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Help Solving Integration $int_-infty^infty (textStandard Normal times textSigmoid) $
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Does anyone know how to analytically solve the below expression?
$$int_-infty^infty p(x) times sigma(ax + b) , dx $$
where $ p(x) = N(0, 1) = frac1sqrt2pie^frac-x^22$ and $sigma(z) = frac11+e^-z$.
I know that the integration of standard Gaussian is $pi$. Moreover, the integration of sigmoid alone is,
$$int sigma(x) , df = int frace^x1+e^x , dx = x + c$$
But I cant figure out how to integrate the first equation), when they are multiplied together.
Can we use integration by part and solve above equation?
Any help will be appreciated.
Thanks
Edit 1:
Sorry about the complicated notations.
If we simplify the above expression, then we get following equation.
$$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-x dx $$
integration gaussian-integral
asked Aug 2 at 12:06
Nadheesh
165
 |Â
show 16 more comments
up vote
2
down vote
favorite
Does anyone know how to analytically solve the below expression?
$$int_-infty^infty p(x) times sigma(ax + b) , dx $$
where $ p(x) = N(0, 1) = frac1sqrt2pie^frac-x^22$ and $sigma(z) = frac11+e^-z$.
I know that the integration of standard Gaussian is $pi$. Moreover, the integration of sigmoid alone is,
$$int sigma(x) , df = int frace^x1+e^x , dx = x + c$$
But I cant figure out how to integrate the first equation), when they are multiplied together.
Can we use integration by part and solve above equation?
Any help will be appreciated.
Thanks
Edit 1:
Sorry about the complicated notations.
If we simplify the above expression, then we get following equation.
$$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-x dx $$
integration gaussian-integral
asked Aug 2 at 12:06
Nadheesh
165
What is $p(x)$ ? Is it the probability density function ?
– MysteryGuy
Aug 2 at 12:11
yes. $p(x) = frac1sqrt2pie^frac-x^22$
– Nadheesh
Aug 2 at 12:17
If $x$ is between $pminfty,$ as opposed to $x$ having an imaginary part, then $ln|e^x|$ is redundant and one can just write $ln e^x,$ and $ln e^x$ is the same as $x,$ so $ln e^x + c = x + c. qquad$
– Michael Hardy
Aug 2 at 12:26
@MichaelHardy thanks for the suggestion. edited the post accordingly. However, that is not what I'm interested in. I want to solve the first expression.
– Nadheesh
Aug 2 at 12:33
Sorry I’m not familiar with this topic...but what is $N(0,1)$?
– Szeto
Aug 2 at 12:35
 |Â
show 16 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Does anyone know how to analytically solve the below expression?
$$int_-infty^infty p(x) times sigma(ax + b) , dx $$
where $ p(x) = N(0, 1) = frac1sqrt2pie^frac-x^22$ and $sigma(z) = frac11+e^-z$.
I know that the integration of standard Gaussian is $pi$. Moreover, the integration of sigmoid alone is,
$$int sigma(x) , df = int frace^x1+e^x , dx = x + c$$
But I cant figure out how to integrate the first equation), when they are multiplied together.
Can we use integration by part and solve above equation?
Any help will be appreciated.
Thanks
Edit 1:
Sorry about the complicated notations.
If we simplify the above expression, then we get following equation.
$$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-x dx $$
integration gaussian-integral
asked Aug 2 at 12:06
Nadheesh
165
Does anyone know how to analytically solve the below expression?
$$int_-infty^infty p(x) times sigma(ax + b) , dx $$
where $ p(x) = N(0, 1) = frac1sqrt2pie^frac-x^22$ and $sigma(z) = frac11+e^-z$.
I know that the integration of standard Gaussian is $pi$. Moreover, the integration of sigmoid alone is,
$$int sigma(x) , df = int frace^x1+e^x , dx = x + c$$
But I cant figure out how to integrate the first equation), when they are multiplied together.
Can we use integration by part and solve above equation?
Any help will be appreciated.
Thanks
Edit 1:
Sorry about the complicated notations.
