Mixing
How can we randomize a mapping/function?
$begingroup$
Suppose we have a set $A={a_1,a_2,ldots,a_n}$, where $ninmathbb{N}$. Suppose also that we have a injective function $f:Ato[0,1]$ which maps each element of $A$ to a real number between $0$ and $1$.
I want to write an explicit formula of $f$ that randomly assigns elements of $A$ to the closed interval$[0,1]$ But my issue is that I am not sure how to randomize the function $f$.
Is it possible to define an explicit formula of $f$ which randomizes the mapping of the elements of $A$ to $[0,1]$? My guess is that because of the closed interval $[0,1]$, function $f$ is common in mathematics, though I couldn't find anything.
I'd appreciate any hints.
probability functions
asked Feb 1 at 17:08
johnny09johnny09
705624
$endgroup$
add a comment |
$begingroup$
Suppose we have a set $A={a_1,a_2,ldots,a_n}$, where $ninmathbb{N}$. Suppose also that we have a injective function $f:Ato[0,1]$ which maps each element of $A$ to a real number between $0$ and $1$.
I want to write an explicit formula of $f$ that randomly assigns elements of $A$ to the closed interval$[0,1]$ But my issue is that I am not sure how to randomize the function $f$.
Is it possible to define an explicit formula of $f$ which randomizes the mapping of the elements of $A$ to $[0,1]$? My guess is that because of the closed interval $[0,1]$, function $f$ is common in mathematics, though I couldn't find anything.
I'd appreciate any hints.
probability functions
asked Feb 1 at 17:08
johnny09johnny09
705624
$endgroup$
1
$begingroup$
Do you mean injective functions? Since there is no bijection from a finite set to $[0,1]$. Also, what kind of randomness do you want? Even for coin tossing, where you have only two possible outcomes, you have freedom to choose the probability at which the head appears. Then in this case, you have infinitely many possible outcomes...
$endgroup$
– Sangchul Lee
Feb 1 at 17:33
$begingroup$
@SangchulLee Yes, my bad! I edited my question to correct the mistake! Thanks. I am not sure how to describe the "randomness." But as a "designer", say, it doesn't make a lot of sense to explicitly define the formula of function $f$, right?
$endgroup$
– johnny09
Feb 1 at 17:39
$begingroup$
@SangchulLee For example, if $f$ is a social preference weight function then ideally we'd want to randomize the mapping of the elements of $A$ to the interval $[0,1]$. The output will give you a scale of preference then.
$endgroup$
– johnny09
Feb 1 at 17:41
2
$begingroup$
What I wanted to say was that there are enormously many different ways to create a random function $f : A to [0,1]$, hence the word 'randomize' is not so well-defined. One simple choice is to assign each $f(a_i)$ a value uniformly chosen from $[0, 1]$, independently of all the others. But this means that any profile is equally likely, which is very unlikely if you want to model a real situation. In full generality, you may regard $f=(f(a_1),cdots,f(a_n))$ as random vector, and in order to generate a random vector, you need to specify the joint distribution.
$endgroup$
– Sangchul Lee
Feb 1 at 18:02
add a comment |
$begingroup$
Suppose we have a set $A={a_1,a_2,ldots,a_n}$, where $ninmathbb{N}$. Suppose also that we have a injective function $f:Ato[0,1]$ which maps each element of $A$ to a real number between $0$ and $1$.
I want to write an explicit formula of $f$ that randomly assigns elements of $A$ to the closed interval$[0,1]$ But my issue is that I am not sure how to randomize the function $f$.
Is it possible to define an explicit formula of $f$ which randomizes the mapping of the elements of $A$ to $[0,1]$? My guess is that because of the closed interval $[0,1]$, function $f$ is common in mathematics, though I couldn't find anything.
I'd appreciate any hints.
probability functions
asked Feb 1 at 17:08
johnny09johnny09
705624
$endgroup$
Suppose we have a set $A={a_1,a_2,ldots,a_n}$, where $ninmathbb{N}$. Suppose also that we have a injective function $f:Ato[0,1]$ which maps each element of $A$ to a real number between $0$ and $1$.
I want to write an explicit formula of $f$ that randomly assigns elements of $A$ to the closed interval$[0,1]$ But my issue is that I am not sure how to randomize the function $f$.
