$mathrm {Var}(X)=mathrm E(X|mathcal B)+mathrm E[X-mathrm E(X|mathcal B)]^2Rightarrow mathrm {Var}(E(X|mathcal...












0












$begingroup$


Let $X, Y in L^2$ on the probability space $(Omega, mathcal A, P)$ with the sub-$sigma$-algebra $mathcal B subseteq mathcal A$.



How can I show $mathrm {Var}(X)=mathrm E(X|mathcal B)+mathrm E[X-mathrm E(X|mathcal B)]^2Rightarrow mathrm {Var}(E(X|mathcal B)))lemathrm {Var}(X)$






measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can $mathbf{V}mathrm{ar}(X) = mathbf{E}(X mid mathscr{B}) + mathbf{E}(X - mathbf{E}(X mid mathscr{B}))^2$? I mean, the variance is a number and it is not random.
    $endgroup$
    – Will M.
    Dec 18 '18 at 19:28
















0












$begingroup$


Let $X, Y in L^2$ on the probability space $(Omega, mathcal A, P)$ with the sub-$sigma$-algebra $mathcal B subseteq mathcal A$.



How can I show $mathrm {Var}(X)=mathrm E(X|mathcal B)+mathrm E[X-mathrm E(X|mathcal B)]^2Rightarrow mathrm {Var}(E(X|mathcal B)))lemathrm {Var}(X)$






measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can $mathbf{V}mathrm{ar}(X) = mathbf{E}(X mid mathscr{B}) + mathbf{E}(X - mathbf{E}(X mid mathscr{B}))^2$? I mean, the variance is a number and it is not random.
    $endgroup$
    – Will M.
    Dec 18 '18 at 19:28














0












0








0


0



$begingroup$


Let $X, Y in L^2$ on the probability space $(Omega, mathcal A, P)$ with the sub-$sigma$-algebra $mathcal B subseteq mathcal A$.



How can I show $mathrm {Var}(X)=mathrm E(X|mathcal B)+mathrm E[X-mathrm E(X|mathcal B)]^2Rightarrow mathrm {Var}(E(X|mathcal B)))lemathrm {Var}(X)$






measure-theory







share|cite|improve this question









$endgroup$




Let $X, Y in L^2$ on the probability space $(Omega, mathcal A, P)$ with the sub-$sigma$-algebra $mathcal B subseteq mathcal A$.



How can I show $mathrm {Var}(X)=mathrm E(X|mathcal B)+mathrm E[X-mathrm E(X|mathcal B)]^2Rightarrow mathrm {Var}(E(X|mathcal B)))lemathrm {Var}(X)$






measure-theory




measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 19:25









user626880user626880

204




204












  • $begingroup$
    Can $mathbf{V}mathrm{ar}(X) = mathbf{E}(X mid mathscr{B}) + mathbf{E}(X - mathbf{E}(X mid mathscr{B}))^2$? I mean, the variance is a number and it is not random.
    $endgroup$
    – Will M.
    Dec 18 '18 at 19:28


















  • $begingroup$
    Can $mathbf{V}mathrm{ar}(X) = mathbf{E}(X mid mathscr{B}) + mathbf{E}(X - mathbf{E}(X mid mathscr{B}))^2$? I mean, the variance is a number and it is not random.
    $endgroup$
    – Will M.
    Dec 18 '18 at 19:28
















$begingroup$
Can $mathbf{V}mathrm{ar}(X) = mathbf{E}(X mid mathscr{B}) + mathbf{E}(X - mathbf{E}(X mid mathscr{B}))^2$? I mean, the variance is a number and it is not random.
$endgroup$
– Will M.
Dec 18 '18 at 19:28




$begingroup$
Can $mathbf{V}mathrm{ar}(X) = mathbf{E}(X mid mathscr{B}) + mathbf{E}(X - mathbf{E}(X mid mathscr{B}))^2$? I mean, the variance is a number and it is not random.
$endgroup$
– Will M.
Dec 18 '18 at 19:28










1 Answer
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1












$begingroup$

I am not sure what you meant in your expression for $Var[X]$. But to show $Var[E[X|mathcal{B}]]leq Var[X]$ you can use the law of total variance. That is
$$Var[X]=Var[E[X|mathcal{B}]]+E[Var[X|mathcal{B}]].$$
The above implies
$$Var[X]geq Var[E[X|mathcal{B}]] iff E[Var[X|mathcal{B}]]geq 0.$$



Now to show the latter, let us write the definition of conditional variance: $Var[X|mathcal{B}] = E[(X-E[X|mathcal{B}])^2|mathcal{B}]$ which is the expectation of a non-negative random variable and hence must be $P$-a.s. non-negative. Hence, $Var[X|mathcal{B}]$ is a $P$-a.s. non-negative random variable which implies that its expectation must be non-negative.



If you were not allowed to use the law of total variance I provide you here with a proof of it: On one hand,
$$Var[E[X|mathcal{B}]]=E[E(X|mathcal{B})^2]+E[E[X|mathcal{B}]]^2 = E[(X|mathcal{B})^2]+E[X]^2.$$
On the other hand,
$$E[Var[X|mathcal{B}]]=Eleft[ E[(X-E[X|mathcal{B}])^2|mathcal{B}]right]=dots=E[X^2]-E[E[X|mathcal{B}]^2].$$
Then both together gives you the desired identity.






share|cite|improve this answer











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    $begingroup$

    I am not sure what you meant in your expression for $Var[X]$. But to show $Var[E[X|mathcal{B}]]leq Var[X]$ you can use the law of total variance. That is
    $$Var[X]=Var[E[X|mathcal{B}]]+E[Var[X|mathcal{B}]].$$
    The above implies
    $$Var[X]geq Var[E[X|mathcal{B}]] iff E[Var[X|mathcal{B}]]geq 0.$$



    Now to show the latter, let us write the definition of conditional variance: $Var[X|mathcal{B}] = E[(X-E[X|mathcal{B}])^2|mathcal{B}]$ which is the expectation of a non-negative random variable and hence must be $P$-a.s. non-negative. Hence, $Var[X|mathcal{B}]$ is a $P$-a.s. non-negative random variable which implies that its expectation must be non-negative.



