Joint PDF P[X+Y<=0.5]












1












$begingroup$


I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



I need to find the $P(x+yleq 0.5)$.



For the double integration, I have the following bounds:




  • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


  • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



Here is the initial set up:



$int_0^.5 int_0^{.5-x}(x +y) dydx$.



My steps of integration:



1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






probability







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



    Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



    I need to find the $P(x+yleq 0.5)$.



    For the double integration, I have the following bounds:




    • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


    • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



    Here is the initial set up:



    $int_0^.5 int_0^{.5-x}(x +y) dydx$.



    My steps of integration:



    1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



    2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



    3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



    I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






    probability







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



      Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



      I need to find the $P(x+yleq 0.5)$.



      For the double integration, I have the following bounds:




      • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


      • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



      Here is the initial set up:



      $int_0^.5 int_0^{.5-x}(x +y) dydx$.



      My steps of integration:



      1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



      2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



      3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



      I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






      probability







      share|cite|improve this question











      $endgroup$




      I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



      Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



      I need to find the $P(x+yleq 0.5)$.



      For the double integration, I have the following bounds:




      • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


      • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



      Here is the initial set up:



      $int_0^.5 int_0^{.5-x}(x +y) dydx$.



      My steps of integration:



      1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



      2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



      3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



      I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






      probability




      probability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 10:42







      MitterHai

















      asked Dec 5 '18 at 10:19









      MitterHaiMitterHai

      84




      84






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43



















          1












          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43
















          0












          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43














          0












          0








          0





          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$



          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 11:22









          farruhotafarruhota

          20.2k2738




          20.2k2738












          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43


















          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43
















          $begingroup$
          Farruhota, thank you for the awesome feedback. I really appreciate your help.
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:42




          $begingroup$
          Farruhota, thank you for the awesome feedback. I really appreciate your help.
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:42












          $begingroup$
          You are welcome. Good luck.
          $endgroup$
          – farruhota
          Dec 5 '18 at 11:43




          $begingroup$
          You are welcome. Good luck.
          $endgroup$
          – farruhota
          Dec 5 '18 at 11:43











          1












          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40
















          1












          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40














          1












          1








          1





          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$



          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 10:22









          Kavi Rama MurthyKavi Rama Murthy

          57.7k42160




          57.7k42160












          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40


















          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40
















          $begingroup$
          Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
          $endgroup$
          – MitterHai
          Dec 5 '18 at 10:38




          $begingroup$
          Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
          $endgroup$
          – MitterHai
          Dec 5 '18 at 10:38












          $begingroup$
          @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
          $endgroup$
          – littleO
          Dec 5 '18 at 11:22




          $begingroup$
          @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
          $endgroup$
          – littleO
          Dec 5 '18 at 11:22












          $begingroup$
          What a face palm moment. Thanks Pro..fe.. I mean littleO!
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:40




          $begingroup$
          What a face palm moment. Thanks Pro..fe.. I mean littleO!
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:40


















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          Musa

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          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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