What value of the constant $alpha$ makes $Y$ and $Z$ uncorrelated?












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$begingroup$



The random variables $X_1$ and $X_2$ have means 1 and 2 respectively, with same variance 4 and covariance 0.8. Let $Y = X_1 − 3X_2$ and $Z = alpha X_1 + X_2$. What value of the constant α makes Y and Z uncorrelated?







probability







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$endgroup$

















    1












    $begingroup$



    The random variables $X_1$ and $X_2$ have means 1 and 2 respectively, with same variance 4 and covariance 0.8. Let $Y = X_1 − 3X_2$ and $Z = alpha X_1 + X_2$. What value of the constant α makes Y and Z uncorrelated?







    probability







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$



      The random variables $X_1$ and $X_2$ have means 1 and 2 respectively, with same variance 4 and covariance 0.8. Let $Y = X_1 − 3X_2$ and $Z = alpha X_1 + X_2$. What value of the constant α makes Y and Z uncorrelated?







      probability







      share|cite|improve this question











      $endgroup$





      The random variables $X_1$ and $X_2$ have means 1 and 2 respectively, with same variance 4 and covariance 0.8. Let $Y = X_1 − 3X_2$ and $Z = alpha X_1 + X_2$. What value of the constant α makes Y and Z uncorrelated?







      probability




      probability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 25 at 8:35







      Mark Jacon

















      asked Jan 24 at 12:07









      Mark JaconMark Jacon

      1127




      1127






















          2 Answers
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          $begingroup$

          It is given that covariance of $X_1$ and $X_2$ is $0.8$ Hence $EX_1X_2-(EX_1) (EX_2)=0.8$. So $EX_1X_2 =0.8+(1)(2)=2.8$. Now can you proceed?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
            $endgroup$
            – Mark Jacon
            Jan 24 at 13:36






          • 1




            $begingroup$
            $EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 24 at 23:13












          • $begingroup$
            Ok thank you, now the last question, how can I calculate $E[X_1^2]$ and $E[X_2^2]$, then I can complete the question :)
            $endgroup$
            – Mark Jacon
            Jan 25 at 8:17






          • 1




            $begingroup$
            $EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 25 at 8:19



















          0












          $begingroup$

          Solve:



          $E[X_1]=1$, $E[X_2]=2$, $operatorname{Var}(X_1)=4$, $operatorname{Var}(X_2)=4$ and $operatorname{Cov}(X_1,X_2)=0.8$



          $Y, Z$ are uncorrelated if $operatorname{Cov}(Y,Z)=0$ where $operatorname{Cov}(Y,Z) = E[YZ] − E[Y]E[Z]$.



          $E[Y]=E[X_1]-3E[X_2]=1-3*2=-5$



          $E[Z]=alpha E[X_1]+E[X_2]=alpha+2$



          $Cov(X_1,X_2)=E[X_1X_2]-E[X_1]*E[X_2]=0.8$ so $E[X_1X_2]=0.8+2*1=2.8$



          $E[X_1^2]=Var(X_1)+(E[X_1])^2=4+1^2=5$ and $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$



          $E[YZ]=E[(X_1 − 3X_2)(alpha X_1 + X_2)]=alpha E[X_1^2]+(1-3 alpha)E[X_1X_2]-3E[X_2^2]=5alpha+(1-3alpha)*2.8-3*8$



          $Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5alpha+(1-3alpha)*2.8-3*8-(-5)*(2+alpha)=0$, $alpha=7$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
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            $begingroup$

            It is given that covariance of $X_1$ and $X_2$ is $0.8$ Hence $EX_1X_2-(EX_1) (EX_2)=0.8$. So $EX_1X_2 =0.8+(1)(2)=2.8$. Now can you proceed?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
              $endgroup$
              – Mark Jacon
              Jan 24 at 13:36






            • 1




              $begingroup$
              $EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
              $endgroup$
              – Kavi Rama Murthy
              Jan 24 at 23:13












            • $begingroup$
              Ok thank you, now the last question, how can I calculate $E[X_1^2]$ and $E[X_2^2]$, then I can complete the question :)
              $endgroup$
              – Mark Jacon
              Jan 25 at 8:17






            • 1




              $begingroup$
              $EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
              $endgroup$
              – Kavi Rama Murthy
              Jan 25 at 8:19
















            2












            $begingroup$

            It is given that covariance of $X_1$ and $X_2$ is $0.8$ Hence $EX_1X_2-(EX_1) (EX_2)=0.8$. So $EX_1X_2 =0.8+(1)(2)=2.8$. Now can you proceed?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
              $endgroup$
              – Mark Jacon
              Jan 24 at 13:36






