Question on redefinition of an expectation












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$begingroup$


There is a probability space $(Omega,mathrm{B},P)$. Let $Y$ be a positive random variable with $E(Y)=1$. Define $P^*(A) = int_A Y,dP.$ Show that



a. $E^*(X)=E(XY)$ for any random variable $Xgeq 0$.



b. For any $Xgeq 0$ that $$ E^*(X|Z_1,cdots,Z_n) = frac{E(XY|Z_1,cdots,Z_n)}{E(Y|Z_1,cdots,Z_n)} $$ for any random variables $Z_1,cdots,Z_n$.



My attempt:



a. $E^*(X)=int_Omega X,dP^* = int_Omega X int_{dw}Y,dP = int_Omega XY,P(dw)$.



$E(XY)=int_Omega XY,P(dw)$.



This proves part a. My question: Why do we require $X$ to be non-negative? Please provide any hints if my approach has some flaws in it.



b. I was attempting to solve this using Question on conditional expectation like chain rule, but alas, this approach seems to be wrong.



So, any hints to solve this problem will be appreciated! Thanks in advance.










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$endgroup$

















    0












    $begingroup$


    There is a probability space $(Omega,mathrm{B},P)$. Let $Y$ be a positive random variable with $E(Y)=1$. Define $P^*(A) = int_A Y,dP.$ Show that



    a. $E^*(X)=E(XY)$ for any random variable $Xgeq 0$.



    b. For any $Xgeq 0$ that $$ E^*(X|Z_1,cdots,Z_n) = frac{E(XY|Z_1,cdots,Z_n)}{E(Y|Z_1,cdots,Z_n)} $$ for any random variables $Z_1,cdots,Z_n$.



    My attempt:



    a. $E^*(X)=int_Omega X,dP^* = int_Omega X int_{dw}Y,dP = int_Omega XY,P(dw)$.



    $E(XY)=int_Omega XY,P(dw)$.



    This proves part a. My question: Why do we require $X$ to be non-negative? Please provide any hints if my approach has some flaws in it.



    b. I was attempting to solve this using Question on conditional expectation like chain rule, but alas, this approach seems to be wrong.



    So, any hints to solve this problem will be appreciated! Thanks in advance.










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      There is a probability space $(Omega,mathrm{B},P)$. Let $Y$ be a positive random variable with $E(Y)=1$. Define $P^*(A) = int_A Y,dP.$ Show that



      a. $E^*(X)=E(XY)$ for any random variable $Xgeq 0$.



      b. For any $Xgeq 0$ that $$ E^*(X|Z_1,cdots,Z_n) = frac{E(XY|Z_1,cdots,Z_n)}{E(Y|Z_1,cdots,Z_n)} $$ for any random variables $Z_1,cdots,Z_n$.



      My attempt:



      a. $E^*(X)=int_Omega X,dP^* = int_Omega X int_{dw}Y,dP = int_Omega XY,P(dw)$.



      $E(XY)=int_Omega XY,P(dw)$.



      This proves part a. My question: Why do we require $X$ to be non-negative? Please provide any hints if my approach has some flaws in it.



      b. I was attempting to solve this using Question on conditional expectation like chain rule, but alas, this approach seems to be wrong.



      So, any hints to solve this problem will be appreciated! Thanks in advance.










      share|cite|improve this question











      $endgroup$




      There is a probability space $(Omega,mathrm{B},P)$. Let $Y$ be a positive random variable with $E(Y)=1$. Define $P^*(A) = int_A Y,dP.$ Show that



      a. $E^*(X)=E(XY)$ for any random variable $Xgeq 0$.



      b. For any $Xgeq 0$ that $$ E^*(X|Z_1,cdots,Z_n) = frac{E(XY|Z_1,cdots,Z_n)}{E(Y|Z_1,cdots,Z_n)} $$ for any random variables $Z_1,cdots,Z_n$.



