Mixing
Operator in the presence of noise
$begingroup$
I have a vector $x in mathbb{R}^d$ with norm one. I also have random noise vector $eta in mathbb{R}^d$ with norm $1/k$ taken at random from the sphere $S^{d-1}$ of radius 1/k, where k is a large natural number.
Let's now define the softmax function $sigma: mathbb{R}^d rightarrow mathbb{R}^d$ defined in the following way:
$sigma(x)_i = frac{e^{x_i}}{sum_j e^{x_j}}$
I need to compute the following expected value, but in the comments people made me realize that it's not possible.
$$ mathbb{E}[||sigma(x) - sigma(x+eta)||]?$$
Which means i have to look for a reasonable upper bound. Any suggestions?
Thank you!
probability norm
asked Jan 21 at 17:40
AlfredAlfred
357
$endgroup$
add a comment |
$begingroup$
I have a vector $x in mathbb{R}^d$ with norm one. I also have random noise vector $eta in mathbb{R}^d$ with norm $1/k$ taken at random from the sphere $S^{d-1}$ of radius 1/k, where k is a large natural number.
Let's now define the softmax function $sigma: mathbb{R}^d rightarrow mathbb{R}^d$ defined in the following way:
$sigma(x)_i = frac{e^{x_i}}{sum_j e^{x_j}}$
I need to compute the following expected value, but in the comments people made me realize that it's not possible.
$$ mathbb{E}[||sigma(x) - sigma(x+eta)||]?$$
Which means i have to look for a reasonable upper bound. Any suggestions?
Thank you!
probability norm
asked Jan 21 at 17:40
AlfredAlfred
357
$endgroup$
add a comment |
$begingroup$
I have a vector $x in mathbb{R}^d$ with norm one. I also have random noise vector $eta in mathbb{R}^d$ with norm $1/k$ taken at random from the sphere $S^{d-1}$ of radius 1/k, where k is a large natural number.
Let's now define the softmax function $sigma: mathbb{R}^d rightarrow mathbb{R}^d$ defined in the following way:
$sigma(x)_i = frac{e^{x_i}}{sum_j e^{x_j}}$
I need to compute the following expected value, but in the comments people made me realize that it's not possible.
$$ mathbb{E}[||sigma(x) - sigma(x+eta)||]?$$
Which means i have to look for a reasonable upper bound. Any suggestions?
Thank you!
probability norm
asked Jan 21 at 17:40
AlfredAlfred
357
$endgroup$
I have a vector $x in mathbb{R}^d$ with norm one. I also have random noise vector $eta in mathbb{R}^d$ with norm $1/k$ taken at random from the sphere $S^{d-1}$ of radius 1/k, where k is a large natural number.
Let's now define the softmax function $sigma: mathbb{R}^d rightarrow mathbb{R}^d$ defined in the following way:
$sigma(x)_i = frac{e^{x_i}}{sum_j e^{x_j}}$
I need to compute the following expected value, but in the comments people made me realize that it's not possible.
$$ mathbb{E}[||sigma(x) - sigma(x+eta)||]?$$
Which means i have to look for a reasonable upper bound. Any suggestions?
Thank you!
probability norm
probability norm
asked Jan 21 at 17:40
AlfredAlfred
357
asked Jan 21 at 17:40
AlfredAlfred
357
edited Feb 14 at 0:25
Alfred
asked Jan 21 at 17:40
AlfredAlfred
357
asked Jan 21 at 17:40
AlfredAlfred
357
asked Jan 21 at 17:40
AlfredAlfred
357
357
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think it is not possible analytically.
As a very simplified version of your problem, I have tried to calculate the expected value of the exponent of your random variable in 2D.
$$langle e^{eta_1} rangle = frac{1}{2 pi} int_0^{2pi} e^{frac{cos(theta)}{k}}dtheta$$
According to WolframAlpha, there is no closed-form solution to this integral in terms of basic functions. In order to solve your problem analytically, an integral of a more complicated integrand needs to be performed over $d-1$ dimensional sphere.
