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Showing $2int_-infty^infty x f(x) F(x) dx = frac1sqrtpi$ for standard normal pdf and cfd
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i am trying to prove the identity in the title. I strongly think I need to use the error function $$texterf(x) = frac2sqrtpi int_0^x exp-t^2 dt$$ in some way. Best I have so far is replacing $F(x) = frac1+texterfleft(fracxsqrt2right)2$ to end up with
$$2int_-infty^infty x , f(x) , F(x) dx = int_-infty^infty texterfleft(fracxsqrt2right) , x, frac1sqrt2pi expleft-frac12x^2right dx.$$
My attempts on partial integration have failed, any other ideas or does anyone succeed?
I have a document stating that the equality holds without any further remarks or calculations and I have confirmed it via integration by quadrature and am looking for an analytic proof.
Very thankful for any help.
probability integration
asked Jul 18 at 17:17
InterestedStudent
306
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up vote
5
down vote
favorite
i am trying to prove the identity in the title. I strongly think I need to use the error function $$texterf(x) = frac2sqrtpi int_0^x exp-t^2 dt$$ in some way. Best I have so far is replacing $F(x) = frac1+texterfleft(fracxsqrt2right)2$ to end up with
$$2int_-infty^infty x , f(x) , F(x) dx = int_-infty^infty texterfleft(fracxsqrt2right) , x, frac1sqrt2pi expleft-frac12x^2right dx.$$
My attempts on partial integration have failed, any other ideas or does anyone succeed?
I have a document stating that the equality holds without any further remarks or calculations and I have confirmed it via integration by quadrature and am looking for an analytic proof.
Very thankful for any help.
probability integration
asked Jul 18 at 17:17
InterestedStudent
306
1
This is basically $E(XPhi(X))$ previously shown here.
– StubbornAtom
Jul 18 at 20:34
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
i am trying to prove the identity in the title. I strongly think I need to use the error function $$texterf(x) = frac2sqrtpi int_0^x exp-t^2 dt$$ in some way. Best I have so far is replacing $F(x) = frac1+texterfleft(fracxsqrt2right)2$ to end up with
$$2int_-infty^infty x , f(x) , F(x) dx = int_-infty^infty texterfleft(fracxsqrt2right) , x, frac1sqrt2pi expleft-frac12x^2right dx.$$
My attempts on partial integration have failed, any other ideas or does anyone succeed?
I have a document stating that the equality holds without any further remarks or calculations and I have confirmed it via integration by quadrature and am looking for an analytic proof.
Very thankful for any help.
probability integration
asked Jul 18 at 17:17
InterestedStudent
306
i am trying to prove the identity in the title. I strongly think I need to use the error function $$texterf(x) = frac2sqrtpi int_0^x exp-t^2 dt$$ in some way. Best I have so far is replacing $F(x) = frac1+texterfleft(fracxsqrt2right)2$ to end up with
$$2int_-infty^infty x , f(x) , F(x) dx = int_-infty^infty texterfleft(fracxsqrt2right) , x, frac1sqrt2pi expleft-frac12x^2right dx.$$
My attempts on partial integration have failed, any other ideas or does anyone succeed?
I have a document stating that the equality holds without any further remarks or calculations and I have confirmed it via integration by quadrature and am looking for an analytic proof.
Very thankful for any help.
probability integration
asked Jul 18 at 17:17
InterestedStudent
306
edited Jul 18 at 17:20
asked Jul 18 at 17:17
InterestedStudent
306
asked Jul 18 at 17:17
InterestedStudent
306
asked Jul 18 at 17:17
InterestedStudent
306
306
1
This is basically $E(XPhi(X))$ previously shown here.
– StubbornAtom
Jul 18 at 20:34
add a comment |Â
1
This is basically $E(XPhi(X))$ previously shown here.
– StubbornAtom
Jul 18 at 20:34
1
1
This is basically $E(XPhi(X))$ previously shown here.
– StubbornAtom
Jul 18 at 20:34
This is basically $E(XPhi(X))$ previously shown here.
– StubbornAtom
Jul 18 at 20:34
add a comment |Â
2 Answers
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We have
$$ int x f(x) , dx = frac1sqrt2piint x e^-x^2/2 , dx = -frac1sqrt2pie^-x^2/2+C = -f(x)+C, $$
so integrating by parts gives
$$ 2int_-infty^infty xf(x)F(x) , dx = [-2f(x)F(x)]_-infty^infty + int_-infty^infty 2f(x)^2 , dx $$
since $F'=f$. But $f(x)^2 = e^-x^2$, which is only a scaling away from the standard normal, by putting $x=y/sqrt2$, so $dx=dy/sqrt2$, and hence
$$ 2int_-infty^infty xf(x)F(x) , dx = frac2sqrt2sqrt2pi = frac1sqrtpi. $$
answered Jul 18 at 17:28
Chappers
55k74190
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up vote
2
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Hint: Your integral is $$frac2sqrt2 pi int_-infty^infty x e^-x^2 / 2 F(x) ,dx,.$$
Try integration by parts with $u = F(x)$ and $dv = xe^-x^2 / 2 ,dx$.
