Integrate $int_{-infty}^infty {rm erfc} left( frac{x}{sqrt{2}} right) e^{-frac{(x-mu)^2}{2 sigma^2}} dx$
$begingroup$
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 $mu neq 0$?
calculus integration improper-integrals error-function
edited Dec 19 '18 at 11:44
Namaste
1
asked Jun 8 '17 at 18:32
BobyBoby
1,0691930
$endgroup$
add a comment |
$begingroup$
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 $mu neq 0$?
calculus integration improper-integrals error-function
edited Dec 19 '18 at 11:44
Namaste
1
asked Jun 8 '17 at 18:32
BobyBoby
1,0691930
$endgroup$
add a comment |
$begingroup$
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 $mu neq 0$?
calculus integration improper-integrals error-function
edited Dec 19 '18 at 11:44
Namaste
1
asked Jun 8 '17 at 18:32
BobyBoby
1,0691930
$endgroup$
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
calculus integration improper-integrals error-function
edited Dec 19 '18 at 11:44
Namaste
1
asked Jun 8 '17 at 18:32
BobyBoby
1,0691930
edited Dec 19 '18 at 11:44
Namaste
1
asked Jun 8 '17 at 18:32
BobyBoby
1,0691930
edited Dec 19 '18 at 11:44
Namaste
1
edited Dec 19 '18 at 11:44
Namaste
1
edited Dec 19 '18 at 11:44
Namaste
1
1
asked Jun 8 '17 at 18:32
BobyBoby
1,0691930
asked Jun 8 '17 at 18:32
BobyBoby
1,0691930
asked Jun 8 '17 at 18:32
BobyBoby
1,0691930
1,0691930
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$ sigma sqrt{2pi}; text{erfc}left(frac{mu}{sqrt{2 + 2 sigma^2}}right)$$
answered Jun 8 '17 at 19:20
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
Very nice proof. Thanks.
$endgroup$
– Boby
Jun 8 '17 at 19:45
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$ sigma sqrt{2pi}; text{erfc}left(frac{mu}{sqrt{2 + 2 sigma^2}}right)$$
answered Jun 8 '17 at 19:20
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
Very nice proof. Thanks.
$endgroup$
– Boby
Jun 8 '17 at 19:45
add a comment |
$begingroup$
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
$$ sigma sqrt{2pi}; text{erfc}left(frac{mu}{sqrt{2 + 2 sigma^2}}right)$$
answered Jun 8 '17 at 19:20
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
Very nice proof. Thanks.
$endgroup$
– Boby
Jun 8 '17 at 19:45
add a comment |
$begingroup$
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
$$ sigma sqrt{2pi}; text{erfc}left(frac{mu}{sqrt{2 + 2 sigma^2}}right)$$
answered Jun 8 '17 at 19:20
Robert IsraelRobert Israel
330k23219473
$endgroup$
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
$$ sigma sqrt{2pi}; text{erfc}left(frac{mu}{sqrt{2 + 2 sigma^2}}right)$$
answered Jun 8 '17 at 19:20
Robert IsraelRobert Israel
330k23219473
answered Jun 8 '17 at 19:20
Robert IsraelRobert Israel
330k23219473
answered Jun 8 '17 at 19:20
Robert IsraelRobert Israel
330k23219473
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
add a comment |
$begingroup$
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
add a comment |
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