Let $ X sim operatorname{Exp}(lambda =1), Ysim U(1,2) $ be independent continuous variables. What is...












0












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I'm struggling to understand how to start the following:




Let $ X sim operatorname{Exp}(lambda =1), Ysim U(1,2) $ be independent
continuous variables. What is $E(frac{x}{y})$?




Thanks :)






probability expected-value







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I'm struggling to understand how to start the following:




    Let $ X sim operatorname{Exp}(lambda =1), Ysim U(1,2) $ be independent
    continuous variables. What is $E(frac{x}{y})$?




    Thanks :)






    probability expected-value







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I'm struggling to understand how to start the following:




      Let $ X sim operatorname{Exp}(lambda =1), Ysim U(1,2) $ be independent
      continuous variables. What is $E(frac{x}{y})$?




      Thanks :)






      probability expected-value







      share|cite|improve this question











      $endgroup$




      I'm struggling to understand how to start the following:




      Let $ X sim operatorname{Exp}(lambda =1), Ysim U(1,2) $ be independent
      continuous variables. What is $E(frac{x}{y})$?




      Thanks :)






      probability expected-value




      probability expected-value






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 10:07









      Bernard

      125k743118




      125k743118










      asked Jan 15 at 10:02









      superuser123superuser123

      48628




      48628






















          2 Answers
          2






          active

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          1












          $begingroup$

          $E(frac X Y |Y)=frac 1 Y EX=frac 1 Y$. Taking expectation we get $Efrac X Y=Efrac 1 Y=int_1^{2} frac 1 y , dy=ln, 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Note that this only holds because $Y$ is never zero and thus the division expression is well-defined.
            $endgroup$
            – Thomas Lang
            Jan 15 at 10:08






          • 1




            $begingroup$
            Thanks, but why is the condition taking place? ($E(frac{X}{Y} | Y) $?
            $endgroup$
            – superuser123
            Jan 15 at 10:19










          • $begingroup$
            @superuser123 You can also do this using a double integral but using conditional expectation makes it easier.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 10:20










          • $begingroup$
            @KaviRamaMurthy how could it be done by a double integral? thanks!
            $endgroup$
            – superuser123
            Jan 15 at 10:22






          • 1




            $begingroup$
            @superuser123 Calculate (by integrating w.r.t $x$ and then w.r.t. $y$ or the other way around) $int_0^{infty} int_1^{2} frac x y e^{-x} ,dy, dx$
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 10:25





















          0












          $begingroup$

          Guide:



          If $X$ and $Y$ are independent then so are $f(X)$ and $g(Y)$ where $f$ and $g$ are measurable functions.



          Consequence for expectation (if it exists):$$mathbb Ef(X)g(Y)=mathbb Ef(X)mathbb Eg(Y)$$



          This can be applied to find: $$mathbb E[XY^{-1}]=mathbb EXmathbb EY^{-1}$$
          It only remains now to find $mathbb EX$ and $mathbb EY^{-1}$.






          share|cite|improve this answer











          $endgroup$














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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            $E(frac X Y |Y)=frac 1 Y EX=frac 1 Y$. Taking expectation we get $Efrac X Y=Efrac 1 Y=int_1^{2} frac 1 y , dy=ln, 2$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Note that this only holds because $Y$ is never zero and thus the division expression is well-defined.
              $endgroup$
              – Thomas Lang
              Jan 15 at 10:08






            • 1




              $begingroup$
              Thanks, but why is the condition taking place? ($E(frac{X}{Y} | Y) $?
              $endgroup$
              – superuser123
              Jan 15 at 10:19










            • $begingroup$
              @superuser123 You can also do this using a double integral but using conditional expectation makes it easier.
              $endgroup$
              – Kavi Rama Murthy
              Jan 15 at 10:20










            • $begingroup$
              @KaviRamaMurthy how could it be done by a double integral? thanks!
              $endgroup$
              – superuser123
              Jan 15 at 10:22






