Is there an analytic form for the squared error of the difference of two univariate Gaussians?
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Using anchored ensembling it is possible to estimate the mean $mu$ and the variance $sigma^2$ of a output. I had the insight than if I then sampled from $N(mu,sigma^2)$, I could estimate both the aleatoric (data) and epistemic (model) uncertainty in one shot and use it to guide exploration. I could do this optimization using samples, but it has occurred to me that propagating uncertainty through the TD error might result in much less variance in the gradients.
Thanks to the uncertainty propagation page I understand how that might be done, but I am missing the last step which is the cost function.
At first glance it might seem that the sixth example formula ($f=aA^b$) should be enough to derive the squared error, but it is obvious to me that this would just pull the variance downward if I used it as cost directly.
How do I integrate: $E_{a sim N({mu_1},{sigma_1}),b sim N({mu_2},{sigma_2})}(a - b)^2$? I am hoping for something neat that I could backprop through.
calculus probability integration gaussian-integral
asked Dec 24 '18 at 10:18
Marko GrdinicMarko Grdinic
1419
$endgroup$
|
show 2 more comments
$begingroup$
Using anchored ensembling it is possible to estimate the mean $mu$ and the variance $sigma^2$ of a output. I had the insight than if I then sampled from $N(mu,sigma^2)$, I could estimate both the aleatoric (data) and epistemic (model) uncertainty in one shot and use it to guide exploration. I could do this optimization using samples, but it has occurred to me that propagating uncertainty through the TD error might result in much less variance in the gradients.
Thanks to the uncertainty propagation page I understand how that might be done, but I am missing the last step which is the cost function.
At first glance it might seem that the sixth example formula ($f=aA^b$) should be enough to derive the squared error, but it is obvious to me that this would just pull the variance downward if I used it as cost directly.
How do I integrate: $E_{a sim N({mu_1},{sigma_1}),b sim N({mu_2},{sigma_2})}(a - b)^2$? I am hoping for something neat that I could backprop through.
calculus probability integration gaussian-integral
asked Dec 24 '18 at 10:18
Marko GrdinicMarko Grdinic
1419
$endgroup$
$begingroup$
No. I am assuming that p(a) and p(b) are univariate Gaussian distributions, but have different means and variances. I've changed the question to better reflect what I am asking.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 10:49
3
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Then $mathbb{E}(a-b)^2=(mu_1-mu_2)^2+sigma_1^2+sigma_2^2$ (if $a,b$ are independent).
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– metamorphy
Dec 24 '18 at 11:00
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That seems really convenient, but it can't be right. It is obvious to me that to minimize that function you'd just set the two ${sigma}$s to zero. This would be the case even if the two ${mu}$s were held constant. If I sampled instead and optimized through those then the variances could grow. It is not a matter of them being independent either. Am I asking something wrong here?
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– Marko Grdinic
Dec 24 '18 at 11:34
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By can't be right, I mean it can't be right as an optimization target for the task I am trying to solve. Not that I do not believe that the equation is right.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:41
1
$begingroup$
The example formula you reference ($f=a A^b$) is in the "nonlinear" section, you describe something that is linear and there is an exact formula as @metamorphy has given you.
$endgroup$
– JimB
Dec 24 '18 at 18:05
|
show 2 more comments
$begingroup$
Using anchored ensembling it is possible to estimate the mean $mu$ and the variance $sigma^2$ of a output. I had the insight than if I then sampled from $N(mu,sigma^2)$, I could estimate both the aleatoric (data) and epistemic (model) uncertainty in one shot and use it to guide exploration. I could do this optimization using samples, but it has occurred to me that propagating uncertainty through the TD error might result in much less variance in the gradients.
Thanks to the uncertainty propagation page I understand how that might be done, but I am missing the last step which is the cost function.
At first glance it might seem that the sixth example formula ($f=aA^b$) should be enough to derive the squared error, but it is obvious to me that this would just pull the variance downward if I used it as cost directly.
How do I integrate: $E_{a sim N({mu_1},{sigma_1}),b sim N({mu_2},{sigma_2})}(a - b)^2$? I am hoping for something neat that I could backprop through.
calculus probability integration gaussian-integral
asked Dec 24 '18 at 10:18
Marko GrdinicMarko Grdinic
1419
$endgroup$
Using anchored ensembling it is possible to estimate the mean $mu$ and the variance $sigma^2$ of a output. I had the insight than if I then sampled from $N(mu,sigma^2)$, I could estimate both the aleatoric (data) and epistemic (model) uncertainty in one shot and use it to guide exploration. I could do this optimization using samples, but it has occurred to me that propagating uncertainty through the TD error might result in much less variance in the gradients.
Thanks to the uncertainty propagation page I understand how that might be done, but I am missing the last step which is the cost function.
At first glance it might seem that the sixth example formula ($f=aA^b$) should be enough to derive the squared error, but it is obvious to me that this would just pull the variance downward if I used it as cost directly.
How do I integrate: $E_{a sim N({mu_1},{sigma_1}),b sim N({mu_2},{sigma_2})}(a - b)^2$? I am hoping for something neat that I could backprop through.
calculus probability integration gaussian-integral
calculus probability integration gaussian-integral
asked Dec 24 '18 at 10:18
Marko GrdinicMarko Grdinic
1419
asked Dec 24 '18 at 10:18
Marko GrdinicMarko Grdinic
1419
edited Dec 24 '18 at 11:46
Marko Grdinic
asked Dec 24 '18 at 10:18
Marko GrdinicMarko Grdinic
1419
asked Dec 24 '18 at 10:18
Marko GrdinicMarko Grdinic
1419
asked Dec 24 '18 at 10:18
Marko GrdinicMarko Grdinic
1419
1419
$begingroup$
No. I am assuming that p(a) and p(b) are univariate Gaussian distributions, but have different means and variances. I've changed the question to better reflect what I am asking.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 10:49
3
$begingroup$
Then $mathbb{E}(a-b)^2=(mu_1-mu_2)^2+sigma_1^2+sigma_2^2$ (if $a,b$ are independent).
