Analytical approximation for logit-normal-binomial distribution
$begingroup$
As I understand, there is no closed form expression for
$$f(x, mu, sigma) = int_0^1 p^{(x-1)}(1-p)^{n-x-1}expleft(-{(text{logit}(p) -mu)^2 over 2sigma^2}right)dp.$$
Is it possible to obtain an analytical approximation for this?
integration numerical-methods approximation approximate-integration
asked Dec 17 '18 at 5:00
zipzapboingzipzapboing
979
$endgroup$
|
show 3 more comments
$begingroup$
As I understand, there is no closed form expression for
$$f(x, mu, sigma) = int_0^1 p^{(x-1)}(1-p)^{n-x-1}expleft(-{(text{logit}(p) -mu)^2 over 2sigma^2}right)dp.$$
Is it possible to obtain an analytical approximation for this?
integration numerical-methods approximation approximate-integration
asked Dec 17 '18 at 5:00
zipzapboingzipzapboing
979
$endgroup$
$begingroup$
Do you mean "asymptotic"? There are numerous analytic approximations to any continuous function.
$endgroup$
– user14717
Dec 17 '18 at 7:28
$begingroup$
I suppose there must be some that are more computationally efficient/reach small error with a smaller number of terms? I need to insert this as part of a statistical model so a tradeoff between accuracy and efficiency is needed.
$endgroup$
– zipzapboing
Dec 17 '18 at 11:19
$begingroup$
You could use quadrature to evaluate it at many arguments and then do interpolation.
$endgroup$
– user14717
Dec 17 '18 at 16:12
$begingroup$
I suppose that can work. I've never attempted something similar so I had something like a multivariable Taylor series in mind. I guess that theoretical guarantees are irrelevant when there's enough computational power to brute force the approximation and ensure it works over the range of things I care about.
$endgroup$
– zipzapboing
Dec 17 '18 at 16:23
1
$begingroup$
The theoretical guarantees will be much stronger with an interpolator than with multivariate Taylor series.
$endgroup$
– user14717
Dec 17 '18 at 16:27
|
show 3 more comments
$begingroup$
As I understand, there is no closed form expression for
$$f(x, mu, sigma) = int_0^1 p^{(x-1)}(1-p)^{n-x-1}expleft(-{(text{logit}(p) -mu)^2 over 2sigma^2}right)dp.$$
Is it possible to obtain an analytical approximation for this?
integration numerical-methods approximation approximate-integration
asked Dec 17 '18 at 5:00
zipzapboingzipzapboing
979
$endgroup$
As I understand, there is no closed form expression for
$$f(x, mu, sigma) = int_0^1 p^{(x-1)}(1-p)^{n-x-1}expleft(-{(text{logit}(p) -mu)^2 over 2sigma^2}right)dp.$$
Is it possible to obtain an analytical approximation for this?
integration numerical-methods approximation approximate-integration
integration numerical-methods approximation approximate-integration
asked Dec 17 '18 at 5:00
zipzapboingzipzapboing
979
asked Dec 17 '18 at 5:00
zipzapboingzipzapboing
979
asked Dec 17 '18 at 5:00
zipzapboingzipzapboing
979
asked Dec 17 '18 at 5:00
zipzapboingzipzapboing
979
asked Dec 17 '18 at 5:00
zipzapboingzipzapboing
979
979
$begingroup$
Do you mean "asymptotic"? There are numerous analytic approximations to any continuous function.
$endgroup$
– user14717
Dec 17 '18 at 7:28
$begingroup$
I suppose there must be some that are more computationally efficient/reach small error with a smaller number of terms? I need to insert this as part of a statistical model so a tradeoff between accuracy and efficiency is needed.
$endgroup$
– zipzapboing
Dec 17 '18 at 11:19
$begingroup$
You could use quadrature to evaluate it at many arguments and then do interpolation.
$endgroup$
– user14717
Dec 17 '18 at 16:12
$begingroup$
I suppose that can work. I've never attempted something similar so I had something like a multivariable Taylor series in mind. I guess that theoretical guarantees are irrelevant when there's enough computational power to brute force the approximation and ensure it works over the range of things I care about.
$endgroup$
– zipzapboing
Dec 17 '18 at 16:23
1
$begingroup$
The theoretical guarantees will be much stronger with an interpolator than with multivariate Taylor series.
$endgroup$
– user14717
Dec 17 '18 at 16:27
|
show 3 more comments
$begingroup$
Do you mean "asymptotic"? There are numerous analytic approximations to any continuous function.
$endgroup$
– user14717
Dec 17 '18 at 7:28
$begingroup$
I suppose there must be some that are more computationally efficient/reach small error with a smaller number of terms? I need to insert this as part of a statistical model so a tradeoff between accuracy and efficiency is needed.
$endgroup$
– zipzapboing
Dec 17 '18 at 11:19
$begingroup$
You could use quadrature to evaluate it at many arguments and then do interpolation.
