Integration of PDF's to find valid constant











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I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:






Let $X$ and $Y$ have joint PDF



$$f_{X, Y}(x, y) = cxy$$
for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.






It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.






My approach



Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,



begin{align}
int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
& = int_{y=x}^{y=1}frac{c}{2}y^3dy \
& = frac{c(1-x^4)}{8}
end{align}



Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.






Correct solution



begin{align}
int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
& = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
& = frac{c}{4} \
end{align}



This gives us $c = 4$.






I have two specific questions:




  1. Why did the correct solution switch the order of integration?


  2. The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?






Any feedback is appreciated. Thank you.






EDIT



The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.










share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:






    Let $X$ and $Y$ have joint PDF



    $$f_{X, Y}(x, y) = cxy$$
    for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.






    It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.






    My approach



    Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,



    begin{align}
    int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
    & = int_{y=x}^{y=1}frac{c}{2}y^3dy \
    & = frac{c(1-x^4)}{8}
    end{align}



    Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.






    Correct solution



    begin{align}
    int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
    & = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
    & = frac{c}{4} \
    end{align}



    This gives us $c = 4$.






    I have two specific questions:




    1. Why did the correct solution switch the order of integration?


    2. The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?






    Any feedback is appreciated. Thank you.






    EDIT



    The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:






      Let $X$ and $Y$ have joint PDF



      $$f_{X, Y}(x, y) = cxy$$
      for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.






      It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.






      My approach



      Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,



      begin{align}
      int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
      & = int_{y=x}^{y=1}frac{c}{2}y^3dy \
      & = frac{c(1-x^4)}{8}
      end{align}



      Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.






      Correct solution



      begin{align}
      int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
      & = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
      & = frac{c}{4} \
      end{align}



      This gives us $c = 4$.






      I have two specific questions:




      1. Why did the correct solution switch the order of integration?


      2. The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?






      Any feedback is appreciated. Thank you.






      EDIT



      The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.










      share|cite|improve this question















      I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:






      Let $X$ and $Y$ have joint PDF



      $$f_{X, Y}(x, y) = cxy$$
      for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.






      It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.






      My approach



      Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,



      begin{align}
      int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
      & = int_{y=x}^{y=1}frac{c}{2}y^3dy \
      & = frac{c(1-x^4)}{8}
      end{align}



      Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.






      Correct solution



      begin{align}
      int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
      & = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
      & = frac{c}{4} \
      end{align}



      This gives us $c = 4$.






      I have two specific questions:




      1. Why did the correct solution switch the order of integration?


      2. The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?






      Any feedback is appreciated. Thank you.






      EDIT



      The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.







      probability integration density-function






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      edited yesterday

























      asked 2 days ago









      Sean

      21510




      21510






















          1 Answer
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          up vote
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          down vote



          accepted










          Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
          In your solution you are fixing x, but what even is the value of x you are fixing it to?






          share|cite|improve this answer





















          • That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
            – Sean
            2 days ago










          • Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
            – Sean
            2 days ago






          • 1




            You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
            – Sorin Tirc
            2 days ago








          • 1




            What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
            – Sorin Tirc
            2 days ago






          • 1




            Yes, absolutely ;)
            – Sorin Tirc
            2 days ago











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
          In your solution you are fixing x, but what even is the value of x you are fixing it to?






          share|cite|improve this answer





















          • That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
            – Sean
            2 days ago










          • Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
            – Sean
            2 days ago






          • 1




            You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
            – Sorin Tirc
            2 days ago








          • 1




            What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
            – Sorin Tirc
            2 days ago






          • 1




            Yes, absolutely ;)
            – Sorin Tirc
            2 days ago















          up vote
          1
          down vote



          accepted










          Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
          In your solution you are fixing x, but what even is the value of x you are fixing it to?






          share|cite|improve this answer





















          • That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
            – Sean
            2 days ago










          • Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
            – Sean
            2 days ago






          • 1




            You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
            – Sorin Tirc
            2 days ago








          • 1




            What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
            – Sorin Tirc
            2 days ago






          • 1




            Yes, absolutely ;)
            – Sorin Tirc
            2 days ago













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
          In your solution you are fixing x, but what even is the value of x you are fixing it to?






          share|cite|improve this answer












          Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
          In your solution you are fixing x, but what even is the value of x you are fixing it to?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Sorin Tirc

          5039




          5039












          • That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
            – Sean
            2 days ago










          • Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
            – Sean
            2 days ago






          • 1




            You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
            – Sorin Tirc
            2 days ago








          • 1




            What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
            – Sorin Tirc
            2 days ago






          • 1




            Yes, absolutely ;)
            – Sorin Tirc
            2 days ago


















          • That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
            – Sean
            2 days ago










          • Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
            – Sean
            2 days ago






          • 1




            You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
            – Sorin Tirc
            2 days ago








          • 1




            What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
            – Sorin Tirc
            2 days ago






          • 1




            Yes, absolutely ;)
            – Sorin Tirc
            2 days ago
















          That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
          – Sean
          2 days ago




          That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
          – Sean
          2 days ago












          Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
          – Sean
          2 days ago




          Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
          – Sean
          2 days ago




          1




          1




          You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
          – Sorin Tirc
          2 days ago






          You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
          – Sorin Tirc
          2 days ago






          1




          1




          What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
          – Sorin Tirc
          2 days ago




          What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
          – Sorin Tirc
          2 days ago




          1




          1




          Yes, absolutely ;)
          – Sorin Tirc
          2 days ago




          Yes, absolutely ;)
          – Sorin Tirc
          2 days ago


















           

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