answered Jul 18 at 17:26
Marcus M
8,1731847
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We have
$$ int x f(x) , dx = frac1sqrt2piint x e^-x^2/2 , dx = -frac1sqrt2pie^-x^2/2+C = -f(x)+C, $$
so integrating by parts gives
$$ 2int_-infty^infty xf(x)F(x) , dx = [-2f(x)F(x)]_-infty^infty + int_-infty^infty 2f(x)^2 , dx $$
since $F'=f$. But $f(x)^2 = e^-x^2$, which is only a scaling away from the standard normal, by putting $x=y/sqrt2$, so $dx=dy/sqrt2$, and hence
$$ 2int_-infty^infty xf(x)F(x) , dx = frac2sqrt2sqrt2pi = frac1sqrtpi. $$
answered Jul 18 at 17:28
Chappers
55k74190
add a comment |Â
up vote
3
down vote
accepted
We have
$$ int x f(x) , dx = frac1sqrt2piint x e^-x^2/2 , dx = -frac1sqrt2pie^-x^2/2+C = -f(x)+C, $$
so integrating by parts gives
$$ 2int_-infty^infty xf(x)F(x) , dx = [-2f(x)F(x)]_-infty^infty + int_-infty^infty 2f(x)^2 , dx $$
since $F'=f$. But $f(x)^2 = e^-x^2$, which is only a scaling away from the standard normal, by putting $x=y/sqrt2$, so $dx=dy/sqrt2$, and hence
$$ 2int_-infty^infty xf(x)F(x) , dx = frac2sqrt2sqrt2pi = frac1sqrtpi. $$
answered Jul 18 at 17:28
Chappers
55k74190
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We have
$$ int x f(x) , dx = frac1sqrt2piint x e^-x^2/2 , dx = -frac1sqrt2pie^-x^2/2+C = -f(x)+C, $$
so integrating by parts gives
$$ 2int_-infty^infty xf(x)F(x) , dx = [-2f(x)F(x)]_-infty^infty + int_-infty^infty 2f(x)^2 , dx $$
since $F'=f$. But $f(x)^2 = e^-x^2$, which is only a scaling away from the standard normal, by putting $x=y/sqrt2$, so $dx=dy/sqrt2$, and hence
$$ 2int_-infty^infty xf(x)F(x) , dx = frac2sqrt2sqrt2pi = frac1sqrtpi. $$
answered Jul 18 at 17:28
Chappers
55k74190
We have
$$ int x f(x) , dx = frac1sqrt2piint x e^-x^2/2 , dx = -frac1sqrt2pie^-x^2/2+C = -f(x)+C, $$
so integrating by parts gives
$$ 2int_-infty^infty xf(x)F(x) , dx = [-2f(x)F(x)]_-infty^infty + int_-infty^infty 2f(x)^2 , dx $$
since $F'=f$. But $f(x)^2 = e^-x^2$, which is only a scaling away from the standard normal, by putting $x=y/sqrt2$, so $dx=dy/sqrt2$, and hence
$$ 2int_-infty^infty xf(x)F(x) , dx = frac2sqrt2sqrt2pi = frac1sqrtpi. $$
answered Jul 18 at 17:28
Chappers
55k74190
answered Jul 18 at 17:28
Chappers
55k74190
answered Jul 18 at 17:28
Chappers
55k74190
answered Jul 18 at 17:28
Chappers
55k74190
55k74190
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint: Your integral is $$frac2sqrt2 pi int_-infty^infty x e^-x^2 / 2 F(x) ,dx,.$$
Try integration by parts with $u = F(x)$ and $dv = xe^-x^2 / 2 ,dx$.
answered Jul 18 at 17:26
Marcus M
8,1731847
add a comment |Â
up vote
2
down vote
Hint: Your integral is $$frac2sqrt2 pi int_-infty^infty x e^-x^2 / 2 F(x) ,dx,.$$
Try integration by parts with $u = F(x)$ and $dv = xe^-x^2 / 2 ,dx$.
answered Jul 18 at 17:26
Marcus M
8,1731847
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: Your integral is $$frac2sqrt2 pi int_-infty^infty x e^-x^2 / 2 F(x) ,dx,.$$
Try integration by parts with $u = F(x)$ and $dv = xe^-x^2 / 2 ,dx$.
answered Jul 18 at 17:26
Marcus M
8,1731847
Hint: Your integral is $$frac2sqrt2 pi int_-infty^infty x e^-x^2 / 2 F(x) ,dx,.$$
Try integration by parts with $u = F(x)$ and $dv = xe^-x^2 / 2 ,dx$.
answered Jul 18 at 17:26
Marcus M
8,1731847
answered Jul 18 at 17:26
Marcus M
8,1731847
answered Jul 18 at 17:26
Marcus M
8,1731847
answered Jul 18 at 17:26
Marcus M
8,1731847
8,1731847
add a comment |Â
add a comment |Â
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1
This is basically $E(XPhi(X))$ previously shown here.
– StubbornAtom
Jul 18 at 20:34