      My attempt:



      a. $E^*(X)=int_Omega X,dP^* = int_Omega X int_{dw}Y,dP = int_Omega XY,P(dw)$.



      $E(XY)=int_Omega XY,P(dw)$.



      This proves part a. My question: Why do we require $X$ to be non-negative? Please provide any hints if my approach has some flaws in it.



      b. I was attempting to solve this using Question on conditional expectation like chain rule, but alas, this approach seems to be wrong.



      So, any hints to solve this problem will be appreciated! Thanks in advance.







      measure-theory






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      edited Dec 4 '18 at 2:33







      Lohith-kumar

















      asked Dec 4 '18 at 0:30









      Lohith-kumarLohith-kumar

      493111




      493111






















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          $begingroup$

          a. Seems just flawed. For instance, what does $dw$ even mean? Try the following. Consider $mathscr{H}_+$ the set of random variables $X geq 0$ for which $mathbf{E}^*(X) = mathbf{E}(XY).$ Then $mathscr{H}_+$ contains all indicator functions of events (by definition), is a closed under addition, positive-scalars multiplication and closed under monotone limits. Hence, $mathscr{H}_+$ contains all non-negative random variables (for we recall if $X$ is such a random variable, then $X$ equals, a.s., a monotone limit of simple positive r.v.). Now, in regards as to why $X$ has to be non-negative, it is just for the expectations to exist. You could also ask $X$ to be bounded.



          b. Consider now the sigma algebra $mathscr{X}$ generated by $Z_1, ldots, Z_n.$ By definition, $mathbf{E}^*(X mid Z_1, ldots, Z_n)$ is any random variable, measurable with respect to $mathscr{X},$ for which the following relation holds: if $mathrm{E} in mathscr{X}$ then $$intlimits_{mathrm{E}} mathbf{E}^*(X mid Z_1, ldots, Z_n) dmathbf{P}^* = intlimits_{mathrm{E}} X dmathbf{P}^*.$$
          However, by virtue of a., we know that the right hand side above equals $intlimits_mathrm{E} XY dmathbf{P}$ which, in turn, equals $$intlimits_mathrm{E} mathbf{E}(XY mid Z_1, ldots, Z_n) dmathbf{P} = intlimits_mathrm{E} dfrac{mathbf{E}(XY mid Z_1, ldots, Z_n)}{mathbf{E}(Y mid Z_1, ldots, Z_n)} mathbf{E}(Y mid Z_1, ldots, Z_n) dmathbf{P}.$$
          Apply the result of a. to the probability space $(Omega, mathscr{X}, mathbf{P})$ using $mathbf{E}(Y mid Z_1, ldots, Z_n)$ in lieu of $Y$ (recall $mathbf{E}(mathbf{E}(Y mid Z_1, ldots, Z_n)) = mathbf{E}(Y)=1$ and $mathbf{E}(Y mid Z_1, ldots, Z_n) geq 0$) to get $$intlimits_{mathrm{E}} mathbf{E}^*(X mid Z_1, ldots, Z_n) dmathbf{P}^* = intlimits_mathrm{E} dfrac{mathbf{E}(XY mid Z_1, ldots, Z_n)}{mathbf{E}(Y mid Z_1, ldots, Z_n)} dmathbf{P}^{**},$$ where $mathbf{P}^{**}$ is the probability measure on $mathscr{X}$ such that $mathbf{P}^{**}(mathrm{A}) = intlimits_mathrm{A} mathbf{E}(Y mid Z_1, ldots, Z_n) dmathbf{P}$; by definition of conditional expectation, $mathbf{P}^{**}(mathrm{A})=mathbf{P}^*(mathrm{A})$ for all $mathrm{A} in mathscr{X},$ that is $mathbf{P}^{**} = mathbf{P}^*.$ This substantiates b. Q.E.D.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for such an elegant proof idea! There is a typo at the end I believe. It should be $mathbf{P}^{**}(mathrm{A})=mathbf{P^*}(mathrm{A})$ right?
            $endgroup$
            – Lohith-kumar
            Dec 4 '18 at 1:40










