what is the expectation of this random variable?












0












$begingroup$


Suppose X is an Hyper Geometric random variable with parameters w,b and n, that is, X ~ HGeom(w,b,n).



And suppose Y = $Xchoose2$



, what is E(Y)? Using the sum of probability times the value of Y yields a very complicated formula, there should be some clever way to get the result. Could some one help with this?



Thanks.






probability expected-value







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$endgroup$

















    0












    $begingroup$


    Suppose X is an Hyper Geometric random variable with parameters w,b and n, that is, X ~ HGeom(w,b,n).



    And suppose Y = $Xchoose2$



    , what is E(Y)? Using the sum of probability times the value of Y yields a very complicated formula, there should be some clever way to get the result. Could some one help with this?



    Thanks.






    probability expected-value







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose X is an Hyper Geometric random variable with parameters w,b and n, that is, X ~ HGeom(w,b,n).



      And suppose Y = $Xchoose2$



      , what is E(Y)? Using the sum of probability times the value of Y yields a very complicated formula, there should be some clever way to get the result. Could some one help with this?



      Thanks.






      probability expected-value







      share|cite|improve this question









      $endgroup$




      Suppose X is an Hyper Geometric random variable with parameters w,b and n, that is, X ~ HGeom(w,b,n).



      And suppose Y = $Xchoose2$



      , what is E(Y)? Using the sum of probability times the value of Y yields a very complicated formula, there should be some clever way to get the result. Could some one help with this?



      Thanks.






      probability expected-value




      probability expected-value






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 4 '18 at 0:53









      wangshuaijiewangshuaijie

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      1638






















          2 Answers
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          0












          $begingroup$

          First, $Y={ X choose 2}=frac{X!}{2!(X-2)!}=frac{X(X-1)(X-2)!}{2!(X-2)!}=frac{X(X-1)}{2}$.



          Second, use the fact that for any random variable $X$ and $a, b in mathbb{R}$, $E(aX+b)=aE(X)+b$.



          Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.



          So, $E(Y)=E(frac{X(X-1)}{2})=E(frac{X^{2}-X}{2})=frac{E(X^{2})}{2}-frac{E(X)}{2}$



          All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Guide:




            • The mean and variace of Hypergeometric should be well known, that is we know the first two moments.


            • Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.







            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              0












              $begingroup$

              First, $Y={ X choose 2}=frac{X!}{2!(X-2)!}=frac{X(X-1)(X-2)!}{2!(X-2)!}=frac{X(X-1)}{2}$.



              Second, use the fact that for any random variable $X$ and $a, b in mathbb{R}$, $E(aX+b)=aE(X)+b$.



              Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.



              So, $E(Y)=E(frac{X(X-1)}{2})=E(frac{X^{2}-X}{2})=frac{E(X^{2})}{2}-frac{E(X)}{2}$



              All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                First, $Y={ X choose 2}=frac{X!}{2!(X-2)!}=frac{X(X-1)(X-2)!}{2!(X-2)!}=frac{X(X-1)}{2}$.



                Second, use the fact that for any random variable $X$ and $a, b in mathbb{R}$, $E(aX+b)=aE(X)+b$.



                Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.



                So, $E(Y)=E(frac{X(X-1)}{2})=E(frac{X^{2}-X}{2})=frac{E(X^{2})}{2}-frac{E(X)}{2}$



                All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  First, $Y={ X choose 2}=frac{X!}{2!(X-2)!}=frac{X(X-1)(X-2)!}{2!(X-2)!}=frac{X(X-1)}{2}$.



                  Second, use the fact that for any random variable $X$ and $a, b in mathbb{R}$, $E(aX+b)=aE(X)+b$.



                  Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.



                  So, $E(Y)=E(frac{X(X-1)}{2})=E(frac{X^{2}-X}{2})=frac{E(X^{2})}{2}-frac{E(X)}{2}$



                  All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).






                  share|cite|improve this answer











                  $endgroup$



                  First, $Y={ X choose 2}=frac{X!}{2!(X-2)!}=frac{X(X-1)(X-2)!}{2!(X-2)!}=frac{X(X-1)}{2}$.



                  Second, use the fact that for any random variable $X$ and $a, b in mathbb{R}$, $E(aX+b)=aE(X)+b$.



                  Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.



                  So, $E(Y)=E(frac{X(X-1)}{2})=E(frac{X^{2}-X}{2})=frac{E(X^{2})}{2}-frac{E(X)}{2}$



                  All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 4 '18 at 2:37

























                  answered Dec 4 '18 at 2:31









                  Live Free or π HardLive Free or π Hard

                  479213




                  479213























                      0












                      $begingroup$

                      Guide:




                      • The mean and variace of Hypergeometric should be well known, that is we know the first two moments.


                      • Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.







                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Guide:




                        • The mean and variace of Hypergeometric should be well known, that is we know the first two moments.


                        • Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.







                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Guide:




                          • The mean and variace of Hypergeometric should be well known, that is we know the first two moments.


                          • Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.







                          share|cite|improve this answer









                          $endgroup$



                          Guide:




                          • The mean and variace of Hypergeometric should be well known, that is we know the first two moments.


                          • Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.








                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 4 '18 at 0:57









                          Siong Thye GohSiong Thye Goh

                          100k1466117




                          100k1466117






























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