If we simplify the above expression, then we get following equation.
$$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-x dx $$
integration gaussian-integral
asked Aug 2 at 12:06
Nadheesh
165
edited 21 hours ago
asked Aug 2 at 12:06
Nadheesh
165
asked Aug 2 at 12:06
Nadheesh
165
asked Aug 2 at 12:06
Nadheesh
165
165
What is $p(x)$ ? Is it the probability density function ?
– MysteryGuy
Aug 2 at 12:11
yes. $p(x) = frac1sqrt2pie^frac-x^22$
– Nadheesh
Aug 2 at 12:17
If $x$ is between $pminfty,$ as opposed to $x$ having an imaginary part, then $ln|e^x|$ is redundant and one can just write $ln e^x,$ and $ln e^x$ is the same as $x,$ so $ln e^x + c = x + c. qquad$
– Michael Hardy
Aug 2 at 12:26
@MichaelHardy thanks for the suggestion. edited the post accordingly. However, that is not what I'm interested in. I want to solve the first expression.
– Nadheesh
Aug 2 at 12:33
Sorry I’m not familiar with this topic...but what is $N(0,1)$?
– Szeto
Aug 2 at 12:35
 |Â
show 16 more comments
What is $p(x)$ ? Is it the probability density function ?
– MysteryGuy
Aug 2 at 12:11
yes. $p(x) = frac1sqrt2pie^frac-x^22$
– Nadheesh
Aug 2 at 12:17
If $x$ is between $pminfty,$ as opposed to $x$ having an imaginary part, then $ln|e^x|$ is redundant and one can just write $ln e^x,$ and $ln e^x$ is the same as $x,$ so $ln e^x + c = x + c. qquad$
– Michael Hardy
Aug 2 at 12:26
@MichaelHardy thanks for the suggestion. edited the post accordingly. However, that is not what I'm interested in. I want to solve the first expression.
– Nadheesh
Aug 2 at 12:33
Sorry I’m not familiar with this topic...but what is $N(0,1)$?
– Szeto
Aug 2 at 12:35
What is $p(x)$ ? Is it the probability density function ?
– MysteryGuy
Aug 2 at 12:11
What is $p(x)$ ? Is it the probability density function ?
– MysteryGuy
Aug 2 at 12:11
yes. $p(x) = frac1sqrt2pie^frac-x^22$
– Nadheesh
Aug 2 at 12:17
yes. $p(x) = frac1sqrt2pie^frac-x^22$
– Nadheesh
Aug 2 at 12:17
If $x$ is between $pminfty,$ as opposed to $x$ having an imaginary part, then $ln|e^x|$ is redundant and one can just write $ln e^x,$ and $ln e^x$ is the same as $x,$ so $ln e^x + c = x + c. qquad$
– Michael Hardy
Aug 2 at 12:26
If $x$ is between $pminfty,$ as opposed to $x$ having an imaginary part, then $ln|e^x|$ is redundant and one can just write $ln e^x,$ and $ln e^x$ is the same as $x,$ so $ln e^x + c = x + c. qquad$
– Michael Hardy
Aug 2 at 12:26
@MichaelHardy thanks for the suggestion. edited the post accordingly. However, that is not what I'm interested in. I want to solve the first expression.
– Nadheesh
Aug 2 at 12:33
@MichaelHardy thanks for the suggestion. edited the post accordingly. However, that is not what I'm interested in. I want to solve the first expression.
– Nadheesh
Aug 2 at 12:33
Sorry I’m not familiar with this topic...but what is $N(0,1)$?
– Szeto
Aug 2 at 12:35
Sorry I’m not familiar with this topic...but what is $N(0,1)$?
