What is the domain of my marginal PDF function











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Consider the joint pdf of $(X, Y)$ given by



$$f(x, y) = begin{cases}
25x^{4}y^{4} & text{ if } |x| leq y, 0 < y < 1 \
0, & text{ otherwise.}
end{cases} $$



Then,



$$f_{X}(x) = int_{0}^{1} 25x^{4}y^{4 }mathop{dy}$$



$$= 5x^{4}. $$



But what is the restriction to the domain? Is it just $|x| leq y$? Or is it $x in mathbb{R}$? Why? $y$ is no longer in the function. Does it matter? Also to compute $mathbb{E}[X]$, is the following correct:



$$mathbb{E}[X] = int_{-y}^{y} 5x^{4} mathop{dy} = 2y^{5} $$






probability probability-theory probability-distributions marginal-distribution







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  • 1




    $EX$ has to be real number. How can it be a function of $y$?
    – Kavi Rama Murthy
    Nov 28 at 8:40















up vote
0
down vote

favorite












Consider the joint pdf of $(X, Y)$ given by



$$f(x, y) = begin{cases}
25x^{4}y^{4} & text{ if } |x| leq y, 0 < y < 1 \
0, & text{ otherwise.}
end{cases} $$



Then,



$$f_{X}(x) = int_{0}^{1} 25x^{4}y^{4 }mathop{dy}$$



$$= 5x^{4}. $$



But what is the restriction to the domain? Is it just $|x| leq y$? Or is it $x in mathbb{R}$? Why? $y$ is no longer in the function. Does it matter? Also to compute $mathbb{E}[X]$, is the following correct:



$$mathbb{E}[X] = int_{-y}^{y} 5x^{4} mathop{dy} = 2y^{5} $$






probability probability-theory probability-distributions marginal-distribution







share|cite|improve this question


















  • 1




    $EX$ has to be real number. How can it be a function of $y$?
    – Kavi Rama Murthy
    Nov 28 at 8:40













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Consider the joint pdf of $(X, Y)$ given by



$$f(x, y) = begin{cases}
25x^{4}y^{4} & text{ if } |x| leq y, 0 < y < 1 \
0, & text{ otherwise.}
end{cases} $$



Then,



$$f_{X}(x) = int_{0}^{1} 25x^{4}y^{4 }mathop{dy}$$



$$= 5x^{4}. $$



But what is the restriction to the domain? Is it just $|x| leq y$? Or is it $x in mathbb{R}$? Why? $y$ is no longer in the function. Does it matter? Also to compute $mathbb{E}[X]$, is the following correct:



$$mathbb{E}[X] = int_{-y}^{y} 5x^{4} mathop{dy} = 2y^{5} $$






probability probability-theory probability-distributions marginal-distribution







share|cite|improve this question













Consider the joint pdf of $(X, Y)$ given by



$$f(x, y) = begin{cases}
25x^{4}y^{4} & text{ if } |x| leq y, 0 < y < 1 \
0, & text{ otherwise.}
end{cases} $$



Then,



$$f_{X}(x) = int_{0}^{1} 25x^{4}y^{4 }mathop{dy}$$



$$= 5x^{4}. $$



But what is the restriction to the domain? Is it just $|x| leq y$? Or is it $x in mathbb{R}$? Why? $y$ is no longer in the function. Does it matter? Also to compute $mathbb{E}[X]$, is the following correct:



$$mathbb{E}[X] = int_{-y}^{y} 5x^{4} mathop{dy} = 2y^{5} $$






probability probability-theory probability-distributions marginal-distribution




probability probability-theory probability-distributions marginal-distribution






share|cite|improve this question













share|cite|improve this question











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asked Nov 28 at 8:33









joseph

4159




4159








  • 1




    $EX$ has to be real number. How can it be a function of $y$?
    – Kavi Rama Murthy
    Nov 28 at 8:40














  • 1




    $EX$ has to be real number. How can it be a function of $y$?
    – Kavi Rama Murthy
    Nov 28 at 8:40








1




1




$EX$ has to be real number. How can it be a function of $y$?
– Kavi Rama Murthy
Nov 28 at 8:40




$EX$ has to be real number. How can it be a function of $y$?
– Kavi Rama Murthy
Nov 28 at 8:40










