Double integrals - how are the boundaries chosen?











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I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.



After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get



$P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$



Can someone please explain the last equality
$int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.










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    I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.



    After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get



    $P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$



    Can someone please explain the last equality
    $int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.



      After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get



      $P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$



      Can someone please explain the last equality
      $int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.










      share|cite|improve this question













      I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.



      After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get



      $P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$



      Can someone please explain the last equality
      $int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.







      integration convolution density-function






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      asked Nov 19 at 10:15









      Tesla

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          For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).






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          • Got it, thank you very much! And why am I allowed to take the functions out of the integral?
            – Tesla
            Nov 19 at 10:22






          • 1




            When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
            – Kavi Rama Murthy
            Nov 19 at 10:24











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          up vote
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          accepted










          For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).






          share|cite|improve this answer





















          • Got it, thank you very much! And why am I allowed to take the functions out of the integral?
            – Tesla
            Nov 19 at 10:22






          • 1




            When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
            – Kavi Rama Murthy
            Nov 19 at 10:24















          up vote
          1
          down vote



          accepted










          For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).






          share|cite|improve this answer





















          • Got it, thank you very much! And why am I allowed to take the functions out of the integral?
            – Tesla
            Nov 19 at 10:22






          • 1




            When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
            – Kavi Rama Murthy
            Nov 19 at 10:24













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).






          share|cite|improve this answer












          For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 at 10:18









          Kavi Rama Murthy

          46k31854




          46k31854












          • Got it, thank you very much! And why am I allowed to take the functions out of the integral?
            – Tesla
            Nov 19 at 10:22






          • 1




            When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
            – Kavi Rama Murthy
            Nov 19 at 10:24


















          • Got it, thank you very much! And why am I allowed to take the functions out of the integral?
            – Tesla
            Nov 19 at 10:22






          • 1




            When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
            – Kavi Rama Murthy
            Nov 19 at 10:24
















          Got it, thank you very much! And why am I allowed to take the functions out of the integral?
          – Tesla
          Nov 19 at 10:22




          Got it, thank you very much! And why am I allowed to take the functions out of the integral?
          – Tesla
          Nov 19 at 10:22




          1




          1




          When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
          – Kavi Rama Murthy
          Nov 19 at 10:24




          When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
          – Kavi Rama Murthy
          Nov 19 at 10:24


















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