– Szeto
Aug 2 at 12:35
 |Â
show 16 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
beginalign
frac1sqrt2pi int_-infty^infty e^-fracx^22frace^x1+e^x dx&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_-infty^0 e^-fracx^22frace^x1+e^x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_0^infty e^-fracx^22frace^-x1+e^-x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_0^infty e^-fracx^22frac11+e^x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22 dx\
&=frac12
endalign
answered Aug 2 at 13:35
Math-fun
6,8811225
Thanks for all the work, I would have never thought of this myself. How can we solve the expression if the sigmoid function include some other parameters as show below? $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:54
add a comment |Â
up vote
2
down vote
$$
varphi(x) = frac 1 sqrt2pi e^-x^2/2
$$
$$
sigma (x) = frac 1 1+e^-x
$$
Some simple algebra shows that
$$
sigma(-x) = 1 - sigma(x)
$$
and it is even easier to show that
$$
varphi(-x) = varphi(x),
$$
so we have
$$
int_-infty^0 varphi(x),dx = int_0^infty varphi(x),dx = frac 1 2.
$$
Therefore
beginalign
int_-infty^0 sigma(x) varphi(x) , dx & = int_0^+infty (1-sigma(x)) varphi(x), dx \[10pt]
& = int_0^+infty varphi(x),dx - int_0^+infty sigma(x)varphi(x),dx
endalign
Therefore
$$
int_-infty^+infty sigma(x)varphi(x),dx = int_0^+infty varphi(x), dx = frac 1 2.
$$
edited 21 hours ago
Jean-Claude Arbaut
14.1k63260
answered Aug 2 at 22:25
Michael Hardy
204k23185460
1
Thanks for the solution. This seems pretty simple yet, I could not have think of without your help. However, if we have some other variables within the sigmoid function, then how can we solve the same expression. I tried substituting z = ax+b, then I could not get rid of the sigmoid. $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:57
do you think if we can solve this expression when the sigmoid function is changed as shown above.
– Nadheesh
Aug 4 at 2:48
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
beginalign
frac1sqrt2pi int_-infty^infty e^-fracx^22frace^x1+e^x dx&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_-infty^0 e^-fracx^22frace^x1+e^x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_0^infty e^-fracx^22frace^-x1+e^-x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_0^infty e^-fracx^22frac11+e^x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22 dx\
&=frac12
endalign
answered Aug 2 at 13:35
Math-fun
6,8811225
Thanks for all the work, I would have never thought of this myself. How can we solve the expression if the sigmoid function include some other parameters as show below? $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:54
add a comment |Â
up vote
4
down vote
accepted
beginalign
frac1sqrt2pi int_-infty^infty e^-fracx^22frace^x1+e^x dx&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_-infty^0 e^-fracx^22frace^x1+e^x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_0^infty e^-fracx^22frace^-x1+e^-x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_0^infty e^-fracx^22frac11+e^x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22 dx\
&=frac12
endalign
answered Aug 2 at 13:35
Math-fun
6,8811225
Thanks for all the work, I would have never thought of this myself. How can we solve the expression if the sigmoid function include some other parameters as show below? $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:54
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
beginalign
frac1sqrt2pi int_-infty^infty e^-fracx^22frace^x1+e^x dx&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_-infty^0 e^-fracx^22frace^x1+e^x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_0^infty e^-fracx^22frace^-x1+e^-x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_0^infty e^-fracx^22frac11+e^x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22 dx\
&=frac12
endalign
answered Aug 2 at 13:35
Math-fun
6,8811225
beginalign
frac1sqrt2pi int_-infty^infty e^-fracx^22frace^x1+e^x dx&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_-infty^0 e^-fracx^22frace^x1+e^x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_0^infty e^-fracx^22frace^-x1+e^-x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22frace^x1+e^x dx+frac1sqrt2pi int_0^infty e^-fracx^22frac11+e^x dx\
&=frac1sqrt2pi int_0^infty e^-fracx^22 dx\
&=frac12
endalign
answered Aug 2 at 13:35
Math-fun
6,8811225
answered Aug 2 at 13:35
Math-fun
6,8811225
answered Aug 2 at 13:35
Math-fun
6,8811225
answered Aug 2 at 13:35
Math-fun
6,8811225
6,8811225
Thanks for all the work, I would have never thought of this myself. How can we solve the expression if the sigmoid function include some other parameters as show below? $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:54
add a comment |Â
Thanks for all the work, I would have never thought of this myself. How can we solve the expression if the sigmoid function include some other parameters as show below? $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:54
Thanks for all the work, I would have never thought of this myself. How can we solve the expression if the sigmoid function include some other parameters as show below? $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:54
Thanks for all the work, I would have never thought of this myself. How can we solve the expression if the sigmoid function include some other parameters as show below? $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:54
add a comment |Â
up vote
2
down vote
$$
varphi(x) = frac 1 sqrt2pi e^-x^2/2
$$
$$
sigma (x) = frac 1 1+e^-x
$$
Some simple algebra shows that
$$
sigma(-x) = 1 - sigma(x)
$$
and it is even easier to show that
$$
varphi(-x) = varphi(x),
$$
so we have
$$
int_-infty^0 varphi(x),dx = int_0^infty varphi(x),dx = frac 1 2.
