fint joint and marginal distributions of two uniformy distributed variables over a specified region












0












$begingroup$


This exercise (partially) comes from Rice 3.9




Suppose that (X, Y ) is uniformly distributed over the region defined
by $0 ≤ y ≤ 1 − x^2$ and $−1 ≤ x ≤ 1$.



Find the marginal densities of X and Y.



Find the joint density.



Find the two conditional densities.




$X sim U[-1,1]$ and $Y∼U[0, 1-x^2]$



then $F_x(x)=x+1$ for $−1 ≤ x ≤ 1$ and $F_Y(y)=frac{y}{1-x^2}$ for $0 ≤ y ≤ 1 − x^2$



I am not sure if this is correct? I am confused with how to find the joint distribution also.






probability probability-theory statistics probability-distributions uniform-distribution







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    This exercise (partially) comes from Rice 3.9




    Suppose that (X, Y ) is uniformly distributed over the region defined
    by $0 ≤ y ≤ 1 − x^2$ and $−1 ≤ x ≤ 1$.



    Find the marginal densities of X and Y.



    Find the joint density.



    Find the two conditional densities.




    $X sim U[-1,1]$ and $Y∼U[0, 1-x^2]$



    then $F_x(x)=x+1$ for $−1 ≤ x ≤ 1$ and $F_Y(y)=frac{y}{1-x^2}$ for $0 ≤ y ≤ 1 − x^2$



    I am not sure if this is correct? I am confused with how to find the joint distribution also.






    probability probability-theory statistics probability-distributions uniform-distribution







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      This exercise (partially) comes from Rice 3.9




      Suppose that (X, Y ) is uniformly distributed over the region defined
      by $0 ≤ y ≤ 1 − x^2$ and $−1 ≤ x ≤ 1$.



      Find the marginal densities of X and Y.



      Find the joint density.



      Find the two conditional densities.




      $X sim U[-1,1]$ and $Y∼U[0, 1-x^2]$



      then $F_x(x)=x+1$ for $−1 ≤ x ≤ 1$ and $F_Y(y)=frac{y}{1-x^2}$ for $0 ≤ y ≤ 1 − x^2$



      I am not sure if this is correct? I am confused with how to find the joint distribution also.






      probability probability-theory statistics probability-distributions uniform-distribution







      share|cite|improve this question









      $endgroup$




      This exercise (partially) comes from Rice 3.9




      Suppose that (X, Y ) is uniformly distributed over the region defined
      by $0 ≤ y ≤ 1 − x^2$ and $−1 ≤ x ≤ 1$.



      Find the marginal densities of X and Y.



      Find the joint density.



      Find the two conditional densities.




      $X sim U[-1,1]$ and $Y∼U[0, 1-x^2]$



      then $F_x(x)=x+1$ for $−1 ≤ x ≤ 1$ and $F_Y(y)=frac{y}{1-x^2}$ for $0 ≤ y ≤ 1 − x^2$



      I am not sure if this is correct? I am confused with how to find the joint distribution also.






      probability probability-theory statistics probability-distributions uniform-distribution




      probability probability-theory statistics probability-distributions uniform-distribution






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 11 '18 at 12:45









      user1607user1607

      1718




      1718






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          $X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=frac12(x+1)$ on that interval. But this would not be the marginal density



          Hints for the question:




          • Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k, I[-1 le x le 1, 0 le y le 1-x^2$. You can find the value of $k$ from $intlimits_{x=-infty}^inftyintlimits_{y=-infty}^{infty} f_{X,Y}(x,y) , dy , dx = intlimits_{x=-1}^1intlimits_{y=0}^{1-x^2} k , dy , dx = 1$


          • You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density


          • For the marginal density for $X$ you have $f_X(x)=intlimits_{y=0}^{1-x^2} k , dy$ using the $k$ you found earlier


          • For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 le x le 1, 0 le y le 1-x^2$ as $0 le y le 1, -sqrt{1-y} le x le sqrt{1-y}$ and you then want $f_Y(y)=intlimits_{x=-sqrt{1-y}}^{sqrt{1-y}} k , dx$ again using the $k$ you found earlier







          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
            $endgroup$
            – user1607
            Dec 11 '18 at 15:02












          • $begingroup$
            Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
            $endgroup$
            – Henry
            Dec 11 '18 at 15:31












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          1 Answer
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          1 Answer
          1






          active

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          active

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          active

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          2












          $begingroup$

          $X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=frac12(x+1)$ on that interval. But this would not be the marginal density



          Hints for the question:




          • Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k, I[-1 le x le 1, 0 le y le 1-x^2$. You can find the value of $k$ from $intlimits_{x=-infty}^inftyintlimits_{y=-infty}^{infty} f_{X,Y}(x,y) , dy , dx = intlimits_{x=-1}^1intlimits_{y=0}^{1-x^2} k , dy , dx = 1$