            • 1




              $begingroup$
              $EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
              $endgroup$
              – Kavi Rama Murthy
              Jan 24 at 23:13












            • $begingroup$
              Ok thank you, now the last question, how can I calculate $E[X_1^2]$ and $E[X_2^2]$, then I can complete the question :)
              $endgroup$
              – Mark Jacon
              Jan 25 at 8:17






            • 1




              $begingroup$
              $EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
              $endgroup$
              – Kavi Rama Murthy
              Jan 25 at 8:19














            2












            2








            2





            $begingroup$

            It is given that covariance of $X_1$ and $X_2$ is $0.8$ Hence $EX_1X_2-(EX_1) (EX_2)=0.8$. So $EX_1X_2 =0.8+(1)(2)=2.8$. Now can you proceed?






            share|cite|improve this answer









            $endgroup$



            It is given that covariance of $X_1$ and $X_2$ is $0.8$ Hence $EX_1X_2-(EX_1) (EX_2)=0.8$. So $EX_1X_2 =0.8+(1)(2)=2.8$. Now can you proceed?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 24 at 12:18









            Kavi Rama MurthyKavi Rama Murthy

            65.4k42766




            65.4k42766












            • $begingroup$
              Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
              $endgroup$
              – Mark Jacon
              Jan 24 at 13:36






            • 1




              $begingroup$
              $EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
              $endgroup$
              – Kavi Rama Murthy
              Jan 24 at 23:13












            • $begingroup$
              Ok thank you, now the last question, how can I calculate $E[X_1^2]$ and $E[X_2^2]$, then I can complete the question :)
              $endgroup$
              – Mark Jacon
              Jan 25 at 8:17






            • 1




              $begingroup$
              $EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
              $endgroup$
              – Kavi Rama Murthy
              Jan 25 at 8:19


















            • $begingroup$
              Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
              $endgroup$
              – Mark Jacon
              Jan 24 at 13:36






            • 1




              $begingroup$
              $EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
              $endgroup$
              – Kavi Rama Murthy
              Jan 24 at 23:13












            • $begingroup$
              Ok thank you, now the last question, how can I calculate $E[X_1^2]$ and $E[X_2^2]$, then I can complete the question :)
              $endgroup$
              – Mark Jacon
              Jan 25 at 8:17






            • 1




              $begingroup$
              $EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
              $endgroup$
              – Kavi Rama Murthy
              Jan 25 at 8:19
















            $begingroup$
            Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
            $endgroup$
            – Mark Jacon
            Jan 24 at 13:36




            $begingroup$
            Now, how can I relate $E[X_1X_2]$ with $E[YZ]$?
            $endgroup$
            – Mark Jacon
            Jan 24 at 13:36




            1




            1




            $begingroup$
            $EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 24 at 23:13






            $begingroup$
            $EYZ=E(X_1-3X_2)(alpha X_1+X_2)=alpha EX_1^{2}-3alpha EX_1X_2+EX_1X_2-3EX_2^{2}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 24 at 23:13














            $begingroup$
            Ok thank you, now the last question, how can I calculate $E[X_1^2]$ and $E[X_2^2]$, then I can complete the question :)
            $endgroup$
            – Mark Jacon
            Jan 25 at 8:17




            $begingroup$
            Ok thank you, now the last question, how can I calculate $E[X_1^2]$ and $E[X_2^2]$, then I can complete the question :)
            $endgroup$
            – Mark Jacon
            Jan 25 at 8:17




            1




            1




            $begingroup$
            $EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 25 at 8:19




            $begingroup$
            $EX_1^{2}=var(X_1)+(EX_1)^{2}=4+1=5$. similarly, $EX_2^{2}=4+4=8$.
            $endgroup$
            – Kavi Rama Murthy
            Jan 25 at 8:19











            0












            $begingroup$

            Solve:



            $E[X_1]=1$, $E[X_2]=2$, $operatorname{Var}(X_1)=4$, $operatorname{Var}(X_2)=4$ and $operatorname{Cov}(X_1,X_2)=0.8$



            $Y, Z$ are uncorrelated if $operatorname{Cov}(Y,Z)=0$ where $operatorname{Cov}(Y,Z) = E[YZ] − E[Y]E[Z]$.