$$
Therefore
beginalign
int_-infty^0 sigma(x) varphi(x) , dx & = int_0^+infty (1-sigma(x)) varphi(x), dx \[10pt]
& = int_0^+infty varphi(x),dx - int_0^+infty sigma(x)varphi(x),dx
endalign
Therefore
$$
int_-infty^+infty sigma(x)varphi(x),dx = int_0^+infty varphi(x), dx = frac 1 2.
$$
edited 21 hours ago
Jean-Claude Arbaut
14.1k63260
answered Aug 2 at 22:25
Michael Hardy
204k23185460
1
Thanks for the solution. This seems pretty simple yet, I could not have think of without your help. However, if we have some other variables within the sigmoid function, then how can we solve the same expression. I tried substituting z = ax+b, then I could not get rid of the sigmoid. $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:57
do you think if we can solve this expression when the sigmoid function is changed as shown above.
– Nadheesh
Aug 4 at 2:48
add a comment |Â
up vote
2
down vote
$$
varphi(x) = frac 1 sqrt2pi e^-x^2/2
$$
$$
sigma (x) = frac 1 1+e^-x
$$
Some simple algebra shows that
$$
sigma(-x) = 1 - sigma(x)
$$
and it is even easier to show that
$$
varphi(-x) = varphi(x),
$$
so we have
$$
int_-infty^0 varphi(x),dx = int_0^infty varphi(x),dx = frac 1 2.
$$
Therefore
beginalign
int_-infty^0 sigma(x) varphi(x) , dx & = int_0^+infty (1-sigma(x)) varphi(x), dx \[10pt]
& = int_0^+infty varphi(x),dx - int_0^+infty sigma(x)varphi(x),dx
endalign
Therefore
$$
int_-infty^+infty sigma(x)varphi(x),dx = int_0^+infty varphi(x), dx = frac 1 2.
$$
edited 21 hours ago
Jean-Claude Arbaut
14.1k63260
answered Aug 2 at 22:25
Michael Hardy
204k23185460
1
Thanks for the solution. This seems pretty simple yet, I could not have think of without your help. However, if we have some other variables within the sigmoid function, then how can we solve the same expression. I tried substituting z = ax+b, then I could not get rid of the sigmoid. $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:57
do you think if we can solve this expression when the sigmoid function is changed as shown above.
– Nadheesh
Aug 4 at 2:48
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$
varphi(x) = frac 1 sqrt2pi e^-x^2/2
$$
$$
sigma (x) = frac 1 1+e^-x
$$
Some simple algebra shows that
$$
sigma(-x) = 1 - sigma(x)
$$
and it is even easier to show that
$$
varphi(-x) = varphi(x),
$$
so we have
$$
int_-infty^0 varphi(x),dx = int_0^infty varphi(x),dx = frac 1 2.
$$
Therefore
beginalign
int_-infty^0 sigma(x) varphi(x) , dx & = int_0^+infty (1-sigma(x)) varphi(x), dx \[10pt]
& = int_0^+infty varphi(x),dx - int_0^+infty sigma(x)varphi(x),dx
endalign
Therefore
$$
int_-infty^+infty sigma(x)varphi(x),dx = int_0^+infty varphi(x), dx = frac 1 2.