          • $begingroup$
            Yes. Corrected, thanks.
            $endgroup$
            – Will M.
            Dec 4 '18 at 1:48










          • $begingroup$
            Is it possible for you to take a look at this question math.stackexchange.com/questions/3025024/… as well? Thanks!
            $endgroup$
            – Lohith-kumar
            Dec 4 '18 at 2:09










          • $begingroup$
            Yes, modify my argument here from a.
            $endgroup$
            – Will M.
            Dec 4 '18 at 2:23






          • 1




            $begingroup$
            I am actually not sure what you mean. I am sure, however, that the argument of a. here can be modified to prove that exercise.
            $endgroup$
            – Will M.
            Dec 4 '18 at 2:29











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          $begingroup$

          a. Seems just flawed. For instance, what does $dw$ even mean? Try the following. Consider $mathscr{H}_+$ the set of random variables $X geq 0$ for which $mathbf{E}^*(X) = mathbf{E}(XY).$ Then $mathscr{H}_+$ contains all indicator functions of events (by definition), is a closed under addition, positive-scalars multiplication and closed under monotone limits. Hence, $mathscr{H}_+$ contains all non-negative random variables (for we recall if $X$ is such a random variable, then $X$ equals, a.s., a monotone limit of simple positive r.v.). Now, in regards as to why $X$ has to be non-negative, it is just for the expectations to exist. You could also ask $X$ to be bounded.



          b. Consider now the sigma algebra $mathscr{X}$ generated by $Z_1, ldots, Z_n.$ By definition, $mathbf{E}^*(X mid Z_1, ldots, Z_n)$ is any random variable, measurable with respect to $mathscr{X},$ for which the following relation holds: if $mathrm{E} in mathscr{X}$ then $$intlimits_{mathrm{E}} mathbf{E}^*(X mid Z_1, ldots, Z_n) dmathbf{P}^* = intlimits_{mathrm{E}} X dmathbf{P}^*.$$
          However, by virtue of a., we know that the right hand side above equals $intlimits_mathrm{E} XY dmathbf{P}$ which, in turn, equals $$intlimits_mathrm{E} mathbf{E}(XY mid Z_1, ldots, Z_n) dmathbf{P} = intlimits_mathrm{E} dfrac{mathbf{E}(XY mid Z_1, ldots, Z_n)}{mathbf{E}(Y mid Z_1, ldots, Z_n)} mathbf{E}(Y mid Z_1, ldots, Z_n) dmathbf{P}.$$
          Apply the result of a. to the probability space $(Omega, mathscr{X}, mathbf{P})$ using $mathbf{E}(Y mid Z_1, ldots, Z_n)$ in lieu of $Y$ (recall $mathbf{E}(mathbf{E}(Y mid Z_1, ldots, Z_n)) = mathbf{E}(Y)=1$ and $mathbf{E}(Y mid Z_1, ldots, Z_n) geq 0$) to get $$intlimits_{mathrm{E}} mathbf{E}^*(X mid Z_1, ldots, Z_n) dmathbf{P}^* = intlimits_mathrm{E} dfrac{mathbf{E}(XY mid Z_1, ldots, Z_n)}{mathbf{E}(Y mid Z_1, ldots, Z_n)} dmathbf{P}^{**},$$ where $mathbf{P}^{**}$ is the probability measure on $mathscr{X}$ such that $mathbf{P}^{**}(mathrm{A}) = intlimits_mathrm{A} mathbf{E}(Y mid Z_1, ldots, Z_n) dmathbf{P}$; by definition of conditional expectation, $mathbf{P}^{**}(mathrm{A})=mathbf{P}^*(mathrm{A})$ for all $mathrm{A} in mathscr{X},$ that is $mathbf{P}^{**} = mathbf{P}^*.$ This substantiates b. Q.E.D.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for such an elegant proof idea! There is a typo at the end I believe. It should be $mathbf{P}^{**}(mathrm{A})=mathbf{P^*}(mathrm{A})$ right?
            $endgroup$
            – Lohith-kumar
            Dec 4 '18 at 1:40










