Integrate $int_{-infty}^infty {rm erfc} left( frac{x}{sqrt{2}} right) e^{-frac{(x-mu)^2}{2 sigma^2}} dx$












1












$begingroup$


Does the integral
begin{align}
int_{-infty}^infty {rm erfc} left( frac{x}{sqrt{2}} right) e^{-frac{(x-mu)^2}{2 sigma^2}} dx
end{align}
have a close form expression.



I found that for $mu=0$



begin{align}
int_{-infty}^infty {rm erfc} left( frac{x}{sqrt{2}} right) e^{-frac{x^2}{2 sigma^2}} dx = sigma sqrt{2 pi}
end{align}



Can this be done for $mu neq 0$?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Does the integral
    begin{align}
    int_{-infty}^infty {rm erfc} left( frac{x}{sqrt{2}} right) e^{-frac{(x-mu)^2}{2 sigma^2}} dx
    end{align}
    have a close form expression.



    I found that for $mu=0$



    begin{align}
    int_{-infty}^infty {rm erfc} left( frac{x}{sqrt{2}} right) e^{-frac{x^2}{2 sigma^2}} dx = sigma sqrt{2 pi}
    end{align}



    Can this be done for $mu neq 0$?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Does the integral
      begin{align}
      int_{-infty}^infty {rm erfc} left( frac{x}{sqrt{2}} right) e^{-frac{(x-mu)^2}{2 sigma^2}} dx
      end{align}
      have a close form expression.



      I found that for $mu=0$



      begin{align}
      int_{-infty}^infty {rm erfc} left( frac{x}{sqrt{2}} right) e^{-frac{x^2}{2 sigma^2}} dx = sigma sqrt{2 pi}
      end{align}



      Can this be done for $mu neq 0$?










      share|cite|improve this question











      $endgroup$




      Does the integral
      begin{align}
      int_{-infty}^infty {rm erfc} left( frac{x}{sqrt{2}} right) e^{-frac{(x-mu)^2}{2 sigma^2}} dx
      end{align}
      have a close form expression.



      I found that for $mu=0$



      begin{align}
      int_{-infty}^infty {rm erfc} left( frac{x}{sqrt{2}} right) e^{-frac{x^2}{2 sigma^2}} dx = sigma sqrt{2 pi}
      end{align}



      Can this be done for $mu neq 0$?







      calculus integration improper-integrals error-function






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      edited Dec 19 '18 at 11:44









      Namaste

      1




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      asked Jun 8 '17 at 18:32









      BobyBoby

      1,0691930




      1,0691930






















          1 Answer
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          $begingroup$

          If $Z$ is a standard normal random variable, $text{erfc}(x/sqrt{2}) = 2 mathbb P(Z > x)$. Thus if $X$ is another normal random variable, independent of $Z$, with mean $mu$ and standard deviation $sigma$, your integral is
          $$ 2 sigma sqrt{2pi} mathbb P(Z > X) = 2 sigma sqrt{2pi} mathbb P(Z-X > 0)$$
          Now $Z - X$ is normal with mean $-mu$ and standard deviation $sqrt{1+sigma^2}$, so this should be
          $$ sigma sqrt{2pi}; text{erfc}left(frac{mu}{sqrt{2 + 2 sigma^2}}right)$$






          share|cite|improve this answer









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          • $begingroup$
            Very nice proof. Thanks.
            $endgroup$
            – Boby
            Jun 8 '17 at 19:45












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          1 Answer
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          active

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          4












          $begingroup$

          If $Z$ is a standard normal random variable, $text{erfc}(x/sqrt{2}) = 2 mathbb P(Z > x)$. Thus if $X$ is another normal random variable, independent of $Z$, with mean $mu$ and standard deviation $sigma$, your integral is
          $$ 2 sigma sqrt{2pi} mathbb P(Z > X) = 2 sigma sqrt{2pi} mathbb P(Z-X > 0)$$
          Now $Z - X$ is normal with mean $-mu$ and standard deviation $sqrt{1+sigma^2}$, so this should be
          $$ sigma sqrt{2pi}; text{erfc}left(frac{mu}{sqrt{2 + 2 sigma^2}}right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very nice proof. Thanks.
            $endgroup$
            – Boby
            Jun 8 '17 at 19:45
















          4












          $begingroup$

          If $Z$ is a standard normal random variable, $text{erfc}(x/sqrt{2}) = 2 mathbb P(Z > x)$. Thus if $X$ is another normal random variable, independent of $Z$, with mean $mu$ and standard deviation $sigma$, your integral is
          $$ 2 sigma sqrt{2pi} mathbb P(Z > X) = 2 sigma sqrt{2pi} mathbb P(Z-X > 0)$$
          Now $Z - X$ is normal with mean $-mu$ and standard deviation $sqrt{1+sigma^2}$, so this should be
          $$ sigma sqrt{2pi}; text{erfc}left(frac{mu}{sqrt{2 + 2 sigma^2}}right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very nice proof. Thanks.
            $endgroup$
            – Boby
            Jun 8 '17 at 19:45














          4












          4








          4





          $begingroup$

          If $Z$ is a standard normal random variable, $text{erfc}(x/sqrt{2}) = 2 mathbb P(Z > x)$. Thus if $X$ is another normal random variable, independent of $Z$, with mean $mu$ and standard deviation $sigma$, your integral is
          $$ 2 sigma sqrt{2pi} mathbb P(Z > X) = 2 sigma sqrt{2pi} mathbb P(Z-X > 0)$$
          Now $Z - X$ is normal with mean $-mu$ and standard deviation $sqrt{1+sigma^2}$, so this should be
          $$ sigma sqrt{2pi}; text{erfc}left(frac{mu}{sqrt{2 + 2 sigma^2}}right)$$






          share|cite|improve this answer









          $endgroup$



          If $Z$ is a standard normal random variable, $text{erfc}(x/sqrt{2}) = 2 mathbb P(Z > x)$. Thus if $X$ is another normal random variable, independent of $Z$, with mean $mu$ and standard deviation $sigma$, your integral is
          $$ 2 sigma sqrt{2pi} mathbb P(Z > X) = 2 sigma sqrt{2pi} mathbb P(Z-X > 0)$$
          Now $Z - X$ is normal with mean $-mu$ and standard deviation $sqrt{1+sigma^2}$, so this should be
          $$ sigma sqrt{2pi}; text{erfc}left(frac{mu}{sqrt{2 + 2 sigma^2}}right)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jun 8 '17 at 19:20









          Robert IsraelRobert Israel

          330k23219473




          330k23219473












          • $begingroup$
            Very nice proof. Thanks.
            $endgroup$
            – Boby
            Jun 8 '17 at 19:45


















          • $begingroup$
            Very nice proof. Thanks.
            $endgroup$
            – Boby
            Jun 8 '17 at 19:45
















          $begingroup$
          Very nice proof. Thanks.
          $endgroup$
          – Boby
          Jun 8 '17 at 19:45




          $begingroup$
          Very nice proof. Thanks.
          $endgroup$
          – Boby
          Jun 8 '17 at 19:45


















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