confused about selection of limits of joint distribution.












1












$begingroup$


I am given a joint distribution of two random variables, X and Y, which is as follows:



$$ f(x,y)= 8xy \ 0≤y≤x≤1 $$



Now I need to find :



$$ P[Xleq 1/2] $$



Now I now I need to find marginal pdf first but I am confused between the limits. Will the limits be from $0$ to $1/2$, since we need to substitute the limits of y, if not, then what is the trick behind the selection of limits if we are given of this type?






probability probability-distributions







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    1












    $begingroup$


    I am given a joint distribution of two random variables, X and Y, which is as follows:



    $$ f(x,y)= 8xy \ 0≤y≤x≤1 $$



    Now I need to find :



    $$ P[Xleq 1/2] $$



    Now I now I need to find marginal pdf first but I am confused between the limits. Will the limits be from $0$ to $1/2$, since we need to substitute the limits of y, if not, then what is the trick behind the selection of limits if we are given of this type?






    probability probability-distributions







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am given a joint distribution of two random variables, X and Y, which is as follows:



      $$ f(x,y)= 8xy \ 0≤y≤x≤1 $$



      Now I need to find :



      $$ P[Xleq 1/2] $$



      Now I now I need to find marginal pdf first but I am confused between the limits. Will the limits be from $0$ to $1/2$, since we need to substitute the limits of y, if not, then what is the trick behind the selection of limits if we are given of this type?






      probability probability-distributions







      share|cite|improve this question









      $endgroup$




      I am given a joint distribution of two random variables, X and Y, which is as follows:



      $$ f(x,y)= 8xy \ 0≤y≤x≤1 $$



      Now I need to find :



      $$ P[Xleq 1/2] $$



      Now I now I need to find marginal pdf first but I am confused between the limits. Will the limits be from $0$ to $1/2$, since we need to substitute the limits of y, if not, then what is the trick behind the selection of limits if we are given of this type?






      probability probability-distributions




      probability probability-distributions






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      share|cite|improve this question











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      asked Jan 11 at 15:31









      Ahmad QayyumAhmad Qayyum

      677




      677






















          1 Answer
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          $begingroup$

          That the region common to the support ${(x,y):0le yle xle 1}$ of the joint distribution and the restriction $x<1/2$ is ${(x,y):0le yle xle 1/2}$ should be clear if you draw a picture.



          Then it follows that



          begin{align}
          Pleft(Xle frac{1}{2}right)&=E,[mathbf1_{Xle1/2}]
          \&=iint mathbf1_{xle1/2},f(x,y),mathrm{d}x,mathrm{d}y
          \&=iint mathbf1_{xle 1/2},8xy,mathbf1_{0le yle xle 1},mathrm{d}x,mathrm{d}y
          \&=8iint xy,mathbf1_{0le yle xle 1/2},mathrm{d}x,mathrm{d}y
          end{align}



          You are supposed to write the double integral as an iterated integral using Fubini's theorem, and depending on the order of integration, you have



          $$iint xy,mathbf1_{0le yle xle 1/2},mathrm{d}x,mathrm{d}y=int_0^{1/2} yleft(int_y^{1/2} x,mathrm{d}xright)mathrm{d}y=int_0^{1/2} xleft(int_0^x y,mathrm{d}yright)mathrm{d}x$$



          Since the ranges of $x$ and $y$ are dependent, the inner integral is a function of the variable with respect to which you are evaluating the outer integral, while the outer integral runs free of the other variable, giving you a real value as the final answer.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            then the limits will be $0$ to $1/2$ for outer integral and $y$ to $1/2$ for inner?
            $endgroup$
            – Ahmad Qayyum
            Jan 11 at 17:09










          • $begingroup$
            @AhmadQayyum Yes.
            $endgroup$
            – StubbornAtom
            Jan 11 at 18:01











          Your Answer





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          1 Answer
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          active

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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          That the region common to the support ${(x,y):0le yle xle 1}$ of the joint distribution and the restriction $x<1/2$ is ${(x,y):0le yle xle 1/2}$ should be clear if you draw a picture.



          Then it follows that



          begin{align}
          Pleft(Xle frac{1}{2}right)&=E,[mathbf1_{Xle1/2}]
          \&=iint mathbf1_{xle1/2},f(x,y),mathrm{d}x,mathrm{d}y
          \&=iint mathbf1_{xle 1/2},8xy,mathbf1_{0le yle xle 1},mathrm{d}x,mathrm{d}y
          \&=8iint xy,mathbf1_{0le yle xle 1/2},mathrm{d}x,mathrm{d}y
          end{align}



          You are supposed to write the double integral as an iterated integral using Fubini's theorem, and depending on the order of integration, you have



          $$iint xy,mathbf1_{0le yle xle 1/2},mathrm{d}x,mathrm{d}y=int_0^{1/2} yleft(int_y^{1/2} x,mathrm{d}xright)mathrm{d}y=int_0^{1/2} xleft(int_0^x y,mathrm{d}yright)mathrm{d}x$$



          Since the ranges of $x$ and $y$ are dependent, the inner integral is a function of the variable with respect to which you are evaluating the outer integral, while the outer integral runs free of the other variable, giving you a real value as the final answer.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            then the limits will be $0$ to $1/2$ for outer integral and $y$ to $1/2$ for inner?
            $endgroup$
            – Ahmad Qayyum
            Jan 11 at 17:09










          • $begingroup$
            @AhmadQayyum Yes.
            $endgroup$
            – StubbornAtom
            Jan 11 at 18:01
















          1












          $begingroup$

          That the region common to the support ${(x,y):0le yle xle 1}$ of the joint distribution and the restriction $x<1/2$ is ${(x,y):0le yle xle 1/2}$ should be clear if you draw a picture.



