Transformation method of two random variables












0












$begingroup$


My problem:



Let $Y_1$ and $Y_2$ be two random variables and let $f(y_1,y_2)=e^{-y_2}$ from $0leq y_1leq y_2 < infty$ and $0$ otherwise be the joint density function.



(a) Calculate $E(Y_1|Y_2 = y_2)$ and $E(Y_2 |Y_1 = y_1)$.



(b) Calculate the density functions for $E(Y_1|Y_2 )$ and $E(Y_2 |Y_1 )$.



My solution in (a) were easy. I got $E(Y_1|Y_2 = y_2)=y_2/2$ and $E(Y_2 |Y_1 = y_1)=y_1+1$. For (b) I believe I should use transformation method, i.e. I define two random variables as
begin{equation} U=Y_2/2 hspace 0.5cm and hspace 0.5cm V=Y_1+1 end{equation}
With the transformation method I obtain for $U$ the following density function
begin{equation} f_U(u)=2e^{-2u} end{equation}



My problem is now to find $V=Y_1+1$ because I don't have $Y_1$ in my joint density function...






probability-theory







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$endgroup$

















    0












    $begingroup$


    My problem:



    Let $Y_1$ and $Y_2$ be two random variables and let $f(y_1,y_2)=e^{-y_2}$ from $0leq y_1leq y_2 < infty$ and $0$ otherwise be the joint density function.



    (a) Calculate $E(Y_1|Y_2 = y_2)$ and $E(Y_2 |Y_1 = y_1)$.



    (b) Calculate the density functions for $E(Y_1|Y_2 )$ and $E(Y_2 |Y_1 )$.



    My solution in (a) were easy. I got $E(Y_1|Y_2 = y_2)=y_2/2$ and $E(Y_2 |Y_1 = y_1)=y_1+1$. For (b) I believe I should use transformation method, i.e. I define two random variables as
    begin{equation} U=Y_2/2 hspace 0.5cm and hspace 0.5cm V=Y_1+1 end{equation}
    With the transformation method I obtain for $U$ the following density function
    begin{equation} f_U(u)=2e^{-2u} end{equation}



    My problem is now to find $V=Y_1+1$ because I don't have $Y_1$ in my joint density function...






    probability-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      My problem:



      Let $Y_1$ and $Y_2$ be two random variables and let $f(y_1,y_2)=e^{-y_2}$ from $0leq y_1leq y_2 < infty$ and $0$ otherwise be the joint density function.



      (a) Calculate $E(Y_1|Y_2 = y_2)$ and $E(Y_2 |Y_1 = y_1)$.



      (b) Calculate the density functions for $E(Y_1|Y_2 )$ and $E(Y_2 |Y_1 )$.



      My solution in (a) were easy. I got $E(Y_1|Y_2 = y_2)=y_2/2$ and $E(Y_2 |Y_1 = y_1)=y_1+1$. For (b) I believe I should use transformation method, i.e. I define two random variables as
      begin{equation} U=Y_2/2 hspace 0.5cm and hspace 0.5cm V=Y_1+1 end{equation}
      With the transformation method I obtain for $U$ the following density function
      begin{equation} f_U(u)=2e^{-2u} end{equation}



      My problem is now to find $V=Y_1+1$ because I don't have $Y_1$ in my joint density function...






      probability-theory







      share|cite|improve this question









      $endgroup$




      My problem:



      Let $Y_1$ and $Y_2$ be two random variables and let $f(y_1,y_2)=e^{-y_2}$ from $0leq y_1leq y_2 < infty$ and $0$ otherwise be the joint density function.



      (a) Calculate $E(Y_1|Y_2 = y_2)$ and $E(Y_2 |Y_1 = y_1)$.



      (b) Calculate the density functions for $E(Y_1|Y_2 )$ and $E(Y_2 |Y_1 )$.