    If you were not allowed to use the law of total variance I provide you here with a proof of it: On one hand,
    $$Var[E[X|mathcal{B}]]=E[E(X|mathcal{B})^2]+E[E[X|mathcal{B}]]^2 = E[(X|mathcal{B})^2]+E[X]^2.$$
    On the other hand,
    $$E[Var[X|mathcal{B}]]=Eleft[ E[(X-E[X|mathcal{B}])^2|mathcal{B}]right]=dots=E[X^2]-E[E[X|mathcal{B}]^2].$$
    Then both together gives you the desired identity.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      I am not sure what you meant in your expression for $Var[X]$. But to show $Var[E[X|mathcal{B}]]leq Var[X]$ you can use the law of total variance. That is
      $$Var[X]=Var[E[X|mathcal{B}]]+E[Var[X|mathcal{B}]].$$
      The above implies
      $$Var[X]geq Var[E[X|mathcal{B}]] iff E[Var[X|mathcal{B}]]geq 0.$$



      Now to show the latter, let us write the definition of conditional variance: $Var[X|mathcal{B}] = E[(X-E[X|mathcal{B}])^2|mathcal{B}]$ which is the expectation of a non-negative random variable and hence must be $P$-a.s. non-negative. Hence, $Var[X|mathcal{B}]$ is a $P$-a.s. non-negative random variable which implies that its expectation must be non-negative.



      If you were not allowed to use the law of total variance I provide you here with a proof of it: On one hand,
      $$Var[E[X|mathcal{B}]]=E[E(X|mathcal{B})^2]+E[E[X|mathcal{B}]]^2 = E[(X|mathcal{B})^2]+E[X]^2.$$
      On the other hand,
      $$E[Var[X|mathcal{B}]]=Eleft[ E[(X-E[X|mathcal{B}])^2|mathcal{B}]right]=dots=E[X^2]-E[E[X|mathcal{B}]^2].$$
      Then both together gives you the desired identity.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        I am not sure what you meant in your expression for $Var[X]$. But to show $Var[E[X|mathcal{B}]]leq Var[X]$ you can use the law of total variance. That is
        $$Var[X]=Var[E[X|mathcal{B}]]+E[Var[X|mathcal{B}]].$$
        The above implies
        $$Var[X]geq Var[E[X|mathcal{B}]] iff E[Var[X|mathcal{B}]]geq 0.$$



        Now to show the latter, let us write the definition of conditional variance: $Var[X|mathcal{B}] = E[(X-E[X|mathcal{B}])^2|mathcal{B}]$ which is the expectation of a non-negative random variable and hence must be $P$-a.s. non-negative. Hence, $Var[X|mathcal{B}]$ is a $P$-a.s. non-negative random variable which implies that its expectation must be non-negative.



        If you were not allowed to use the law of total variance I provide you here with a proof of it: On one hand,
        $$Var[E[X|mathcal{B}]]=E[E(X|mathcal{B})^2]+E[E[X|mathcal{B}]]^2 = E[(X|mathcal{B})^2]+E[X]^2.$$
        On the other hand,
        $$E[Var[X|mathcal{B}]]=Eleft[ E[(X-E[X|mathcal{B}])^2|mathcal{B}]right]=dots=E[X^2]-E[E[X|mathcal{B}]^2].$$
        Then both together gives you the desired identity.






        share|cite|improve this answer











        $endgroup$



        I am not sure what you meant in your expression for $Var[X]$. But to show $Var[E[X|mathcal{B}]]leq Var[X]$ you can use the law of total variance. That is
        $$Var[X]=Var[E[X|mathcal{B}]]+E[Var[X|mathcal{B}]].$$
        The above implies
        $$Var[X]geq Var[E[X|mathcal{B}]] iff E[Var[X|mathcal{B}]]geq 0.$$



        Now to show the latter, let us write the definition of conditional variance: $Var[X|mathcal{B}] = E[(X-E[X|mathcal{B}])^2|mathcal{B}]$ which is the expectation of a non-negative random variable and hence must be $P$-a.s. non-negative. Hence, $Var[X|mathcal{B}]$ is a $P$-a.s. non-negative random variable which implies that its expectation must be non-negative.



        If you were not allowed to use the law of total variance I provide you here with a proof of it: On one hand,
        $$Var[E[X|mathcal{B}]]=E[E(X|mathcal{B})^2]+E[E[X|mathcal{B}]]^2 = E[(X|mathcal{B})^2]+E[X]^2.$$
        On the other hand,
        $$E[Var[X|mathcal{B}]]=Eleft[ E[(X-E[X|mathcal{B}])^2|mathcal{B}]right]=dots=E[X^2]-E[E[X|mathcal{B}]^2].$$
        Then both together gives you the desired identity.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 18 '18 at 19:45

























        answered Dec 18 '18 at 19:37









        MartingaloMartingalo

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        706618






























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