$$
edited 21 hours ago
Jean-Claude Arbaut
14.1k63260
answered Aug 2 at 22:25
Michael Hardy
204k23185460
$$
varphi(x) = frac 1 sqrt2pi e^-x^2/2
$$
$$
sigma (x) = frac 1 1+e^-x
$$
Some simple algebra shows that
$$
sigma(-x) = 1 - sigma(x)
$$
and it is even easier to show that
$$
varphi(-x) = varphi(x),
$$
so we have
$$
int_-infty^0 varphi(x),dx = int_0^infty varphi(x),dx = frac 1 2.
$$
Therefore
beginalign
int_-infty^0 sigma(x) varphi(x) , dx & = int_0^+infty (1-sigma(x)) varphi(x), dx \[10pt]
& = int_0^+infty varphi(x),dx - int_0^+infty sigma(x)varphi(x),dx
endalign
Therefore
$$
int_-infty^+infty sigma(x)varphi(x),dx = int_0^+infty varphi(x), dx = frac 1 2.
$$
edited 21 hours ago
Jean-Claude Arbaut
14.1k63260
answered Aug 2 at 22:25
Michael Hardy
204k23185460
edited 21 hours ago
Jean-Claude Arbaut
14.1k63260
edited 21 hours ago
Jean-Claude Arbaut
14.1k63260
edited 21 hours ago
Jean-Claude Arbaut
14.1k63260
14.1k63260
answered Aug 2 at 22:25
Michael Hardy
204k23185460
answered Aug 2 at 22:25
Michael Hardy
204k23185460
answered Aug 2 at 22:25
Michael Hardy
204k23185460
204k23185460
1
Thanks for the solution. This seems pretty simple yet, I could not have think of without your help. However, if we have some other variables within the sigmoid function, then how can we solve the same expression. I tried substituting z = ax+b, then I could not get rid of the sigmoid. $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:57
do you think if we can solve this expression when the sigmoid function is changed as shown above.
– Nadheesh
Aug 4 at 2:48
add a comment |Â
1
Thanks for the solution. This seems pretty simple yet, I could not have think of without your help. However, if we have some other variables within the sigmoid function, then how can we solve the same expression. I tried substituting z = ax+b, then I could not get rid of the sigmoid. $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:57
do you think if we can solve this expression when the sigmoid function is changed as shown above.
– Nadheesh
Aug 4 at 2:48
1
1
Thanks for the solution. This seems pretty simple yet, I could not have think of without your help. However, if we have some other variables within the sigmoid function, then how can we solve the same expression. I tried substituting z = ax+b, then I could not get rid of the sigmoid. $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:57
Thanks for the solution. This seems pretty simple yet, I could not have think of without your help. However, if we have some other variables within the sigmoid function, then how can we solve the same expression. I tried substituting z = ax+b, then I could not get rid of the sigmoid. $$ frac1sqrt2pi int_-infty^infty frace^-fracx^221+e^-(ax+b) dx $$
– Nadheesh
Aug 3 at 1:57
do you think if we can solve this expression when the sigmoid function is changed as shown above.
– Nadheesh
Aug 4 at 2:48
do you think if we can solve this expression when the sigmoid function is changed as shown above.
– Nadheesh
Aug 4 at 2:48
add a comment |Â
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What is $p(x)$ ? Is it the probability density function ?
– MysteryGuy
Aug 2 at 12:11
yes. $p(x) = frac1sqrt2pie^frac-x^22$
– Nadheesh
Aug 2 at 12:17
If $x$ is between $pminfty,$ as opposed to $x$ having an imaginary part, then $ln|e^x|$ is redundant and one can just write $ln e^x,$ and $ln e^x$ is the same as $x,$ so $ln e^x + c = x + c. qquad$
– Michael Hardy
Aug 2 at 12:26
@MichaelHardy thanks for the suggestion. edited the post accordingly. However, that is not what I'm interested in. I want to solve the first expression.
– Nadheesh
Aug 2 at 12:33
Sorry I’m not familiar with this topic...but what is $N(0,1)$?
– Szeto
Aug 2 at 12:35