            • 1




              $begingroup$
              @superuser123 Calculate (by integrating w.r.t $x$ and then w.r.t. $y$ or the other way around) $int_0^{infty} int_1^{2} frac x y e^{-x} ,dy, dx$
              $endgroup$
              – Kavi Rama Murthy
              Jan 15 at 10:25


















            1












            $begingroup$

            $E(frac X Y |Y)=frac 1 Y EX=frac 1 Y$. Taking expectation we get $Efrac X Y=Efrac 1 Y=int_1^{2} frac 1 y , dy=ln, 2$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Note that this only holds because $Y$ is never zero and thus the division expression is well-defined.
              $endgroup$
              – Thomas Lang
              Jan 15 at 10:08






            • 1




              $begingroup$
              Thanks, but why is the condition taking place? ($E(frac{X}{Y} | Y) $?
              $endgroup$
              – superuser123
              Jan 15 at 10:19










            • $begingroup$
              @superuser123 You can also do this using a double integral but using conditional expectation makes it easier.
              $endgroup$
              – Kavi Rama Murthy
              Jan 15 at 10:20










            • $begingroup$
              @KaviRamaMurthy how could it be done by a double integral? thanks!
              $endgroup$
              – superuser123
              Jan 15 at 10:22






            • 1




              $begingroup$
              @superuser123 Calculate (by integrating w.r.t $x$ and then w.r.t. $y$ or the other way around) $int_0^{infty} int_1^{2} frac x y e^{-x} ,dy, dx$
              $endgroup$
              – Kavi Rama Murthy
              Jan 15 at 10:25
















            1












            1








            1





            $begingroup$

            $E(frac X Y |Y)=frac 1 Y EX=frac 1 Y$. Taking expectation we get $Efrac X Y=Efrac 1 Y=int_1^{2} frac 1 y , dy=ln, 2$.






            share|cite|improve this answer









            $endgroup$



            $E(frac X Y |Y)=frac 1 Y EX=frac 1 Y$. Taking expectation we get $Efrac X Y=Efrac 1 Y=int_1^{2} frac 1 y , dy=ln, 2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 15 at 10:07









            Kavi Rama MurthyKavi Rama Murthy

            76.9k53471




            76.9k53471












            • $begingroup$
              Note that this only holds because $Y$ is never zero and thus the division expression is well-defined.
              $endgroup$
              – Thomas Lang
              Jan 15 at 10:08






            • 1




              $begingroup$
              Thanks, but why is the condition taking place? ($E(frac{X}{Y} | Y) $?
              $endgroup$
              – superuser123
              Jan 15 at 10:19










            • $begingroup$
              @superuser123 You can also do this using a double integral but using conditional expectation makes it easier.
              $endgroup$
              – Kavi Rama Murthy
              Jan 15 at 10:20










            • $begingroup$
              @KaviRamaMurthy how could it be done by a double integral? thanks!
              $endgroup$
              – superuser123
              Jan 15 at 10:22






            • 1




              $begingroup$
              @superuser123 Calculate (by integrating w.r.t $x$ and then w.r.t. $y$ or the other way around) $int_0^{infty} int_1^{2} frac x y e^{-x} ,dy, dx$
              $endgroup$
              – Kavi Rama Murthy
              Jan 15 at 10:25




















            • $begingroup$
              Note that this only holds because $Y$ is never zero and thus the division expression is well-defined.
              $endgroup$
              – Thomas Lang
              Jan 15 at 10:08






            • 1




              $begingroup$
              Thanks, but why is the condition taking place? ($E(frac{X}{Y} | Y) $?
              $endgroup$
              – superuser123
              Jan 15 at 10:19










            • $begingroup$
              @superuser123 You can also do this using a double integral but using conditional expectation makes it easier.
              $endgroup$
              – Kavi Rama Murthy
              Jan 15 at 10:20










            • $begingroup$
              @KaviRamaMurthy how could it be done by a double integral? thanks!
              $endgroup$
              – superuser123
              Jan 15 at 10:22






            • 1




              $begingroup$
              @superuser123 Calculate (by integrating w.r.t $x$ and then w.r.t. $y$ or the other way around) $int_0^{infty} int_1^{2} frac x y e^{-x} ,dy, dx$
              $endgroup$
              – Kavi Rama Murthy
              Jan 15 at 10:25


