$endgroup$
– metamorphy
Dec 24 '18 at 11:00
$begingroup$
That seems really convenient, but it can't be right. It is obvious to me that to minimize that function you'd just set the two ${sigma}$s to zero. This would be the case even if the two ${mu}$s were held constant. If I sampled instead and optimized through those then the variances could grow. It is not a matter of them being independent either. Am I asking something wrong here?
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:34
$begingroup$
By can't be right, I mean it can't be right as an optimization target for the task I am trying to solve. Not that I do not believe that the equation is right.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:41
1
$begingroup$
The example formula you reference ($f=a A^b$) is in the "nonlinear" section, you describe something that is linear and there is an exact formula as @metamorphy has given you.
$endgroup$
– JimB
Dec 24 '18 at 18:05
|
show 2 more comments
$begingroup$
No. I am assuming that p(a) and p(b) are univariate Gaussian distributions, but have different means and variances. I've changed the question to better reflect what I am asking.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 10:49
3
$begingroup$
Then $mathbb{E}(a-b)^2=(mu_1-mu_2)^2+sigma_1^2+sigma_2^2$ (if $a,b$ are independent).
$endgroup$
– metamorphy
Dec 24 '18 at 11:00
$begingroup$
That seems really convenient, but it can't be right. It is obvious to me that to minimize that function you'd just set the two ${sigma}$s to zero. This would be the case even if the two ${mu}$s were held constant. If I sampled instead and optimized through those then the variances could grow. It is not a matter of them being independent either. Am I asking something wrong here?
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:34
$begingroup$
By can't be right, I mean it can't be right as an optimization target for the task I am trying to solve. Not that I do not believe that the equation is right.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:41
1
$begingroup$
The example formula you reference ($f=a A^b$) is in the "nonlinear" section, you describe something that is linear and there is an exact formula as @metamorphy has given you.
$endgroup$
– JimB
Dec 24 '18 at 18:05
$begingroup$
No. I am assuming that p(a) and p(b) are univariate Gaussian distributions, but have different means and variances. I've changed the question to better reflect what I am asking.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 10:49
$begingroup$
No. I am assuming that p(a) and p(b) are univariate Gaussian distributions, but have different means and variances. I've changed the question to better reflect what I am asking.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 10:49
3
3
$begingroup$
Then $mathbb{E}(a-b)^2=(mu_1-mu_2)^2+sigma_1^2+sigma_2^2$ (if $a,b$ are independent).
$endgroup$
– metamorphy
Dec 24 '18 at 11:00
$begingroup$
Then $mathbb{E}(a-b)^2=(mu_1-mu_2)^2+sigma_1^2+sigma_2^2$ (if $a,b$ are independent).
$endgroup$
– metamorphy
Dec 24 '18 at 11:00
$begingroup$
That seems really convenient, but it can't be right. It is obvious to me that to minimize that function you'd just set the two ${sigma}$s to zero. This would be the case even if the two ${mu}$s were held constant. If I sampled instead and optimized through those then the variances could grow. It is not a matter of them being independent either. Am I asking something wrong here?
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:34
$begingroup$
That seems really convenient, but it can't be right. It is obvious to me that to minimize that function you'd just set the two ${sigma}$s to zero. This would be the case even if the two ${mu}$s were held constant. If I sampled instead and optimized through those then the variances could grow. It is not a matter of them being independent either. Am I asking something wrong here?
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:34
$begingroup$
By can't be right, I mean it can't be right as an optimization target for the task I am trying to solve. Not that I do not believe that the equation is right.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:41
$begingroup$
By can't be right, I mean it can't be right as an optimization target for the task I am trying to solve. Not that I do not believe that the equation is right.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:41
1
1
$begingroup$
The example formula you reference ($f=a A^b$) is in the "nonlinear" section, you describe something that is linear and there is an exact formula as @metamorphy has given you.
$endgroup$
– JimB
Dec 24 '18 at 18:05
$begingroup$
The example formula you reference ($f=a A^b$) is in the "nonlinear" section, you describe something that is linear and there is an exact formula as @metamorphy has given you.
$endgroup$
– JimB
Dec 24 '18 at 18:05
|
show 2 more comments
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$begingroup$
No. I am assuming that p(a) and p(b) are univariate Gaussian distributions, but have different means and variances. I've changed the question to better reflect what I am asking.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 10:49
3
$begingroup$
Then $mathbb{E}(a-b)^2=(mu_1-mu_2)^2+sigma_1^2+sigma_2^2$ (if $a,b$ are independent).
$endgroup$
– metamorphy
Dec 24 '18 at 11:00
$begingroup$
That seems really convenient, but it can't be right. It is obvious to me that to minimize that function you'd just set the two ${sigma}$s to zero. This would be the case even if the two ${mu}$s were held constant. If I sampled instead and optimized through those then the variances could grow. It is not a matter of them being independent either. Am I asking something wrong here?
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:34
$begingroup$
By can't be right, I mean it can't be right as an optimization target for the task I am trying to solve. Not that I do not believe that the equation is right.
$endgroup$
– Marko Grdinic
Dec 24 '18 at 11:41
1
$begingroup$
The example formula you reference ($f=a A^b$) is in the "nonlinear" section, you describe something that is linear and there is an exact formula as @metamorphy has given you.
$endgroup$
– JimB
Dec 24 '18 at 18:05