          • $begingroup$
            Yes. Corrected, thanks.
            $endgroup$
            – Will M.
            Dec 4 '18 at 1:48










          • $begingroup$
            Is it possible for you to take a look at this question math.stackexchange.com/questions/3025024/… as well? Thanks!
            $endgroup$
            – Lohith-kumar
            Dec 4 '18 at 2:09










          • $begingroup$
            Yes, modify my argument here from a.
            $endgroup$
            – Will M.
            Dec 4 '18 at 2:23






          • 1




            $begingroup$
            I am actually not sure what you mean. I am sure, however, that the argument of a. here can be modified to prove that exercise.
            $endgroup$
            – Will M.
            Dec 4 '18 at 2:29
















          1












          $begingroup$

          a. Seems just flawed. For instance, what does $dw$ even mean? Try the following. Consider $mathscr{H}_+$ the set of random variables $X geq 0$ for which $mathbf{E}^*(X) = mathbf{E}(XY).$ Then $mathscr{H}_+$ contains all indicator functions of events (by definition), is a closed under addition, positive-scalars multiplication and closed under monotone limits. Hence, $mathscr{H}_+$ contains all non-negative random variables (for we recall if $X$ is such a random variable, then $X$ equals, a.s., a monotone limit of simple positive r.v.). Now, in regards as to why $X$ has to be non-negative, it is just for the expectations to exist. You could also ask $X$ to be bounded.



          b. Consider now the sigma algebra $mathscr{X}$ generated by $Z_1, ldots, Z_n.$ By definition, $mathbf{E}^*(X mid Z_1, ldots, Z_n)$ is any random variable, measurable with respect to $mathscr{X},$ for which the following relation holds: if $mathrm{E} in mathscr{X}$ then $$intlimits_{mathrm{E}} mathbf{E}^*(X mid Z_1, ldots, Z_n) dmathbf{P}^* = intlimits_{mathrm{E}} X dmathbf{P}^*.$$
          However, by virtue of a., we know that the right hand side above equals $intlimits_mathrm{E} XY dmathbf{P}$ which, in turn, equals $$intlimits_mathrm{E} mathbf{E}(XY mid Z_1, ldots, Z_n) dmathbf{P} = intlimits_mathrm{E} dfrac{mathbf{E}(XY mid Z_1, ldots, Z_n)}{mathbf{E}(Y mid Z_1, ldots, Z_n)} mathbf{E}(Y mid Z_1, ldots, Z_n) dmathbf{P}.$$
          Apply the result of a. to the probability space $(Omega, mathscr{X}, mathbf{P})$ using $mathbf{E}(Y mid Z_1, ldots, Z_n)$ in lieu of $Y$ (recall $mathbf{E}(mathbf{E}(Y mid Z_1, ldots, Z_n)) = mathbf{E}(Y)=1$ and $mathbf{E}(Y mid Z_1, ldots, Z_n) geq 0$) to get $$intlimits_{mathrm{E}} mathbf{E}^*(X mid Z_1, ldots, Z_n) dmathbf{P}^* = intlimits_mathrm{E} dfrac{mathbf{E}(XY mid Z_1, ldots, Z_n)}{mathbf{E}(Y mid Z_1, ldots, Z_n)} dmathbf{P}^{**},$$ where $mathbf{P}^{**}$ is the probability measure on $mathscr{X}$ such that $mathbf{P}^{**}(mathrm{A}) = intlimits_mathrm{A} mathbf{E}(Y mid Z_1, ldots, Z_n) dmathbf{P}$; by definition of conditional expectation, $mathbf{P}^{**}(mathrm{A})=mathbf{P}^*(mathrm{A})$ for all $mathrm{A} in mathscr{X},$ that is $mathbf{P}^{**} = mathbf{P}^*.$ This substantiates b. Q.E.D.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for such an elegant proof idea! There is a typo at the end I believe. It should be $mathbf{P}^{**}(mathrm{A})=mathbf{P^*}(mathrm{A})$ right?
            $endgroup$
            – Lohith-kumar
            Dec 4 '18 at 1:40