$endgroup$
– user14717
Dec 17 '18 at 16:12
$begingroup$
I suppose that can work. I've never attempted something similar so I had something like a multivariable Taylor series in mind. I guess that theoretical guarantees are irrelevant when there's enough computational power to brute force the approximation and ensure it works over the range of things I care about.
$endgroup$
– zipzapboing
Dec 17 '18 at 16:23
1
$begingroup$
The theoretical guarantees will be much stronger with an interpolator than with multivariate Taylor series.
$endgroup$
– user14717
Dec 17 '18 at 16:27
$begingroup$
Do you mean "asymptotic"? There are numerous analytic approximations to any continuous function.
$endgroup$
– user14717
Dec 17 '18 at 7:28
$begingroup$
Do you mean "asymptotic"? There are numerous analytic approximations to any continuous function.
$endgroup$
– user14717
Dec 17 '18 at 7:28
$begingroup$
I suppose there must be some that are more computationally efficient/reach small error with a smaller number of terms? I need to insert this as part of a statistical model so a tradeoff between accuracy and efficiency is needed.
$endgroup$
– zipzapboing
Dec 17 '18 at 11:19
$begingroup$
I suppose there must be some that are more computationally efficient/reach small error with a smaller number of terms? I need to insert this as part of a statistical model so a tradeoff between accuracy and efficiency is needed.
$endgroup$
– zipzapboing
Dec 17 '18 at 11:19
$begingroup$
You could use quadrature to evaluate it at many arguments and then do interpolation.
$endgroup$
– user14717
Dec 17 '18 at 16:12
$begingroup$
You could use quadrature to evaluate it at many arguments and then do interpolation.
$endgroup$
– user14717
Dec 17 '18 at 16:12
$begingroup$
I suppose that can work. I've never attempted something similar so I had something like a multivariable Taylor series in mind. I guess that theoretical guarantees are irrelevant when there's enough computational power to brute force the approximation and ensure it works over the range of things I care about.
$endgroup$
– zipzapboing
Dec 17 '18 at 16:23
$begingroup$
I suppose that can work. I've never attempted something similar so I had something like a multivariable Taylor series in mind. I guess that theoretical guarantees are irrelevant when there's enough computational power to brute force the approximation and ensure it works over the range of things I care about.
$endgroup$
– zipzapboing
Dec 17 '18 at 16:23
1
1
$begingroup$
The theoretical guarantees will be much stronger with an interpolator than with multivariate Taylor series.
$endgroup$
– user14717
Dec 17 '18 at 16:27
$begingroup$
The theoretical guarantees will be much stronger with an interpolator than with multivariate Taylor series.
$endgroup$
– user14717
Dec 17 '18 at 16:27
|
show 3 more comments
1 Answer
1
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oldest
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$begingroup$
Here's what you need to do:
Decide on an interpolator. I suggest a tricubic b-spline, but finding software for this is going to be painful. To understand this interpolant, start in Rainer Kress's Numerical Analysis which introduces it in 1D, learn about the bicubic b-splines in 2D, and then you'll be able to understand the tricubic. If you don't like tricubic b-splines, as an alternative, you might also be able to use multivariate Chebyshev series.
Interpolators require data at a particular geometry of points; figure out what those points are for your given interpolator and then evaluate the integral by quadrature at each point. (For tricubic b-splines it's easy: A uniform grid.) It looks like tanh-sinh quadrature is probably the best for this integral but Gaussian or Gauss-Kronrod will also work fine.
Another alternative is just to use quadrature to evaluate $f$ at any point $(x, mu, sigma)$, and ditch the interpolator. This will reduce the speed by a factor of 10 to 100, but since a quadrature takes about 500ns-1$mu$s, you might not really care.
If you've never done anything like this get ready for some effort shock.
answered Dec 17 '18 at 16:57
user14717user14717
3,8631120
$endgroup$
add a comment |
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$begingroup$
Here's what you need to do:
Decide on an interpolator. I suggest a tricubic b-spline, but finding software for this is going to be painful. To understand this interpolant, start in Rainer Kress's Numerical Analysis which introduces it in 1D, learn about the bicubic b-splines in 2D, and then you'll be able to understand the tricubic. If you don't like tricubic b-splines, as an alternative, you might also be able to use multivariate Chebyshev series.
Interpolators require data at a particular geometry of points; figure out what those points are for your given interpolator and then evaluate the integral by quadrature at each point. (For tricubic b-splines it's easy: A uniform grid.) It looks like tanh-sinh quadrature is probably the best for this integral but Gaussian or Gauss-Kronrod will also work fine.
Another alternative is just to use quadrature to evaluate $f$ at any point $(x, mu, sigma)$, and ditch the interpolator. This will reduce the speed by a factor of 10 to 100, but since a quadrature takes about 500ns-1$mu$s, you might not really care.