      My solution in (a) were easy. I got $E(Y_1|Y_2 = y_2)=y_2/2$ and $E(Y_2 |Y_1 = y_1)=y_1+1$. For (b) I believe I should use transformation method, i.e. I define two random variables as
      begin{equation} U=Y_2/2 hspace 0.5cm and hspace 0.5cm V=Y_1+1 end{equation}
      With the transformation method I obtain for $U$ the following density function
      begin{equation} f_U(u)=2e^{-2u} end{equation}



      My problem is now to find $V=Y_1+1$ because I don't have $Y_1$ in my joint density function...






      probability-theory




      probability-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 9 at 20:55









      Joey AdamsJoey Adams

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      457






















          1 Answer
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          $begingroup$

          Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=int f(y_1,y_2)dy_2 =int_{y_1}^{infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<infty$; $f_{Y_2}(y_2)=int f(y_1,y_2)dy_1 =int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
            $endgroup$
            – Joey Adams
            Jan 10 at 0:05






          • 1




            $begingroup$
            @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 10 at 0:36










          • $begingroup$
            Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
            $endgroup$
            – Joey Adams
            Jan 10 at 0:39













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          1 Answer
          1






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          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=int f(y_1,y_2)dy_2 =int_{y_1}^{infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<infty$; $f_{Y_2}(y_2)=int f(y_1,y_2)dy_1 =int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
            $endgroup$
            – Joey Adams
            Jan 10 at 0:05






          • 1




            $begingroup$
            @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 10 at 0:36










          • $begingroup$
            Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
            $endgroup$
            – Joey Adams
            Jan 10 at 0:39


















          1












          $begingroup$

          Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=int f(y_1,y_2)dy_2 =int_{y_1}^{infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<infty$; $f_{Y_2}(y_2)=int f(y_1,y_2)dy_1 =int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
            $endgroup$
            – Joey Adams
            Jan 10 at 0:05






          • 1




            $begingroup$
            @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 10 at 0:36










          • $begingroup$
            Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
            $endgroup$
            – Joey Adams
            Jan 10 at 0:39
















          1












          1








          1





          $begingroup$

          Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=int f(y_1,y_2)dy_2 =int_{y_1}^{infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<infty$; $f_{Y_2}(y_2)=int f(y_1,y_2)dy_1 =int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<infty$.






          share|cite|improve this answer











          $endgroup$



          Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=int f(y_1,y_2)dy_2 =int_{y_1}^{infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<infty$; $f_{Y_2}(y_2)=int f(y_1,y_2)dy_1 =int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<infty$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 10 at 0:35

























          answered Jan 9 at 23:48









          Kavi Rama MurthyKavi Rama Murthy

          54.4k32055




          54.4k32055












          • $begingroup$
            So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
            $endgroup$
            – Joey Adams
            Jan 10 at 0:05






          • 1




            $begingroup$
            @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 10 at 0:36










          • $begingroup$
            Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
            $endgroup$
            – Joey Adams
            Jan 10 at 0:39




















          • $begingroup$
            So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
            $endgroup$
            – Joey Adams
            Jan 10 at 0:05






          • 1




            $begingroup$
            @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 10 at 0:36










          • $begingroup$
            Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
            $endgroup$
            – Joey Adams
            Jan 10 at 0:39


















          $begingroup$
          So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
          $endgroup$
          – Joey Adams
          Jan 10 at 0:05




          $begingroup$
          So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
          $endgroup$
          – Joey Adams
          Jan 10 at 0:05




          1




          1




          $begingroup$
          @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
          $endgroup$
          – Kavi Rama Murthy
          Jan 10 at 0:36




          $begingroup$
          @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
          $endgroup$
          – Kavi Rama Murthy
          Jan 10 at 0:36












          $begingroup$
          Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
          $endgroup$
          – Joey Adams
          Jan 10 at 0:39






          $begingroup$
          Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
          $endgroup$
          – Joey Adams
          Jan 10 at 0:39




















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