          • $begingroup$
            Yes. Corrected, thanks.
            $endgroup$
            – Will M.
            Dec 4 '18 at 1:48










          • $begingroup$
            Is it possible for you to take a look at this question math.stackexchange.com/questions/3025024/… as well? Thanks!
            $endgroup$
            – Lohith-kumar
            Dec 4 '18 at 2:09










          • $begingroup$
            Yes, modify my argument here from a.
            $endgroup$
            – Will M.
            Dec 4 '18 at 2:23






          • 1




            $begingroup$
            I am actually not sure what you mean. I am sure, however, that the argument of a. here can be modified to prove that exercise.
            $endgroup$
            – Will M.
            Dec 4 '18 at 2:29














          1












          1








          1





          $begingroup$

          a. Seems just flawed. For instance, what does $dw$ even mean? Try the following. Consider $mathscr{H}_+$ the set of random variables $X geq 0$ for which $mathbf{E}^*(X) = mathbf{E}(XY).$ Then $mathscr{H}_+$ contains all indicator functions of events (by definition), is a closed under addition, positive-scalars multiplication and closed under monotone limits. Hence, $mathscr{H}_+$ contains all non-negative random variables (for we recall if $X$ is such a random variable, then $X$ equals, a.s., a monotone limit of simple positive r.v.). Now, in regards as to why $X$ has to be non-negative, it is just for the expectations to exist. You could also ask $X$ to be bounded.



          b. Consider now the sigma algebra $mathscr{X}$ generated by $Z_1, ldots, Z_n.$ By definition, $mathbf{E}^*(X mid Z_1, ldots, Z_n)$ is any random variable, measurable with respect to $mathscr{X},$ for which the following relation holds: if $mathrm{E} in mathscr{X}$ then $$intlimits_{mathrm{E}} mathbf{E}^*(X mid Z_1, ldots, Z_n) dmathbf{P}^* = intlimits_{mathrm{E}} X dmathbf{P}^*.$$
          However, by virtue of a., we know that the right hand side above equals $intlimits_mathrm{E} XY dmathbf{P}$ which, in turn, equals $$intlimits_mathrm{E} mathbf{E}(XY mid Z_1, ldots, Z_n) dmathbf{P} = intlimits_mathrm{E} dfrac{mathbf{E}(XY mid Z_1, ldots, Z_n)}{mathbf{E}(Y mid Z_1, ldots, Z_n)} mathbf{E}(Y mid Z_1, ldots, Z_n) dmathbf{P}.$$
          Apply the result of a. to the probability space $(Omega, mathscr{X}, mathbf{P})$ using $mathbf{E}(Y mid Z_1, ldots, Z_n)$ in lieu of $Y$ (recall $mathbf{E}(mathbf{E}(Y mid Z_1, ldots, Z_n)) = mathbf{E}(Y)=1$ and $mathbf{E}(Y mid Z_1, ldots, Z_n) geq 0$) to get $$intlimits_{mathrm{E}} mathbf{E}^*(X mid Z_1, ldots, Z_n) dmathbf{P}^* = intlimits_mathrm{E} dfrac{mathbf{E}(XY mid Z_1, ldots, Z_n)}{mathbf{E}(Y mid Z_1, ldots, Z_n)} dmathbf{P}^{**},$$ where $mathbf{P}^{**}$ is the probability measure on $mathscr{X}$ such that $mathbf{P}^{**}(mathrm{A}) = intlimits_mathrm{A} mathbf{E}(Y mid Z_1, ldots, Z_n) dmathbf{P}$; by definition of conditional expectation, $mathbf{P}^{**}(mathrm{A})=mathbf{P}^*(mathrm{A})$ for all $mathrm{A} in mathscr{X},$ that is $mathbf{P}^{**} = mathbf{P}^*.$ This substantiates b. Q.E.D.