This is not unusual. Expected value is just an integral, and not all integrals have nice analytic solutions. However, it should be rather simple to compute your expected value numerically by uniformly sampling points on a sphere and taking the average of the computed norm.
$$ mathbb{E}[||sigma(vec{x}) - sigma(vec{x}+vec{eta})||] approx frac{1}{n}sum_{j=1}^n ||sigma(vec{x}) - sigma(vec{x}+vec{eta}_j)||$$
where $vec{eta}_j$ is a sample from your spherical distribution.
answered Jan 22 at 12:16
Aleksejs FominsAleksejs Fomins
555211
$endgroup$
$begingroup$
Oh thank you very much. Which means i have to find an upper bound to the expected value. I'll edit the question in that direction. Thank you again!
$endgroup$
– Alfred
Jan 22 at 13:55
$begingroup$
Finding an upper bound is a different question to finding an analytic or numeric solution. I suggest that you mark this question as complete and ask a new question. Asking a new question is beneficial for you, because it will attract more attention, as it will appear at the beginning of the question list.
$endgroup$
– Aleksejs Fomins
Jan 22 at 15:38
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%2f3082160%2foperator-in-the-presence-of-noise%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
$begingroup$
I think it is not possible analytically.
As a very simplified version of your problem, I have tried to calculate the expected value of the exponent of your random variable in 2D.
$$langle e^{eta_1} rangle = frac{1}{2 pi} int_0^{2pi} e^{frac{cos(theta)}{k}}dtheta$$
According to WolframAlpha, there is no closed-form solution to this integral in terms of basic functions. In order to solve your problem analytically, an integral of a more complicated integrand needs to be performed over $d-1$ dimensional sphere.
This is not unusual. Expected value is just an integral, and not all integrals have nice analytic solutions. However, it should be rather simple to compute your expected value numerically by uniformly sampling points on a sphere and taking the average of the computed norm.
$$ mathbb{E}[||sigma(vec{x}) - sigma(vec{x}+vec{eta})||] approx frac{1}{n}sum_{j=1}^n ||sigma(vec{x}) - sigma(vec{x}+vec{eta}_j)||$$
where $vec{eta}_j$ is a sample from your spherical distribution.
answered Jan 22 at 12:16
Aleksejs FominsAleksejs Fomins
555211
$endgroup$
$begingroup$
Oh thank you very much. Which means i have to find an upper bound to the expected value. I'll edit the question in that direction. Thank you again!
$endgroup$
– Alfred
Jan 22 at 13:55
$begingroup$
Finding an upper bound is a different question to finding an analytic or numeric solution. I suggest that you mark this question as complete and ask a new question. Asking a new question is beneficial for you, because it will attract more attention, as it will appear at the beginning of the question list.
$endgroup$
– Aleksejs Fomins
Jan 22 at 15:38
add a comment |
$begingroup$
I think it is not possible analytically.
As a very simplified version of your problem, I have tried to calculate the expected value of the exponent of your random variable in 2D.
$$langle e^{eta_1} rangle = frac{1}{2 pi} int_0^{2pi} e^{frac{cos(theta)}{k}}dtheta$$
According to WolframAlpha, there is no closed-form solution to this integral in terms of basic functions. In order to solve your problem analytically, an integral of a more complicated integrand needs to be performed over $d-1$ dimensional sphere.
This is not unusual. Expected value is just an integral, and not all integrals have nice analytic solutions. However, it should be rather simple to compute your expected value numerically by uniformly sampling points on a sphere and taking the average of the computed norm.
$$ mathbb{E}[||sigma(vec{x}) - sigma(vec{x}+vec{eta})||] approx frac{1}{n}sum_{j=1}^n ||sigma(vec{x}) - sigma(vec{x}+vec{eta}_j)||$$
where $vec{eta}_j$ is a sample from your spherical distribution.
answered Jan 22 at 12:16
Aleksejs FominsAleksejs Fomins
555211
$endgroup$
$begingroup$
Oh thank you very much. Which means i have to find an upper bound to the expected value. I'll edit the question in that direction. Thank you again!
$endgroup$
– Alfred
Jan 22 at 13:55
$begingroup$
Finding an upper bound is a different question to finding an analytic or numeric solution. I suggest that you mark this question as complete and ask a new question. Asking a new question is beneficial for you, because it will attract more attention, as it will appear at the beginning of the question list.
$endgroup$
– Aleksejs Fomins
Jan 22 at 15:38
add a comment |
$begingroup$
I think it is not possible analytically.
