Why is the conditional expectation of an $L^{p}$-function again in $L^{p}$?












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Let $(Omega, mathcal{A},P)$ be a probability space and let $X,Ycolon Omega rightarrow mathbb{R}$ be random variables. Furthermore, let $Y$ be $p$-integrable. Then why is the conditional expectation $mathbb{E}[Y vert X]$ again necessarily $p$-integrable?



Kind regards and thank you very much!






probability probability-theory measure-theory lp-spaces conditional-expectation







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    $begingroup$


    Let $(Omega, mathcal{A},P)$ be a probability space and let $X,Ycolon Omega rightarrow mathbb{R}$ be random variables. Furthermore, let $Y$ be $p$-integrable. Then why is the conditional expectation $mathbb{E}[Y vert X]$ again necessarily $p$-integrable?



    Kind regards and thank you very much!






    probability probability-theory measure-theory lp-spaces conditional-expectation







    share|cite|improve this question











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      1








      1





      $begingroup$


      Let $(Omega, mathcal{A},P)$ be a probability space and let $X,Ycolon Omega rightarrow mathbb{R}$ be random variables. Furthermore, let $Y$ be $p$-integrable. Then why is the conditional expectation $mathbb{E}[Y vert X]$ again necessarily $p$-integrable?



      Kind regards and thank you very much!






      probability probability-theory measure-theory lp-spaces conditional-expectation







      share|cite|improve this question











      $endgroup$




      Let $(Omega, mathcal{A},P)$ be a probability space and let $X,Ycolon Omega rightarrow mathbb{R}$ be random variables. Furthermore, let $Y$ be $p$-integrable. Then why is the conditional expectation $mathbb{E}[Y vert X]$ again necessarily $p$-integrable?



      Kind regards and thank you very much!






      probability probability-theory measure-theory lp-spaces conditional-expectation




      probability probability-theory measure-theory lp-spaces conditional-expectation






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      edited Jan 25 at 18:52









      Davide Giraudo

      127k17154268




      127k17154268










      asked Jan 25 at 15:21









      Joker123Joker123

      537212




      537212






















          1 Answer
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          $begingroup$

          By conditional Jensen's inequality and from the fact that $xmapsto |x|^p$ is convex for $pge 1$, we find that
          $$
          left|Bbb Eleft[Ybig|Xright]right|^pleBbb Eleft[|Y|^pbig|Xright],quadtext{a.s.}
          $$
          It follows that
          $$
          Bbb Eleft[left|Bbb Eleft[Ybig|Xright]right|^pright]le Bbb Eleft[Bbb Eleft[|Y|^pbig|Xright]right]=Bbb Eleft[|Y|^pright]<infty.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Fantastic! Thanks.
            $endgroup$
            – Joker123
            Jan 25 at 15:45










          • $begingroup$
            Does this also hold if $X$ and $Y$ map into $mathbb{R}^n$? In this case, I can only see how it holds for $p = 2$.
            $endgroup$
            – Joker123
            Jan 25 at 16:11












          • $begingroup$
            I can make sure that the Jensen's inequality holds for any convex(concave) function $psi:Bbb R^n to Bbb R$, but I cannot find a reference to give you ... But we can prove it componentwise: i.e. for $mathbf{Y}=(Y_1,Y_2,ldots , Y_n)in L^p$, we find that each $Y_iin L^p Longrightarrow Bbb E[Y_i |mathbf{X}]in L^p$ and thus $Bbb E[mathbf{Y}|mathbf{X}]=(Bbb E[Y_1|mathbf{X}],ldots,Bbb E[Y_n|mathbf{X}])in L^p$.
            $endgroup$
            – Song
            Jan 25 at 16:31











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          $begingroup$

          By conditional Jensen's inequality and from the fact that $xmapsto |x|^p$ is convex for $pge 1$, we find that
          $$
          left|Bbb Eleft[Ybig|Xright]right|^pleBbb Eleft[|Y|^pbig|Xright],quadtext{a.s.}
          $$
          It follows that
          $$
          Bbb Eleft[left|Bbb Eleft[Ybig|Xright]right|^pright]le Bbb Eleft[Bbb Eleft[|Y|^pbig|Xright]right]=Bbb Eleft[|Y|^pright]<infty.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Fantastic! Thanks.
            $endgroup$
            – Joker123
            Jan 25 at 15:45










