y-b) & =mathbb{E}[X<min(a,Y-b)]\
          & =intint[x<min(a,y-b)]dF(x)dG(y)\
          & =int F(min(a,y-b))dG(y)\
          & =int_{-infty}^{a+b}F(a)dG(y)+int_{a+b}^{infty}F(y-b)dG(y)\
          & =Fleft(aright)Gleft(a+bright)+int_{a+b}^{infty}F(y-b)dG(y)
          end{aligned}
          $$



          In second equality independence is used.



          In third equality it is used that $F$ is a continous CDF.




          Apply this to your mentioned special case.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 11:11









          drhab

          97.6k544128




          97.6k544128












          • Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
            – Steve
            Nov 21 at 1:23












          • Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
            – drhab
            Nov 21 at 7:18


















          • Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
            – Steve
            Nov 21 at 1:23












          • Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
            – drhab
            Nov 21 at 7:18
















          Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
          – Steve
          Nov 21 at 1:23






          Hi drhab, thank you! Can I ask why the upper bound is $a+b$?
          – Steve
          Nov 21 at 1:23














          Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
          – drhab
          Nov 21 at 7:18




          Two cases must be discerned: $min (a,y-b)=a$ and $min(a,y-b)=y-b$. Equivalently the cases $yleq a+b$ and $y>a+b$.
          – drhab
          Nov 21 at 7:18


















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