          Then it follows that



          begin{align}
          Pleft(Xle frac{1}{2}right)&=E,[mathbf1_{Xle1/2}]
          \&=iint mathbf1_{xle1/2},f(x,y),mathrm{d}x,mathrm{d}y
          \&=iint mathbf1_{xle 1/2},8xy,mathbf1_{0le yle xle 1},mathrm{d}x,mathrm{d}y
          \&=8iint xy,mathbf1_{0le yle xle 1/2},mathrm{d}x,mathrm{d}y
          end{align}



          You are supposed to write the double integral as an iterated integral using Fubini's theorem, and depending on the order of integration, you have



          $$iint xy,mathbf1_{0le yle xle 1/2},mathrm{d}x,mathrm{d}y=int_0^{1/2} yleft(int_y^{1/2} x,mathrm{d}xright)mathrm{d}y=int_0^{1/2} xleft(int_0^x y,mathrm{d}yright)mathrm{d}x$$



          Since the ranges of $x$ and $y$ are dependent, the inner integral is a function of the variable with respect to which you are evaluating the outer integral, while the outer integral runs free of the other variable, giving you a real value as the final answer.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            then the limits will be $0$ to $1/2$ for outer integral and $y$ to $1/2$ for inner?
            $endgroup$
            – Ahmad Qayyum
            Jan 11 at 17:09










          • $begingroup$
            @AhmadQayyum Yes.
            $endgroup$
            – StubbornAtom
            Jan 11 at 18:01














          1












          1








          1





          $begingroup$

          That the region common to the support ${(x,y):0le yle xle 1}$ of the joint distribution and the restriction $x<1/2$ is ${(x,y):0le yle xle 1/2}$ should be clear if you draw a picture.



          Then it follows that



          begin{align}
          Pleft(Xle frac{1}{2}right)&=E,[mathbf1_{Xle1/2}]
          \&=iint mathbf1_{xle1/2},f(x,y),mathrm{d}x,mathrm{d}y
          \&=iint mathbf1_{xle 1/2},8xy,mathbf1_{0le yle xle 1},mathrm{d}x,mathrm{d}y
          \&=8iint xy,mathbf1_{0le yle xle 1/2},mathrm{d}x,mathrm{d}y
          end{align}



          You are supposed to write the double integral as an iterated integral using Fubini's theorem, and depending on the order of integration, you have



          $$iint xy,mathbf1_{0le yle xle 1/2},mathrm{d}x,mathrm{d}y=int_0^{1/2} yleft(int_y^{1/2} x,mathrm{d}xright)mathrm{d}y=int_0^{1/2} xleft(int_0^x y,mathrm{d}yright)mathrm{d}x$$



          Since the ranges of $x$ and $y$ are dependent, the inner integral is a function of the variable with respect to which you are evaluating the outer integral, while the outer integral runs free of the other variable, giving you a real value as the final answer.






          share|cite|improve this answer











          $endgroup$



          That the region common to the support ${(x,y):0le yle xle 1}$ of the joint distribution and the restriction $x<1/2$ is ${(x,y):0le yle xle 1/2}$ should be clear if you draw a picture.



          Then it follows that



          begin{align}
          Pleft(Xle frac{1}{2}right)&=E,[mathbf1_{Xle1/2}]
          \&=iint mathbf1_{xle1/2},f(x,y),mathrm{d}x,mathrm{d}y
          \&=iint mathbf1_{xle 1/2},8xy,mathbf1_{0le yle xle 1},mathrm{d}x,mathrm{d}y
          \&=8iint xy,mathbf1_{0le yle xle 1/2},mathrm{d}x,mathrm{d}y
          end{align}



          You are supposed to write the double integral as an iterated integral using Fubini's theorem, and depending on the order of integration, you have



          $$iint xy,mathbf1_{0le yle xle 1/2},mathrm{d}x,mathrm{d}y=int_0^{1/2} yleft(int_y^{1/2} x,mathrm{d}xright)mathrm{d}y=int_0^{1/2} xleft(int_0^x y,mathrm{d}yright)mathrm{d}x$$



          Since the ranges of $x$ and $y$ are dependent, the inner integral is a function of the variable with respect to which you are evaluating the outer integral, while the outer integral runs free of the other variable, giving you a real value as the final answer.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 11 at 18:17

























          answered Jan 11 at 16:43









          StubbornAtomStubbornAtom

          5,63411138




          5,63411138












          • $begingroup$
            then the limits will be $0$ to $1/2$ for outer integral and $y$ to $1/2$ for inner?
            $endgroup$
            – Ahmad Qayyum
            Jan 11 at 17:09










          • $begingroup$
            @AhmadQayyum Yes.
            $endgroup$
            – StubbornAtom
            Jan 11 at 18:01


















          • $begingroup$
            then the limits will be $0$ to $1/2$ for outer integral and $y$ to $1/2$ for inner?
            $endgroup$
            – Ahmad Qayyum
            Jan 11 at 17:09










          • $begingroup$
            @AhmadQayyum Yes.
            $endgroup$
            – StubbornAtom
            Jan 11 at 18:01
















          $begingroup$
          then the limits will be $0$ to $1/2$ for outer integral and $y$ to $1/2$ for inner?
          $endgroup$
          – Ahmad Qayyum
          Jan 11 at 17:09




          $begingroup$
          then the limits will be $0$ to $1/2$ for outer integral and $y$ to $1/2$ for inner?
          $endgroup$
          – Ahmad Qayyum
          Jan 11 at 17:09












          $begingroup$
          @AhmadQayyum Yes.
          $endgroup$
          – StubbornAtom
          Jan 11 at 18:01




          $begingroup$
          @AhmadQayyum Yes.
          $endgroup$
          – StubbornAtom
          Jan 11 at 18:01


















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