          • $begingroup$
            Does this also hold if $X$ and $Y$ map into $mathbb{R}^n$? In this case, I can only see how it holds for $p = 2$.
            $endgroup$
            – Joker123
            Jan 25 at 16:11












          • $begingroup$
            I can make sure that the Jensen's inequality holds for any convex(concave) function $psi:Bbb R^n to Bbb R$, but I cannot find a reference to give you ... But we can prove it componentwise: i.e. for $mathbf{Y}=(Y_1,Y_2,ldots , Y_n)in L^p$, we find that each $Y_iin L^p Longrightarrow Bbb E[Y_i |mathbf{X}]in L^p$ and thus $Bbb E[mathbf{Y}|mathbf{X}]=(Bbb E[Y_1|mathbf{X}],ldots,Bbb E[Y_n|mathbf{X}])in L^p$.
            $endgroup$
            – Song
            Jan 25 at 16:31
















          4












          $begingroup$

          By conditional Jensen's inequality and from the fact that $xmapsto |x|^p$ is convex for $pge 1$, we find that
          $$
          left|Bbb Eleft[Ybig|Xright]right|^pleBbb Eleft[|Y|^pbig|Xright],quadtext{a.s.}
          $$
          It follows that
          $$
          Bbb Eleft[left|Bbb Eleft[Ybig|Xright]right|^pright]le Bbb Eleft[Bbb Eleft[|Y|^pbig|Xright]right]=Bbb Eleft[|Y|^pright]<infty.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Fantastic! Thanks.
            $endgroup$
            – Joker123
            Jan 25 at 15:45










          • $begingroup$
            Does this also hold if $X$ and $Y$ map into $mathbb{R}^n$? In this case, I can only see how it holds for $p = 2$.
            $endgroup$
            – Joker123
            Jan 25 at 16:11












          • $begingroup$
            I can make sure that the Jensen's inequality holds for any convex(concave) function $psi:Bbb R^n to Bbb R$, but I cannot find a reference to give you ... But we can prove it componentwise: i.e. for $mathbf{Y}=(Y_1,Y_2,ldots , Y_n)in L^p$, we find that each $Y_iin L^p Longrightarrow Bbb E[Y_i |mathbf{X}]in L^p$ and thus $Bbb E[mathbf{Y}|mathbf{X}]=(Bbb E[Y_1|mathbf{X}],ldots,Bbb E[Y_n|mathbf{X}])in L^p$.
            $endgroup$
            – Song
            Jan 25 at 16:31














          4












          4








          4





          $begingroup$

          By conditional Jensen's inequality and from the fact that $xmapsto |x|^p$ is convex for $pge 1$, we find that
          $$
          left|Bbb Eleft[Ybig|Xright]right|^pleBbb Eleft[|Y|^pbig|Xright],quadtext{a.s.}
          $$
          It follows that
          $$
          Bbb Eleft[left|Bbb Eleft[Ybig|Xright]right|^pright]le Bbb Eleft[Bbb Eleft[|Y|^pbig|Xright]right]=Bbb Eleft[|Y|^pright]<infty.
          $$






          share|cite|improve this answer









          $endgroup$



          By conditional Jensen's inequality and from the fact that $xmapsto |x|^p$ is convex for $pge 1$, we find that
          $$
          left|Bbb Eleft[Ybig|Xright]right|^pleBbb Eleft[|Y|^pbig|Xright],quadtext{a.s.}
          $$
          It follows that
          $$
          Bbb Eleft[left|Bbb Eleft[Ybig|Xright]right|^pright]le Bbb Eleft[Bbb Eleft[|Y|^pbig|Xright]right]=Bbb Eleft[|Y|^pright]<infty.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 25 at 15:37









          SongSong

          17.2k21246




          17.2k21246












          • $begingroup$
            Fantastic! Thanks.
            $endgroup$
            – Joker123
            Jan 25 at 15:45










