Conditional mean and variance of $X$ given that $Y=6$ for $X$ normal $N(0,1)$ and $Y$ conditionally on $X=x$...












-1












$begingroup$



Assume that $Xsim N(0,1)$ and that, given $X=x$, the conditional distribution of $Ymid X=x$ is $N(x,1)$.
Find the conditional mean and variance of $Xmid Y=6$.




I proceeded as follows:
$$f_{X|Y}(X|Y=6)= frac{f_X(x)f_{Y|X}(Y|X=x)}{int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}$$
I know that $$f_X(x)f_{Y|X}(Y|X=x)=frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}$$
But
$${int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}=int^{infty}_{-infty}frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}dx$$



And that is where I am stuck.






probability-theory probability-distributions conditional-expectation variance expected-value







share|cite|improve this question











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    -1












    $begingroup$



    Assume that $Xsim N(0,1)$ and that, given $X=x$, the conditional distribution of $Ymid X=x$ is $N(x,1)$.
    Find the conditional mean and variance of $Xmid Y=6$.




    I proceeded as follows:
    $$f_{X|Y}(X|Y=6)= frac{f_X(x)f_{Y|X}(Y|X=x)}{int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}$$
    I know that $$f_X(x)f_{Y|X}(Y|X=x)=frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}$$
    But
    $${int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}=int^{infty}_{-infty}frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}dx$$



    And that is where I am stuck.






    probability-theory probability-distributions conditional-expectation variance expected-value







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1


      0



      $begingroup$



      Assume that $Xsim N(0,1)$ and that, given $X=x$, the conditional distribution of $Ymid X=x$ is $N(x,1)$.
      Find the conditional mean and variance of $Xmid Y=6$.




      I proceeded as follows:
      $$f_{X|Y}(X|Y=6)= frac{f_X(x)f_{Y|X}(Y|X=x)}{int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}$$
      I know that $$f_X(x)f_{Y|X}(Y|X=x)=frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}$$
      But
      $${int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}=int^{infty}_{-infty}frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}dx$$



      And that is where I am stuck.






      probability-theory probability-distributions conditional-expectation variance expected-value







      share|cite|improve this question











      $endgroup$





      Assume that $Xsim N(0,1)$ and that, given $X=x$, the conditional distribution of $Ymid X=x$ is $N(x,1)$.
      Find the conditional mean and variance of $Xmid Y=6$.




      I proceeded as follows:
      $$f_{X|Y}(X|Y=6)= frac{f_X(x)f_{Y|X}(Y|X=x)}{int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}$$
      I know that $$f_X(x)f_{Y|X}(Y|X=x)=frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}$$
      But
      $${int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}=int^{infty}_{-infty}frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}dx$$



      And that is where I am stuck.






      probability-theory probability-distributions conditional-expectation variance expected-value




      probability-theory probability-distributions conditional-expectation variance expected-value






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 14:29









      Did

      248k23224463




      248k23224463










      asked Dec 5 '18 at 5:08









      Mahamad A. KanoutéMahamad A. Kanouté

      371113




      371113






















          1 Answer
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          $begingroup$

          One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=frac12Y+T$ where $T=frac12(X-Z)$.



          Since $(Y,T)$ is jointly normal, the fact that $mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.



          Now, $T$ is centered normal with variance $frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $frac12y$ and variance $frac12$.



          This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
            $endgroup$
            – Mahamad A. Kanouté
            Dec 14 '18 at 2:25






          • 1




            $begingroup$
            This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
            $endgroup$
            – Did
            Dec 14 '18 at 6:10











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          1 Answer
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          active

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          2












          $begingroup$

          One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=frac12Y+T$ where $T=frac12(X-Z)$.



          Since $(Y,T)$ is jointly normal, the fact that $mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.



          Now, $T$ is centered normal with variance $frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $frac12y$ and variance $frac12$.



          This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
            $endgroup$
            – Mahamad A. Kanouté
            Dec 14 '18 at 2:25






          • 1




            $begingroup$
            This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
            $endgroup$
            – Did
            Dec 14 '18 at 6:10
















          2












          $begingroup$

          One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=frac12Y+T$ where $T=frac12(X-Z)$.



          Since $(Y,T)$ is jointly normal, the fact that $mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.



          Now, $T$ is centered normal with variance $frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $frac12y$ and variance $frac12$.



          This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
            $endgroup$
            – Mahamad A. Kanouté
            Dec 14 '18 at 2:25






          • 1




            $begingroup$
            This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
            $endgroup$
            – Did
            Dec 14 '18 at 6:10














          2












          2








          2





          $begingroup$

          One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=frac12Y+T$ where $T=frac12(X-Z)$.



          Since $(Y,T)$ is jointly normal, the fact that $mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.



          Now, $T$ is centered normal with variance $frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $frac12y$ and variance $frac12$.



          This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.






          share|cite|improve this answer









          $endgroup$



          One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=frac12Y+T$ where $T=frac12(X-Z)$.



          Since $(Y,T)$ is jointly normal, the fact that $mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.



          Now, $T$ is centered normal with variance $frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $frac12y$ and variance $frac12$.



          This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 14:24









          DidDid

          248k23224463




          248k23224463












          • $begingroup$
            Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
            $endgroup$
            – Mahamad A. Kanouté
            Dec 14 '18 at 2:25






          • 1




            $begingroup$
            This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
            $endgroup$
            – Did
            Dec 14 '18 at 6:10


















          • $begingroup$
            Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
            $endgroup$
            – Mahamad A. Kanouté
            Dec 14 '18 at 2:25






          • 1




            $begingroup$
            This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
            $endgroup$
            – Did
            Dec 14 '18 at 6:10
















          $begingroup$
          Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
          $endgroup$
          – Mahamad A. Kanouté
          Dec 14 '18 at 2:25




          $begingroup$
          Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
          $endgroup$
          – Mahamad A. Kanouté
          Dec 14 '18 at 2:25




          1




          1




          $begingroup$
          This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
          $endgroup$
          – Did
          Dec 14 '18 at 6:10




          $begingroup$
          This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
          $endgroup$
          – Did
          Dec 14 '18 at 6:10


















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