            $begingroup$
            Note that this only holds because $Y$ is never zero and thus the division expression is well-defined.
            $endgroup$
            – Thomas Lang
            Jan 15 at 10:08




            $begingroup$
            Note that this only holds because $Y$ is never zero and thus the division expression is well-defined.
            $endgroup$
            – Thomas Lang
            Jan 15 at 10:08




            1




            1




            $begingroup$
            Thanks, but why is the condition taking place? ($E(frac{X}{Y} | Y) $?
            $endgroup$
            – superuser123
            Jan 15 at 10:19




            $begingroup$
            Thanks, but why is the condition taking place? ($E(frac{X}{Y} | Y) $?
            $endgroup$
            – superuser123
            Jan 15 at 10:19












            $begingroup$
            @superuser123 You can also do this using a double integral but using conditional expectation makes it easier.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 10:20




            $begingroup$
            @superuser123 You can also do this using a double integral but using conditional expectation makes it easier.
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 10:20












            $begingroup$
            @KaviRamaMurthy how could it be done by a double integral? thanks!
            $endgroup$
            – superuser123
            Jan 15 at 10:22




            $begingroup$
            @KaviRamaMurthy how could it be done by a double integral? thanks!
            $endgroup$
            – superuser123
            Jan 15 at 10:22




            1




            1




            $begingroup$
            @superuser123 Calculate (by integrating w.r.t $x$ and then w.r.t. $y$ or the other way around) $int_0^{infty} int_1^{2} frac x y e^{-x} ,dy, dx$
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 10:25






            $begingroup$
            @superuser123 Calculate (by integrating w.r.t $x$ and then w.r.t. $y$ or the other way around) $int_0^{infty} int_1^{2} frac x y e^{-x} ,dy, dx$
            $endgroup$
            – Kavi Rama Murthy
            Jan 15 at 10:25













            0












            $begingroup$

            Guide:



            If $X$ and $Y$ are independent then so are $f(X)$ and $g(Y)$ where $f$ and $g$ are measurable functions.



            Consequence for expectation (if it exists):$$mathbb Ef(X)g(Y)=mathbb Ef(X)mathbb Eg(Y)$$



            This can be applied to find: $$mathbb E[XY^{-1}]=mathbb EXmathbb EY^{-1}$$
            It only remains now to find $mathbb EX$ and $mathbb EY^{-1}$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Guide:



              If $X$ and $Y$ are independent then so are $f(X)$ and $g(Y)$ where $f$ and $g$ are measurable functions.



              Consequence for expectation (if it exists):$$mathbb Ef(X)g(Y)=mathbb Ef(X)mathbb Eg(Y)$$



              This can be applied to find: $$mathbb E[XY^{-1}]=mathbb EXmathbb EY^{-1}$$
              It only remains now to find $mathbb EX$ and $mathbb EY^{-1}$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Guide:



                If $X$ and $Y$ are independent then so are $f(X)$ and $g(Y)$ where $f$ and $g$ are measurable functions.



                Consequence for expectation (if it exists):$$mathbb Ef(X)g(Y)=mathbb Ef(X)mathbb Eg(Y)$$



                This can be applied to find: $$mathbb E[XY^{-1}]=mathbb EXmathbb EY^{-1}$$
                It only remains now to find $mathbb EX$ and $mathbb EY^{-1}$.






                share|cite|improve this answer











                $endgroup$



                Guide:



                If $X$ and $Y$ are independent then so are $f(X)$ and $g(Y)$ where $f$ and $g$ are measurable functions.



                Consequence for expectation (if it exists):$$mathbb Ef(X)g(Y)=mathbb Ef(X)mathbb Eg(Y)$$



                This can be applied to find: $$mathbb E[XY^{-1}]=mathbb EXmathbb EY^{-1}$$
                It only remains now to find $mathbb EX$ and $mathbb EY^{-1}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 15 at 10:29

























                answered Jan 15 at 10:22









                drhabdrhab

                104k545136




                104k545136






























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