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Your calculation of $f_X$ is wrong. The correct value is $int_{|x|}^{1} 25x^{4}y^{4}dy=5x^{4}(1-|x|^{5})$ for $|x| leq 1$. Also $EX=int_{-1}^{1}5x^{5}(1-|x|^{5})dx=0$ because the integral of an odd function from $-1$ to $1$ is $0$.






share|cite|improve this answer





















  • Okay. Then I think I calculated the constant (currently $25$) wrong. Replace $25$ with just a constant $C$. Then, how can I find $C$? Is this right? $int_{0}^{1} int_{-y}^{y} Cx^{4}y^{4} mathop{dx dy} = 1$?
    – joseph
    Nov 28 at 8:40












  • The constant $25$ in $f(x,y)$ is correct.
    – Kavi Rama Murthy
    Nov 28 at 8:44










  • Is my setup to find the constant $C$ correct? Or should the bounds for $x$ be $0$ to $y$?
    – joseph
    Nov 29 at 5:37












  • @joseph Yes, that is correct and you should get $C=25$ if you carry out the integration.
    – Kavi Rama Murthy
    Nov 29 at 5:41


















up vote
1
down vote













Just a way of working that might prevent you from making mistakes in cases like this.




In general let $left[text{conditions on }x,yright]$ be a function
that takes value $1$ if the conditions are satisfied and takes value
$0$ otherwise.



Then $fleft(x,yright)=left[left|xright|leq y,0<y<1right]25x^{4}y^{4}$
so that



$$f_{X}left(xright)=intleft[left|xright|leq y,0<y<1right]25x^{4}y^{4}dy=int_{0}^{1}left[left|xright|leq yright]25x^{4}y^{4}dy$$



Now discern two cases:




  • If $left|xright|geq1$ then $f_{X}left(xright)=0$


  • If $left|xright|<1$ then $f_{X}left(xright)=int_{left|xright|}^{1}25x^{4}y^{4}dy=left[5x^{4}y^{5}right]_{left|xright|}^{1}=5x^{4}left(1-left|xright|^{5}right)$




Use this if you like it. If you think it is too cumbersome then just leave it aside.






share|cite|improve this answer





















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    2 Answers
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    2 Answers
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    up vote
    1
    down vote



    accepted










    Your calculation of $f_X$ is wrong. The correct value is $int_{|x|}^{1} 25x^{4}y^{4}dy=5x^{4}(1-|x|^{5})$ for $|x| leq 1$. Also $EX=int_{-1}^{1}5x^{5}(1-|x|^{5})dx=0$ because the integral of an odd function from $-1$ to $1$ is $0$.






    share|cite|improve this answer





















    • Okay. Then I think I calculated the constant (currently $25$) wrong. Replace $25$ with just a constant $C$. Then, how can I find $C$? Is this right? $int_{0}^{1} int_{-y}^{y} Cx^{4}y^{4} mathop{dx dy} = 1$?
      – joseph
      Nov 28 at 8:40












    • The constant $25$ in $f(x,y)$ is correct.
      – Kavi Rama Murthy
      Nov 28 at 8:44










    • Is my setup to find the constant $C$ correct? Or should the bounds for $x$ be $0$ to $y$?
      – joseph
      Nov 29 at 5:37












    • @joseph Yes, that is correct and you should get $C=25$ if you carry out the integration.
      – Kavi Rama Murthy
      Nov 29 at 5:41















    up vote
    1
    down vote



    accepted










    Your calculation of $f_X$ is wrong. The correct value is $int_{|x|}^{1} 25x^{4}y^{4}dy=5x^{4}(1-|x|^{5})$ for $|x| leq 1$. Also $EX=int_{-1}^{1}5x^{5}(1-|x|^{5})dx=0$ because the integral of an odd function from $-1$ to $1$ is $0$.






    share|cite|improve this answer





















    • Okay. Then I think I calculated the constant (currently $25$) wrong. Replace $25$ with just a constant $C$. Then, how can I find $C$? Is this right? $int_{0}^{1} int_{-y}^{y} Cx^{4}y^{4} mathop{dx dy} = 1$?
      – joseph
      Nov 28 at 8:40