If you've never done anything like this get ready for some effort shock.
answered Dec 17 '18 at 16:57
user14717user14717
3,8631120
$endgroup$
add a comment |
$begingroup$
Here's what you need to do:
Decide on an interpolator. I suggest a tricubic b-spline, but finding software for this is going to be painful. To understand this interpolant, start in Rainer Kress's Numerical Analysis which introduces it in 1D, learn about the bicubic b-splines in 2D, and then you'll be able to understand the tricubic. If you don't like tricubic b-splines, as an alternative, you might also be able to use multivariate Chebyshev series.
Interpolators require data at a particular geometry of points; figure out what those points are for your given interpolator and then evaluate the integral by quadrature at each point. (For tricubic b-splines it's easy: A uniform grid.) It looks like tanh-sinh quadrature is probably the best for this integral but Gaussian or Gauss-Kronrod will also work fine.
Another alternative is just to use quadrature to evaluate $f$ at any point $(x, mu, sigma)$, and ditch the interpolator. This will reduce the speed by a factor of 10 to 100, but since a quadrature takes about 500ns-1$mu$s, you might not really care.
If you've never done anything like this get ready for some effort shock.
answered Dec 17 '18 at 16:57
user14717user14717
3,8631120
$endgroup$
add a comment |
$begingroup$
Here's what you need to do:
Decide on an interpolator. I suggest a tricubic b-spline, but finding software for this is going to be painful. To understand this interpolant, start in Rainer Kress's Numerical Analysis which introduces it in 1D, learn about the bicubic b-splines in 2D, and then you'll be able to understand the tricubic. If you don't like tricubic b-splines, as an alternative, you might also be able to use multivariate Chebyshev series.
Interpolators require data at a particular geometry of points; figure out what those points are for your given interpolator and then evaluate the integral by quadrature at each point. (For tricubic b-splines it's easy: A uniform grid.) It looks like tanh-sinh quadrature is probably the best for this integral but Gaussian or Gauss-Kronrod will also work fine.
Another alternative is just to use quadrature to evaluate $f$ at any point $(x, mu, sigma)$, and ditch the interpolator. This will reduce the speed by a factor of 10 to 100, but since a quadrature takes about 500ns-1$mu$s, you might not really care.
If you've never done anything like this get ready for some effort shock.
answered Dec 17 '18 at 16:57
user14717user14717
3,8631120
$endgroup$
Here's what you need to do:
Decide on an interpolator. I suggest a tricubic b-spline, but finding software for this is going to be painful. To understand this interpolant, start in Rainer Kress's Numerical Analysis which introduces it in 1D, learn about the bicubic b-splines in 2D, and then you'll be able to understand the tricubic. If you don't like tricubic b-splines, as an alternative, you might also be able to use multivariate Chebyshev series.
Interpolators require data at a particular geometry of points; figure out what those points are for your given interpolator and then evaluate the integral by quadrature at each point. (For tricubic b-splines it's easy: A uniform grid.) It looks like tanh-sinh quadrature is probably the best for this integral but Gaussian or Gauss-Kronrod will also work fine.
Another alternative is just to use quadrature to evaluate $f$ at any point $(x, mu, sigma)$, and ditch the interpolator. This will reduce the speed by a factor of 10 to 100, but since a quadrature takes about 500ns-1$mu$s, you might not really care.
If you've never done anything like this get ready for some effort shock.
answered Dec 17 '18 at 16:57
user14717user14717
3,8631120
edited Dec 17 '18 at 17:04
answered Dec 17 '18 at 16:57
user14717user14717
3,8631120
answered Dec 17 '18 at 16:57
user14717user14717
3,8631120
answered Dec 17 '18 at 16:57
user14717user14717
3,8631120
3,8631120
add a comment |
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$begingroup$
Do you mean "asymptotic"? There are numerous analytic approximations to any continuous function.
$endgroup$
– user14717
Dec 17 '18 at 7:28
$begingroup$
I suppose there must be some that are more computationally efficient/reach small error with a smaller number of terms? I need to insert this as part of a statistical model so a tradeoff between accuracy and efficiency is needed.
$endgroup$
– zipzapboing
Dec 17 '18 at 11:19
$begingroup$
You could use quadrature to evaluate it at many arguments and then do interpolation.
$endgroup$
– user14717
Dec 17 '18 at 16:12
$begingroup$
I suppose that can work. I've never attempted something similar so I had something like a multivariable Taylor series in mind. I guess that theoretical guarantees are irrelevant when there's enough computational power to brute force the approximation and ensure it works over the range of things I care about.
$endgroup$
– zipzapboing
Dec 17 '18 at 16:23
1
$begingroup$
The theoretical guarantees will be much stronger with an interpolator than with multivariate Taylor series.
$endgroup$
– user14717
Dec 17 '18 at 16:27