            $E[Y]=E[X_1]-3E[X_2]=1-3*2=-5$



            $E[Z]=alpha E[X_1]+E[X_2]=alpha+2$



            $Cov(X_1,X_2)=E[X_1X_2]-E[X_1]*E[X_2]=0.8$ so $E[X_1X_2]=0.8+2*1=2.8$



            $E[X_1^2]=Var(X_1)+(E[X_1])^2=4+1^2=5$ and $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$



            $E[YZ]=E[(X_1 − 3X_2)(alpha X_1 + X_2)]=alpha E[X_1^2]+(1-3 alpha)E[X_1X_2]-3E[X_2^2]=5alpha+(1-3alpha)*2.8-3*8$



            $Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5alpha+(1-3alpha)*2.8-3*8-(-5)*(2+alpha)=0$, $alpha=7$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Solve:



              $E[X_1]=1$, $E[X_2]=2$, $operatorname{Var}(X_1)=4$, $operatorname{Var}(X_2)=4$ and $operatorname{Cov}(X_1,X_2)=0.8$



              $Y, Z$ are uncorrelated if $operatorname{Cov}(Y,Z)=0$ where $operatorname{Cov}(Y,Z) = E[YZ] − E[Y]E[Z]$.



              $E[Y]=E[X_1]-3E[X_2]=1-3*2=-5$



              $E[Z]=alpha E[X_1]+E[X_2]=alpha+2$



              $Cov(X_1,X_2)=E[X_1X_2]-E[X_1]*E[X_2]=0.8$ so $E[X_1X_2]=0.8+2*1=2.8$



              $E[X_1^2]=Var(X_1)+(E[X_1])^2=4+1^2=5$ and $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$



              $E[YZ]=E[(X_1 − 3X_2)(alpha X_1 + X_2)]=alpha E[X_1^2]+(1-3 alpha)E[X_1X_2]-3E[X_2^2]=5alpha+(1-3alpha)*2.8-3*8$



              $Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5alpha+(1-3alpha)*2.8-3*8-(-5)*(2+alpha)=0$, $alpha=7$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Solve:



                $E[X_1]=1$, $E[X_2]=2$, $operatorname{Var}(X_1)=4$, $operatorname{Var}(X_2)=4$ and $operatorname{Cov}(X_1,X_2)=0.8$



                $Y, Z$ are uncorrelated if $operatorname{Cov}(Y,Z)=0$ where $operatorname{Cov}(Y,Z) = E[YZ] − E[Y]E[Z]$.



                $E[Y]=E[X_1]-3E[X_2]=1-3*2=-5$



                $E[Z]=alpha E[X_1]+E[X_2]=alpha+2$



                $Cov(X_1,X_2)=E[X_1X_2]-E[X_1]*E[X_2]=0.8$ so $E[X_1X_2]=0.8+2*1=2.8$



                $E[X_1^2]=Var(X_1)+(E[X_1])^2=4+1^2=5$ and $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$



                $E[YZ]=E[(X_1 − 3X_2)(alpha X_1 + X_2)]=alpha E[X_1^2]+(1-3 alpha)E[X_1X_2]-3E[X_2^2]=5alpha+(1-3alpha)*2.8-3*8$



                $Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5alpha+(1-3alpha)*2.8-3*8-(-5)*(2+alpha)=0$, $alpha=7$






                share|cite|improve this answer









                $endgroup$



                Solve:



                $E[X_1]=1$, $E[X_2]=2$, $operatorname{Var}(X_1)=4$, $operatorname{Var}(X_2)=4$ and $operatorname{Cov}(X_1,X_2)=0.8$



                $Y, Z$ are uncorrelated if $operatorname{Cov}(Y,Z)=0$ where $operatorname{Cov}(Y,Z) = E[YZ] − E[Y]E[Z]$.



                $E[Y]=E[X_1]-3E[X_2]=1-3*2=-5$



                $E[Z]=alpha E[X_1]+E[X_2]=alpha+2$



                $Cov(X_1,X_2)=E[X_1X_2]-E[X_1]*E[X_2]=0.8$ so $E[X_1X_2]=0.8+2*1=2.8$



                $E[X_1^2]=Var(X_1)+(E[X_1])^2=4+1^2=5$ and $E[X_2^2]=Var(X_2)+(E[X_2])^2=4+2^2=8$



                $E[YZ]=E[(X_1 − 3X_2)(alpha X_1 + X_2)]=alpha E[X_1^2]+(1-3 alpha)E[X_1X_2]-3E[X_2^2]=5alpha+(1-3alpha)*2.8-3*8$



                $Cov(Y,Z)=E[YZ] − E[Y]E[Z]=0$, $5alpha+(1-3alpha)*2.8-3*8-(-5)*(2+alpha)=0$, $alpha=7$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 25 at 8:40









                Mark JaconMark Jacon

                1127




                1127






























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