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
$endgroup$
|
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
$endgroup$
$begingroup$
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
$endgroup$
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
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
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043544%2fanalytical-approximation-for-logit-normal-binomial-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
edited Dec 17 '18 at 17:04
answered Dec 17 '18 at 16:57
user14717user14717
3,8631120
3,8631120
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043544%2fanalytical-approximation-for-logit-normal-binomial-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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