    • The constant $25$ in $f(x,y)$ is correct.
      – Kavi Rama Murthy
      Nov 28 at 8:44










    • Is my setup to find the constant $C$ correct? Or should the bounds for $x$ be $0$ to $y$?
      – joseph
      Nov 29 at 5:37












    • @joseph Yes, that is correct and you should get $C=25$ if you carry out the integration.
      – Kavi Rama Murthy
      Nov 29 at 5:41













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    Your calculation of $f_X$ is wrong. The correct value is $int_{|x|}^{1} 25x^{4}y^{4}dy=5x^{4}(1-|x|^{5})$ for $|x| leq 1$. Also $EX=int_{-1}^{1}5x^{5}(1-|x|^{5})dx=0$ because the integral of an odd function from $-1$ to $1$ is $0$.






    share|cite|improve this answer












    Your calculation of $f_X$ is wrong. The correct value is $int_{|x|}^{1} 25x^{4}y^{4}dy=5x^{4}(1-|x|^{5})$ for $|x| leq 1$. Also $EX=int_{-1}^{1}5x^{5}(1-|x|^{5})dx=0$ because the integral of an odd function from $-1$ to $1$ is $0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 at 8:36









    Kavi Rama Murthy

    47.1k31854




    47.1k31854












    • Okay. Then I think I calculated the constant (currently $25$) wrong. Replace $25$ with just a constant $C$. Then, how can I find $C$? Is this right? $int_{0}^{1} int_{-y}^{y} Cx^{4}y^{4} mathop{dx dy} = 1$?
      – joseph
      Nov 28 at 8:40












    • The constant $25$ in $f(x,y)$ is correct.
      – Kavi Rama Murthy
      Nov 28 at 8:44










    • Is my setup to find the constant $C$ correct? Or should the bounds for $x$ be $0$ to $y$?
      – joseph
      Nov 29 at 5:37












    • @joseph Yes, that is correct and you should get $C=25$ if you carry out the integration.
      – Kavi Rama Murthy
      Nov 29 at 5:41


















    • Okay. Then I think I calculated the constant (currently $25$) wrong. Replace $25$ with just a constant $C$. Then, how can I find $C$? Is this right? $int_{0}^{1} int_{-y}^{y} Cx^{4}y^{4} mathop{dx dy} = 1$?
      – joseph
      Nov 28 at 8:40












    • The constant $25$ in $f(x,y)$ is correct.
      – Kavi Rama Murthy
      Nov 28 at 8:44










    • Is my setup to find the constant $C$ correct? Or should the bounds for $x$ be $0$ to $y$?
      – joseph
      Nov 29 at 5:37












    • @joseph Yes, that is correct and you should get $C=25$ if you carry out the integration.
      – Kavi Rama Murthy
      Nov 29 at 5:41
















    Okay. Then I think I calculated the constant (currently $25$) wrong. Replace $25$ with just a constant $C$. Then, how can I find $C$? Is this right? $int_{0}^{1} int_{-y}^{y} Cx^{4}y^{4} mathop{dx dy} = 1$?
    – joseph
    Nov 28 at 8:40






    Okay. Then I think I calculated the constant (currently $25$) wrong. Replace $25$ with just a constant $C$. Then, how can I find $C$? Is this right? $int_{0}^{1} int_{-y}^{y} Cx^{4}y^{4} mathop{dx dy} = 1$?
    – joseph
    Nov 28 at 8:40














    The constant $25$ in $f(x,y)$ is correct.
    – Kavi Rama Murthy
    Nov 28 at 8:44




    The constant $25$ in $f(x,y)$ is correct.
    – Kavi Rama Murthy
    Nov 28 at 8:44












    Is my setup to find the constant $C$ correct? Or should the bounds for $x$ be $0$ to $y$?
    – joseph
    Nov 29 at 5:37






    Is my setup to find the constant $C$ correct? Or should the bounds for $x$ be $0$ to $y$?
    – joseph
    Nov 29 at 5:37














    @joseph Yes, that is correct and you should get $C=25$ if you carry out the integration.
    – Kavi Rama Murthy
    Nov 29 at 5:41




    @joseph Yes, that is correct and you should get $C=25$ if you carry out the integration.
    – Kavi Rama Murthy
    Nov 29 at 5:41










    up vote
    1
    down vote













    Just a way of working that might prevent you from making mistakes in cases like this.