          • You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density


          • For the marginal density for $X$ you have $f_X(x)=intlimits_{y=0}^{1-x^2} k , dy$ using the $k$ you found earlier


          • For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 le x le 1, 0 le y le 1-x^2$ as $0 le y le 1, -sqrt{1-y} le x le sqrt{1-y}$ and you then want $f_Y(y)=intlimits_{x=-sqrt{1-y}}^{sqrt{1-y}} k , dx$ again using the $k$ you found earlier







          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
            $endgroup$
            – user1607
            Dec 11 '18 at 15:02












          • $begingroup$
            Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
            $endgroup$
            – Henry
            Dec 11 '18 at 15:31
















          2












          $begingroup$

          $X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=frac12(x+1)$ on that interval. But this would not be the marginal density



          Hints for the question:




          • Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k, I[-1 le x le 1, 0 le y le 1-x^2$. You can find the value of $k$ from $intlimits_{x=-infty}^inftyintlimits_{y=-infty}^{infty} f_{X,Y}(x,y) , dy , dx = intlimits_{x=-1}^1intlimits_{y=0}^{1-x^2} k , dy , dx = 1$


          • You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density


          • For the marginal density for $X$ you have $f_X(x)=intlimits_{y=0}^{1-x^2} k , dy$ using the $k$ you found earlier


          • For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 le x le 1, 0 le y le 1-x^2$ as $0 le y le 1, -sqrt{1-y} le x le sqrt{1-y}$ and you then want $f_Y(y)=intlimits_{x=-sqrt{1-y}}^{sqrt{1-y}} k , dx$ again using the $k$ you found earlier







          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
            $endgroup$
            – user1607
            Dec 11 '18 at 15:02












          • $begingroup$
            Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
            $endgroup$
            – Henry
            Dec 11 '18 at 15:31














          2












          2








          2





          $begingroup$

          $X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=frac12(x+1)$ on that interval. But this would not be the marginal density



          Hints for the question:




          • Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k, I[-1 le x le 1, 0 le y le 1-x^2$. You can find the value of $k$ from $intlimits_{x=-infty}^inftyintlimits_{y=-infty}^{infty} f_{X,Y}(x,y) , dy , dx = intlimits_{x=-1}^1intlimits_{y=0}^{1-x^2} k , dy , dx = 1$


          • You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density


          • For the marginal density for $X$ you have $f_X(x)=intlimits_{y=0}^{1-x^2} k , dy$ using the $k$ you found earlier


          • For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 le x le 1, 0 le y le 1-x^2$ as $0 le y le 1, -sqrt{1-y} le x le sqrt{1-y}$ and you then want $f_Y(y)=intlimits_{x=-sqrt{1-y}}^{sqrt{1-y}} k , dx$ again using the $k$ you found earlier







          share|cite|improve this answer











          $endgroup$



          $X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=frac12(x+1)$ on that interval. But this would not be the marginal density



          Hints for the question:




          • Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k, I[-1 le x le 1, 0 le y le 1-x^2$. You can find the value of $k$ from $intlimits_{x=-infty}^inftyintlimits_{y=-infty}^{infty} f_{X,Y}(x,y) , dy , dx = intlimits_{x=-1}^1intlimits_{y=0}^{1-x^2} k , dy , dx = 1$


          • You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density


          • For the marginal density for $X$ you have $f_X(x)=intlimits_{y=0}^{1-x^2} k , dy$ using the $k$ you found earlier


          • For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 le x le 1, 0 le y le 1-x^2$ as $0 le y le 1, -sqrt{1-y} le x le sqrt{1-y}$ and you then want $f_Y(y)=intlimits_{x=-sqrt{1-y}}^{sqrt{1-y}} k , dx$ again using the $k$ you found earlier








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 23:57

























          answered Dec 11 '18 at 14:32









          HenryHenry

          101k482169




          101k482169








          • 1




            $begingroup$
            thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
            $endgroup$
            – user1607
            Dec 11 '18 at 15:02












          • $begingroup$
            Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
            $endgroup$
            – Henry
            Dec 11 '18 at 15:31














          • 1




            $begingroup$
            thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
            $endgroup$
            – user1607
            Dec 11 '18 at 15:02












          • $begingroup$
            Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
            $endgroup$
            – Henry
            Dec 11 '18 at 15:31








          1




          1




          $begingroup$
          thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
          $endgroup$
          – user1607
          Dec 11 '18 at 15:02






          $begingroup$
          thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
          $endgroup$
          – user1607
          Dec 11 '18 at 15:02














          $begingroup$
          Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
          $endgroup$
          – Henry
          Dec 11 '18 at 15:31




          $begingroup$
          Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
          $endgroup$
          – Henry
          Dec 11 '18 at 15:31


















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