How to compute density functions of $E(Y_1|Y_2)$












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$begingroup$


Good afternoon. I've trying to do some exercises. Here is one that I can't continue with.



Here is the question: "Find the density functions of $E(Y_1|Y_2)$". My function that I got is $f(y_1,y_2)=exp(-y_2)$ where $0leq y_1 leq y_2$.



My own idea is that I need to find the $textit{conditional expectation}$ which I did, but then I can't continue from here.






probability-theory







share|cite|improve this question









$endgroup$

















    -1












    $begingroup$


    Good afternoon. I've trying to do some exercises. Here is one that I can't continue with.



    Here is the question: "Find the density functions of $E(Y_1|Y_2)$". My function that I got is $f(y_1,y_2)=exp(-y_2)$ where $0leq y_1 leq y_2$.



    My own idea is that I need to find the $textit{conditional expectation}$ which I did, but then I can't continue from here.






    probability-theory







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1


      0



      $begingroup$


      Good afternoon. I've trying to do some exercises. Here is one that I can't continue with.



      Here is the question: "Find the density functions of $E(Y_1|Y_2)$". My function that I got is $f(y_1,y_2)=exp(-y_2)$ where $0leq y_1 leq y_2$.



      My own idea is that I need to find the $textit{conditional expectation}$ which I did, but then I can't continue from here.






      probability-theory







      share|cite|improve this question









      $endgroup$




      Good afternoon. I've trying to do some exercises. Here is one that I can't continue with.



      Here is the question: "Find the density functions of $E(Y_1|Y_2)$". My function that I got is $f(y_1,y_2)=exp(-y_2)$ where $0leq y_1 leq y_2$.



      My own idea is that I need to find the $textit{conditional expectation}$ which I did, but then I can't continue from here.






      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 Dec 23 '18 at 22:59









      Joey AdamsJoey Adams

      457




      457






















          1 Answer
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          2












          $begingroup$

          The conditional expectation $E(Y_1|Y_2)$ is a function of $Y_2$ that you have presumably figured out. Call it $g$, so $E(Y_1|Y_2)=g(Y_2)$. Since $Y_2$ is a random variable whose distribution you presumably know, you should be able to work out the distribution of the random variable $g(Y_2)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
            $endgroup$
            – Joey Adams
            Dec 23 '18 at 23:19






          • 1




            $begingroup$
            I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
            $endgroup$
            – kimchi lover
            Dec 23 '18 at 23:22










          • $begingroup$
            Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
            $endgroup$
            – Joey Adams
            Dec 23 '18 at 23:52










          • $begingroup$
            @Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
            $endgroup$
            – StubbornAtom
            Dec 24 '18 at 7:11











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          The conditional expectation $E(Y_1|Y_2)$ is a function of $Y_2$ that you have presumably figured out. Call it $g$, so $E(Y_1|Y_2)=g(Y_2)$. Since $Y_2$ is a random variable whose distribution you presumably know, you should be able to work out the distribution of the random variable $g(Y_2)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
            $endgroup$
            – Joey Adams
            Dec 23 '18 at 23:19






          • 1




            $begingroup$
            I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
            $endgroup$
            – kimchi lover
            Dec 23 '18 at 23:22










          • $begingroup$
            Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
            $endgroup$
            – Joey Adams
            Dec 23 '18 at 23:52










          • $begingroup$
            @Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
            $endgroup$
            – StubbornAtom
            Dec 24 '18 at 7:11
















          2












          $begingroup$

          The conditional expectation $E(Y_1|Y_2)$ is a function of $Y_2$ that you have presumably figured out. Call it $g$, so $E(Y_1|Y_2)=g(Y_2)$. Since $Y_2$ is a random variable whose distribution you presumably know, you should be able to work out the distribution of the random variable $g(Y_2)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
            $endgroup$
            – Joey Adams
            Dec 23 '18 at 23:19






          • 1




            $begingroup$
            I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
            $endgroup$
            – kimchi lover
            Dec 23 '18 at 23:22










          • $begingroup$
            Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
            $endgroup$
            – Joey Adams
            Dec 23 '18 at 23:52










          • $begingroup$
            @Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
            $endgroup$
            – StubbornAtom
            Dec 24 '18 at 7:11














          2












          2








          2





          $begingroup$

          The conditional expectation $E(Y_1|Y_2)$ is a function of $Y_2$ that you have presumably figured out. Call it $g$, so $E(Y_1|Y_2)=g(Y_2)$. Since $Y_2$ is a random variable whose distribution you presumably know, you should be able to work out the distribution of the random variable $g(Y_2)$.






          share|cite|improve this answer









          $endgroup$



          The conditional expectation $E(Y_1|Y_2)$ is a function of $Y_2$ that you have presumably figured out. Call it $g$, so $E(Y_1|Y_2)=g(Y_2)$. Since $Y_2$ is a random variable whose distribution you presumably know, you should be able to work out the distribution of the random variable $g(Y_2)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 23:17









          kimchi loverkimchi lover

          10.8k31128




          10.8k31128












          • $begingroup$
            So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
            $endgroup$
            – Joey Adams
            Dec 23 '18 at 23:19






          • 1




            $begingroup$
            I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
            $endgroup$
            – kimchi lover
            Dec 23 '18 at 23:22










          • $begingroup$
            Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
            $endgroup$
            – Joey Adams
            Dec 23 '18 at 23:52










          • $begingroup$
            @Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
            $endgroup$
            – StubbornAtom
            Dec 24 '18 at 7:11


















          • $begingroup$
            So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
            $endgroup$
            – Joey Adams
            Dec 23 '18 at 23:19






          • 1




            $begingroup$
            I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
            $endgroup$
            – kimchi lover
            Dec 23 '18 at 23:22










          • $begingroup$
            Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
            $endgroup$
            – Joey Adams
            Dec 23 '18 at 23:52










          • $begingroup$
            @Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
            $endgroup$
            – StubbornAtom
            Dec 24 '18 at 7:11
















          $begingroup$
          So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
          $endgroup$
          – Joey Adams
          Dec 23 '18 at 23:19




          $begingroup$
          So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
          $endgroup$
          – Joey Adams
          Dec 23 '18 at 23:19




          1




          1




          $begingroup$
          I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
          $endgroup$
          – kimchi lover
          Dec 23 '18 at 23:22




          $begingroup$
          I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
          $endgroup$
          – kimchi lover
          Dec 23 '18 at 23:22












          $begingroup$
          Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
          $endgroup$
          – Joey Adams
          Dec 23 '18 at 23:52




          $begingroup$
          Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
          $endgroup$
          – Joey Adams
          Dec 23 '18 at 23:52












          $begingroup$
          @Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
          $endgroup$
          – StubbornAtom
          Dec 24 '18 at 7:11




          $begingroup$
          @Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
          $endgroup$
          – StubbornAtom
          Dec 24 '18 at 7:11


















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