As a very simplified version of your problem, I have tried to calculate the expected value of the exponent of your random variable in 2D.
$$langle e^{eta_1} rangle = frac{1}{2 pi} int_0^{2pi} e^{frac{cos(theta)}{k}}dtheta$$
According to WolframAlpha, there is no closed-form solution to this integral in terms of basic functions. In order to solve your problem analytically, an integral of a more complicated integrand needs to be performed over $d-1$ dimensional sphere.
This is not unusual. Expected value is just an integral, and not all integrals have nice analytic solutions. However, it should be rather simple to compute your expected value numerically by uniformly sampling points on a sphere and taking the average of the computed norm.
$$ mathbb{E}[||sigma(vec{x}) - sigma(vec{x}+vec{eta})||] approx frac{1}{n}sum_{j=1}^n ||sigma(vec{x}) - sigma(vec{x}+vec{eta}_j)||$$
where $vec{eta}_j$ is a sample from your spherical distribution.
answered Jan 22 at 12:16
Aleksejs FominsAleksejs Fomins
555211
$endgroup$
I think it is not possible analytically.
As a very simplified version of your problem, I have tried to calculate the expected value of the exponent of your random variable in 2D.
$$langle e^{eta_1} rangle = frac{1}{2 pi} int_0^{2pi} e^{frac{cos(theta)}{k}}dtheta$$
According to WolframAlpha, there is no closed-form solution to this integral in terms of basic functions. In order to solve your problem analytically, an integral of a more complicated integrand needs to be performed over $d-1$ dimensional sphere.
This is not unusual. Expected value is just an integral, and not all integrals have nice analytic solutions. However, it should be rather simple to compute your expected value numerically by uniformly sampling points on a sphere and taking the average of the computed norm.
$$ mathbb{E}[||sigma(vec{x}) - sigma(vec{x}+vec{eta})||] approx frac{1}{n}sum_{j=1}^n ||sigma(vec{x}) - sigma(vec{x}+vec{eta}_j)||$$
where $vec{eta}_j$ is a sample from your spherical distribution.
answered Jan 22 at 12:16
Aleksejs FominsAleksejs Fomins
555211
answered Jan 22 at 12:16
Aleksejs FominsAleksejs Fomins
555211
answered Jan 22 at 12:16
Aleksejs FominsAleksejs Fomins
555211
answered Jan 22 at 12:16
Aleksejs FominsAleksejs Fomins
555211
555211
$begingroup$
Oh thank you very much. Which means i have to find an upper bound to the expected value. I'll edit the question in that direction. Thank you again!
$endgroup$
– Alfred
Jan 22 at 13:55
$begingroup$
Finding an upper bound is a different question to finding an analytic or numeric solution. I suggest that you mark this question as complete and ask a new question. Asking a new question is beneficial for you, because it will attract more attention, as it will appear at the beginning of the question list.
$endgroup$
– Aleksejs Fomins
Jan 22 at 15:38
add a comment |
$begingroup$
Oh thank you very much. Which means i have to find an upper bound to the expected value. I'll edit the question in that direction. Thank you again!
$endgroup$
– Alfred
Jan 22 at 13:55
$begingroup$
Finding an upper bound is a different question to finding an analytic or numeric solution. I suggest that you mark this question as complete and ask a new question. Asking a new question is beneficial for you, because it will attract more attention, as it will appear at the beginning of the question list.
$endgroup$
– Aleksejs Fomins
Jan 22 at 15:38
$begingroup$
Oh thank you very much. Which means i have to find an upper bound to the expected value. I'll edit the question in that direction. Thank you again!
$endgroup$
– Alfred
Jan 22 at 13:55
$begingroup$
Oh thank you very much. Which means i have to find an upper bound to the expected value. I'll edit the question in that direction. Thank you again!
$endgroup$
– Alfred
Jan 22 at 13:55
$begingroup$
Finding an upper bound is a different question to finding an analytic or numeric solution. I suggest that you mark this question as complete and ask a new question. Asking a new question is beneficial for you, because it will attract more attention, as it will appear at the beginning of the question list.
$endgroup$
– Aleksejs Fomins
Jan 22 at 15:38
$begingroup$
Finding an upper bound is a different question to finding an analytic or numeric solution. I suggest that you mark this question as complete and ask a new question. Asking a new question is beneficial for you, because it will attract more attention, as it will appear at the beginning of the question list.
$endgroup$
– Aleksejs Fomins
Jan 22 at 15:38
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%2f3082160%2foperator-in-the-presence-of-noise%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