          • $begingroup$
            Does this also hold if $X$ and $Y$ map into $mathbb{R}^n$? In this case, I can only see how it holds for $p = 2$.
            $endgroup$
            – Joker123
            Jan 25 at 16:11












          • $begingroup$
            I can make sure that the Jensen's inequality holds for any convex(concave) function $psi:Bbb R^n to Bbb R$, but I cannot find a reference to give you ... But we can prove it componentwise: i.e. for $mathbf{Y}=(Y_1,Y_2,ldots , Y_n)in L^p$, we find that each $Y_iin L^p Longrightarrow Bbb E[Y_i |mathbf{X}]in L^p$ and thus $Bbb E[mathbf{Y}|mathbf{X}]=(Bbb E[Y_1|mathbf{X}],ldots,Bbb E[Y_n|mathbf{X}])in L^p$.
            $endgroup$
            – Song
            Jan 25 at 16:31


















          • $begingroup$
            Fantastic! Thanks.
            $endgroup$
            – Joker123
            Jan 25 at 15:45










          • $begingroup$
            Does this also hold if $X$ and $Y$ map into $mathbb{R}^n$? In this case, I can only see how it holds for $p = 2$.
            $endgroup$
            – Joker123
            Jan 25 at 16:11












          • $begingroup$
            I can make sure that the Jensen's inequality holds for any convex(concave) function $psi:Bbb R^n to Bbb R$, but I cannot find a reference to give you ... But we can prove it componentwise: i.e. for $mathbf{Y}=(Y_1,Y_2,ldots , Y_n)in L^p$, we find that each $Y_iin L^p Longrightarrow Bbb E[Y_i |mathbf{X}]in L^p$ and thus $Bbb E[mathbf{Y}|mathbf{X}]=(Bbb E[Y_1|mathbf{X}],ldots,Bbb E[Y_n|mathbf{X}])in L^p$.
            $endgroup$
            – Song
            Jan 25 at 16:31
















          $begingroup$
          Fantastic! Thanks.
          $endgroup$
          – Joker123
          Jan 25 at 15:45




          $begingroup$
          Fantastic! Thanks.
          $endgroup$
          – Joker123
          Jan 25 at 15:45












          $begingroup$
          Does this also hold if $X$ and $Y$ map into $mathbb{R}^n$? In this case, I can only see how it holds for $p = 2$.
          $endgroup$
          – Joker123
          Jan 25 at 16:11






          $begingroup$
          Does this also hold if $X$ and $Y$ map into $mathbb{R}^n$? In this case, I can only see how it holds for $p = 2$.
          $endgroup$
          – Joker123
          Jan 25 at 16:11














          $begingroup$
          I can make sure that the Jensen's inequality holds for any convex(concave) function $psi:Bbb R^n to Bbb R$, but I cannot find a reference to give you ... But we can prove it componentwise: i.e. for $mathbf{Y}=(Y_1,Y_2,ldots , Y_n)in L^p$, we find that each $Y_iin L^p Longrightarrow Bbb E[Y_i |mathbf{X}]in L^p$ and thus $Bbb E[mathbf{Y}|mathbf{X}]=(Bbb E[Y_1|mathbf{X}],ldots,Bbb E[Y_n|mathbf{X}])in L^p$.
          $endgroup$
          – Song
          Jan 25 at 16:31




          $begingroup$
          I can make sure that the Jensen's inequality holds for any convex(concave) function $psi:Bbb R^n to Bbb R$, but I cannot find a reference to give you ... But we can prove it componentwise: i.e. for $mathbf{Y}=(Y_1,Y_2,ldots , Y_n)in L^p$, we find that each $Y_iin L^p Longrightarrow Bbb E[Y_i |mathbf{X}]in L^p$ and thus $Bbb E[mathbf{Y}|mathbf{X}]=(Bbb E[Y_1|mathbf{X}],ldots,Bbb E[Y_n|mathbf{X}])in L^p$.
          $endgroup$
          – Song
          Jan 25 at 16:31


















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