          share|cite|improve this answer











          $endgroup$



          a. Seems just flawed. For instance, what does $dw$ even mean? Try the following. Consider $mathscr{H}_+$ the set of random variables $X geq 0$ for which $mathbf{E}^*(X) = mathbf{E}(XY).$ Then $mathscr{H}_+$ contains all indicator functions of events (by definition), is a closed under addition, positive-scalars multiplication and closed under monotone limits. Hence, $mathscr{H}_+$ contains all non-negative random variables (for we recall if $X$ is such a random variable, then $X$ equals, a.s., a monotone limit of simple positive r.v.). Now, in regards as to why $X$ has to be non-negative, it is just for the expectations to exist. You could also ask $X$ to be bounded.



          b. Consider now the sigma algebra $mathscr{X}$ generated by $Z_1, ldots, Z_n.$ By definition, $mathbf{E}^*(X mid Z_1, ldots, Z_n)$ is any random variable, measurable with respect to $mathscr{X},$ for which the following relation holds: if $mathrm{E} in mathscr{X}$ then $$intlimits_{mathrm{E}} mathbf{E}^*(X mid Z_1, ldots, Z_n) dmathbf{P}^* = intlimits_{mathrm{E}} X dmathbf{P}^*.$$
          However, by virtue of a., we know that the right hand side above equals $intlimits_mathrm{E} XY dmathbf{P}$ which, in turn, equals $$intlimits_mathrm{E} mathbf{E}(XY mid Z_1, ldots, Z_n) dmathbf{P} = intlimits_mathrm{E} dfrac{mathbf{E}(XY mid Z_1, ldots, Z_n)}{mathbf{E}(Y mid Z_1, ldots, Z_n)} mathbf{E}(Y mid Z_1, ldots, Z_n) dmathbf{P}.$$
          Apply the result of a. to the probability space $(Omega, mathscr{X}, mathbf{P})$ using $mathbf{E}(Y mid Z_1, ldots, Z_n)$ in lieu of $Y$ (recall $mathbf{E}(mathbf{E}(Y mid Z_1, ldots, Z_n)) = mathbf{E}(Y)=1$ and $mathbf{E}(Y mid Z_1, ldots, Z_n) geq 0$) to get $$intlimits_{mathrm{E}} mathbf{E}^*(X mid Z_1, ldots, Z_n) dmathbf{P}^* = intlimits_mathrm{E} dfrac{mathbf{E}(XY mid Z_1, ldots, Z_n)}{mathbf{E}(Y mid Z_1, ldots, Z_n)} dmathbf{P}^{**},$$ where $mathbf{P}^{**}$ is the probability measure on $mathscr{X}$ such that $mathbf{P}^{**}(mathrm{A}) = intlimits_mathrm{A} mathbf{E}(Y mid Z_1, ldots, Z_n) dmathbf{P}$; by definition of conditional expectation, $mathbf{P}^{**}(mathrm{A})=mathbf{P}^*(mathrm{A})$ for all $mathrm{A} in mathscr{X},$ that is $mathbf{P}^{**} = mathbf{P}^*.$ This substantiates b. Q.E.D.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 1:48

























          answered Dec 4 '18 at 1:08









          Will M.Will M.