    In general let $left[text{conditions on }x,yright]$ be a function
    that takes value $1$ if the conditions are satisfied and takes value
    $0$ otherwise.



    Then $fleft(x,yright)=left[left|xright|leq y,0<y<1right]25x^{4}y^{4}$
    so that



    $$f_{X}left(xright)=intleft[left|xright|leq y,0<y<1right]25x^{4}y^{4}dy=int_{0}^{1}left[left|xright|leq yright]25x^{4}y^{4}dy$$



    Now discern two cases:




    • If $left|xright|geq1$ then $f_{X}left(xright)=0$


    • If $left|xright|<1$ then $f_{X}left(xright)=int_{left|xright|}^{1}25x^{4}y^{4}dy=left[5x^{4}y^{5}right]_{left|xright|}^{1}=5x^{4}left(1-left|xright|^{5}right)$




    Use this if you like it. If you think it is too cumbersome then just leave it aside.






    share|cite|improve this answer

























      up vote
      1
      down vote













      Just a way of working that might prevent you from making mistakes in cases like this.




      In general let $left[text{conditions on }x,yright]$ be a function
      that takes value $1$ if the conditions are satisfied and takes value
      $0$ otherwise.



      Then $fleft(x,yright)=left[left|xright|leq y,0<y<1right]25x^{4}y^{4}$
      so that



      $$f_{X}left(xright)=intleft[left|xright|leq y,0<y<1right]25x^{4}y^{4}dy=int_{0}^{1}left[left|xright|leq yright]25x^{4}y^{4}dy$$



      Now discern two cases:




      • If $left|xright|geq1$ then $f_{X}left(xright)=0$


      • If $left|xright|<1$ then $f_{X}left(xright)=int_{left|xright|}^{1}25x^{4}y^{4}dy=left[5x^{4}y^{5}right]_{left|xright|}^{1}=5x^{4}left(1-left|xright|^{5}right)$




      Use this if you like it. If you think it is too cumbersome then just leave it aside.






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        Just a way of working that might prevent you from making mistakes in cases like this.




        In general let $left[text{conditions on }x,yright]$ be a function
        that takes value $1$ if the conditions are satisfied and takes value
        $0$ otherwise.



        Then $fleft(x,yright)=left[left|xright|leq y,0<y<1right]25x^{4}y^{4}$
        so that



        $$f_{X}left(xright)=intleft[left|xright|leq y,0<y<1right]25x^{4}y^{4}dy=int_{0}^{1}left[left|xright|leq yright]25x^{4}y^{4}dy$$



        Now discern two cases:




        • If $left|xright|geq1$ then $f_{X}left(xright)=0$


        • If $left|xright|<1$ then $f_{X}left(xright)=int_{left|xright|}^{1}25x^{4}y^{4}dy=left[5x^{4}y^{5}right]_{left|xright|}^{1}=5x^{4}left(1-left|xright|^{5}right)$




        Use this if you like it. If you think it is too cumbersome then just leave it aside.






        share|cite|improve this answer












        Just a way of working that might prevent you from making mistakes in cases like this.




        In general let $left[text{conditions on }x,yright]$ be a function
        that takes value $1$ if the conditions are satisfied and takes value
        $0$ otherwise.



        Then $fleft(x,yright)=left[left|xright|leq y,0<y<1right]25x^{4}y^{4}$
        so that



        $$f_{X}left(xright)=intleft[left|xright|leq y,0<y<1right]25x^{4}y^{4}dy=int_{0}^{1}left[left|xright|leq yright]25x^{4}y^{4}dy$$



        Now discern two cases:




        • If $left|xright|geq1$ then $f_{X}left(xright)=0$


        • If $left|xright|<1$ then $f_{X}left(xright)=int_{left|xright|}^{1}25x^{4}y^{4}dy=left[5x^{4}y^{5}right]_{left|xright|}^{1}=5x^{4}left(1-left|xright|^{5}right)$




        Use this if you like it. If you think it is too cumbersome then just leave it aside.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 9:07









        drhab

        96.1k543126




        96.1k543126






























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