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Is expected value $E[X_1;X_1leq X_2]= E[X_1;X_1


Find $E(X_1|X_2leq x_2, X_3leq x_3)$ where $(X_1,X_2,X_3)$ is multivariate normalConditional expectation: $E[X_1 X_2mid X_1 + X_2 X_3]$Probability and expectation of three ordered random variablesFind the value of $mathbbE(X_1+X_2+ldots+X_N)$ of i.i.d random variables $X_i$s.Verifying calculation inside an expected value problemLet $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(fracX_1+X_2X_1+X_2+X_3$)?Throw a dice-expected value.Finding conditional expectation $E[X_1 | X_2 = x_2]$Expected value- $E(1/3)^t$Expected value of $Z=X_1+X_2$ if $X_1<X_3$.and $Z=X_1$ if $X_3leq X_1$













0












$begingroup$


Let $X_1, X_2$ two independent random variables with PDF $f_X_i(x_i)$.
Is this formula is true
$$E[X_1;X_1leq X_2]=
int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$



I am asking if the expected value
$$E[X_1;X_1leq X_2]= E[X_1;X_1< X_2]$$,



or
$$E[X_1;X_1< X_2]=E[X_1]-E[X_1;X_1leq X_2].$$



Thanks










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    Let $X_1, X_2$ two independent random variables with PDF $f_X_i(x_i)$.
    Is this formula is true
    $$E[X_1;X_1leq X_2]=
    int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$



    I am asking if the expected value
    $$E[X_1;X_1leq X_2]= E[X_1;X_1< X_2]$$,



    or
    $$E[X_1;X_1< X_2]=E[X_1]-E[X_1;X_1leq X_2].$$



    Thanks










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      Let $X_1, X_2$ two independent random variables with PDF $f_X_i(x_i)$.
      Is this formula is true
      $$E[X_1;X_1leq X_2]=
      int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$



      I am asking if the expected value
      $$E[X_1;X_1leq X_2]= E[X_1;X_1< X_2]$$,



      or
      $$E[X_1;X_1< X_2]=E[X_1]-E[X_1;X_1leq X_2].$$



      Thanks










      share|cite|improve this question









      $endgroup$




      Let $X_1, X_2$ two independent random variables with PDF $f_X_i(x_i)$.
      Is this formula is true
      $$E[X_1;X_1leq X_2]=
      int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$



      I am asking if the expected value
      $$E[X_1;X_1leq X_2]= E[X_1;X_1< X_2]$$,



      or
      $$E[X_1;X_1< X_2]=E[X_1]-E[X_1;X_1leq X_2].$$



      Thanks







      conditional-expectation expected-value






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 days ago









      MonirMonir

      368




      368




















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          If $X_1$ and $X_2$ are independent random variables with a densities then $PX_1=X_2=0$ so it is true that $E(X_1;X_1 leq X_2)=E(X_1;X_1 < X_2)$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
            $endgroup$
            – Monir
            2 days ago










          • $begingroup$
            @Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
            $endgroup$
            – Kavi Rama Murthy
            2 days ago










          • $begingroup$
            Ok, thank you Prof Kavi Rama Murthy.
            $endgroup$
            – Monir
            2 days ago










          Your Answer





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          active

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          active

          oldest

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          1












          $begingroup$

          If $X_1$ and $X_2$ are independent random variables with a densities then $PX_1=X_2=0$ so it is true that $E(X_1;X_1 leq X_2)=E(X_1;X_1 < X_2)$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
            $endgroup$
            – Monir
            2 days ago










          • $begingroup$
            @Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
            $endgroup$
            – Kavi Rama Murthy
            2 days ago










          • $begingroup$
            Ok, thank you Prof Kavi Rama Murthy.
            $endgroup$
            – Monir
            2 days ago















          1












          $begingroup$

          If $X_1$ and $X_2$ are independent random variables with a densities then $PX_1=X_2=0$ so it is true that $E(X_1;X_1 leq X_2)=E(X_1;X_1 < X_2)$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
            $endgroup$
            – Monir
            2 days ago










          • $begingroup$
            @Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
            $endgroup$
            – Kavi Rama Murthy
            2 days ago










          • $begingroup$
            Ok, thank you Prof Kavi Rama Murthy.
            $endgroup$
            – Monir
            2 days ago













          1












          1








          1





          $begingroup$

          If $X_1$ and $X_2$ are independent random variables with a densities then $PX_1=X_2=0$ so it is true that $E(X_1;X_1 leq X_2)=E(X_1;X_1 < X_2)$.






          share|cite|improve this answer









          $endgroup$



          If $X_1$ and $X_2$ are independent random variables with a densities then $PX_1=X_2=0$ so it is true that $E(X_1;X_1 leq X_2)=E(X_1;X_1 < X_2)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Kavi Rama MurthyKavi Rama Murthy

          66.6k53067




          66.6k53067











          • $begingroup$
            Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
            $endgroup$
            – Monir
            2 days ago










          • $begingroup$
            @Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
            $endgroup$
            – Kavi Rama Murthy
            2 days ago










          • $begingroup$
            Ok, thank you Prof Kavi Rama Murthy.
            $endgroup$
            – Monir
            2 days ago
















          • $begingroup$
            Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
            $endgroup$
            – Monir
            2 days ago










          • $begingroup$
            @Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
            $endgroup$
            – Kavi Rama Murthy
            2 days ago










          • $begingroup$
            Ok, thank you Prof Kavi Rama Murthy.
            $endgroup$
            – Monir
            2 days ago















          $begingroup$
          Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
          $endgroup$
          – Monir
          2 days ago




          $begingroup$
          Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
          $endgroup$
          – Monir
          2 days ago












          $begingroup$
          @Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
          $endgroup$
          – Kavi Rama Murthy
          2 days ago




          $begingroup$
          @Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
          $endgroup$
          – Kavi Rama Murthy
          2 days ago












          $begingroup$
          Ok, thank you Prof Kavi Rama Murthy.
          $endgroup$
          – Monir
          2 days ago




          $begingroup$
          Ok, thank you Prof Kavi Rama Murthy.
          $endgroup$
          – Monir
          2 days ago

















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