          2,440314




          2,440314












          • $begingroup$
            Thanks for such an elegant proof idea! There is a typo at the end I believe. It should be $mathbf{P}^{**}(mathrm{A})=mathbf{P^*}(mathrm{A})$ right?
            $endgroup$
            – Lohith-kumar
            Dec 4 '18 at 1:40










          • $begingroup$
            Yes. Corrected, thanks.
            $endgroup$
            – Will M.
            Dec 4 '18 at 1:48










          • $begingroup$
            Is it possible for you to take a look at this question math.stackexchange.com/questions/3025024/… as well? Thanks!
            $endgroup$
            – Lohith-kumar
            Dec 4 '18 at 2:09










          • $begingroup$
            Yes, modify my argument here from a.
            $endgroup$
            – Will M.
            Dec 4 '18 at 2:23






          • 1




            $begingroup$
            I am actually not sure what you mean. I am sure, however, that the argument of a. here can be modified to prove that exercise.
            $endgroup$
            – Will M.
            Dec 4 '18 at 2:29


















          • $begingroup$
            Thanks for such an elegant proof idea! There is a typo at the end I believe. It should be $mathbf{P}^{**}(mathrm{A})=mathbf{P^*}(mathrm{A})$ right?
            $endgroup$
            – Lohith-kumar
            Dec 4 '18 at 1:40










          • $begingroup$
            Yes. Corrected, thanks.
            $endgroup$
            – Will M.
            Dec 4 '18 at 1:48










          • $begingroup$
            Is it possible for you to take a look at this question math.stackexchange.com/questions/3025024/… as well? Thanks!
            $endgroup$
            – Lohith-kumar
            Dec 4 '18 at 2:09










          • $begingroup$
            Yes, modify my argument here from a.
            $endgroup$
            – Will M.
            Dec 4 '18 at 2:23






          • 1




            $begingroup$
            I am actually not sure what you mean. I am sure, however, that the argument of a. here can be modified to prove that exercise.
            $endgroup$
            – Will M.
            Dec 4 '18 at 2:29
















          $begingroup$
          Thanks for such an elegant proof idea! There is a typo at the end I believe. It should be $mathbf{P}^{**}(mathrm{A})=mathbf{P^*}(mathrm{A})$ right?
          $endgroup$
          – Lohith-kumar
          Dec 4 '18 at 1:40




          $begingroup$
          Thanks for such an elegant proof idea! There is a typo at the end I believe. It should be $mathbf{P}^{**}(mathrm{A})=mathbf{P^*}(mathrm{A})$ right?
          $endgroup$
          – Lohith-kumar
          Dec 4 '18 at 1:40












          $begingroup$
          Yes. Corrected, thanks.
          $endgroup$
          – Will M.
          Dec 4 '18 at 1:48




          $begingroup$
          Yes. Corrected, thanks.
          $endgroup$
          – Will M.
          Dec 4 '18 at 1:48












          $begingroup$
          Is it possible for you to take a look at this question math.stackexchange.com/questions/3025024/… as well? Thanks!
          $endgroup$
          – Lohith-kumar
          Dec 4 '18 at 2:09




          $begingroup$
          Is it possible for you to take a look at this question math.stackexchange.com/questions/3025024/… as well? Thanks!
          $endgroup$
          – Lohith-kumar
          Dec 4 '18 at 2:09












          $begingroup$
          Yes, modify my argument here from a.
          $endgroup$
          – Will M.
          Dec 4 '18 at 2:23




          $begingroup$
          Yes, modify my argument here from a.
          $endgroup$
          – Will M.
          Dec 4 '18 at 2:23




          1




          1




          $begingroup$
          I am actually not sure what you mean. I am sure, however, that the argument of a. here can be modified to prove that exercise.
          $endgroup$
          – Will M.
          Dec 4 '18 at 2:29




          $begingroup$
          I am actually not sure what you mean. I am sure, however, that the argument of a. here can be modified to prove that exercise.
          $endgroup$
          – Will M.
          Dec 4 '18 at 2:29


















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