Is it possible to define an explicit formula of $f$ which randomizes the mapping of the elements of $A$ to $[0,1]$? My guess is that because of the closed interval $[0,1]$, function $f$ is common in mathematics, though I couldn't find anything.
I'd appreciate any hints.
probability functions
probability functions
asked Feb 1 at 17:08
johnny09johnny09
705624
asked Feb 1 at 17:08
johnny09johnny09
705624
edited Feb 1 at 17:36
johnny09
asked Feb 1 at 17:08
johnny09johnny09
705624
asked Feb 1 at 17:08
johnny09johnny09
705624
asked Feb 1 at 17:08
johnny09johnny09
705624
705624
1
$begingroup$
Do you mean injective functions? Since there is no bijection from a finite set to $[0,1]$. Also, what kind of randomness do you want? Even for coin tossing, where you have only two possible outcomes, you have freedom to choose the probability at which the head appears. Then in this case, you have infinitely many possible outcomes...
$endgroup$
– Sangchul Lee
Feb 1 at 17:33
$begingroup$
@SangchulLee Yes, my bad! I edited my question to correct the mistake! Thanks. I am not sure how to describe the "randomness." But as a "designer", say, it doesn't make a lot of sense to explicitly define the formula of function $f$, right?
$endgroup$
– johnny09
Feb 1 at 17:39
$begingroup$
@SangchulLee For example, if $f$ is a social preference weight function then ideally we'd want to randomize the mapping of the elements of $A$ to the interval $[0,1]$. The output will give you a scale of preference then.
$endgroup$
– johnny09
Feb 1 at 17:41
2
$begingroup$
What I wanted to say was that there are enormously many different ways to create a random function $f : A to [0,1]$, hence the word 'randomize' is not so well-defined. One simple choice is to assign each $f(a_i)$ a value uniformly chosen from $[0, 1]$, independently of all the others. But this means that any profile is equally likely, which is very unlikely if you want to model a real situation. In full generality, you may regard $f=(f(a_1),cdots,f(a_n))$ as random vector, and in order to generate a random vector, you need to specify the joint distribution.
$endgroup$
– Sangchul Lee
Feb 1 at 18:02
add a comment |
1
$begingroup$
Do you mean injective functions? Since there is no bijection from a finite set to $[0,1]$. Also, what kind of randomness do you want? Even for coin tossing, where you have only two possible outcomes, you have freedom to choose the probability at which the head appears. Then in this case, you have infinitely many possible outcomes...
$endgroup$
– Sangchul Lee
Feb 1 at 17:33
$begingroup$
@SangchulLee Yes, my bad! I edited my question to correct the mistake! Thanks. I am not sure how to describe the "randomness." But as a "designer", say, it doesn't make a lot of sense to explicitly define the formula of function $f$, right?
$endgroup$
– johnny09
Feb 1 at 17:39
$begingroup$
@SangchulLee For example, if $f$ is a social preference weight function then ideally we'd want to randomize the mapping of the elements of $A$ to the interval $[0,1]$. The output will give you a scale of preference then.
$endgroup$
– johnny09
Feb 1 at 17:41
2
$begingroup$
What I wanted to say was that there are enormously many different ways to create a random function $f : A to [0,1]$, hence the word 'randomize' is not so well-defined. One simple choice is to assign each $f(a_i)$ a value uniformly chosen from $[0, 1]$, independently of all the others. But this means that any profile is equally likely, which is very unlikely if you want to model a real situation. In full generality, you may regard $f=(f(a_1),cdots,f(a_n))$ as random vector, and in order to generate a random vector, you need to specify the joint distribution.
$endgroup$
– Sangchul Lee
Feb 1 at 18:02
1
1
$begingroup$
Do you mean injective functions? Since there is no bijection from a finite set to $[0,1]$. Also, what kind of randomness do you want? Even for coin tossing, where you have only two possible outcomes, you have freedom to choose the probability at which the head appears. Then in this case, you have infinitely many possible outcomes...
$endgroup$
– Sangchul Lee
Feb 1 at 17:33
$begingroup$
Do you mean injective functions? Since there is no bijection from a finite set to $[0,1]$. Also, what kind of randomness do you want? Even for coin tossing, where you have only two possible outcomes, you have freedom to choose the probability at which the head appears. Then in this case, you have infinitely many possible outcomes...
$endgroup$
– Sangchul Lee
Feb 1 at 17:33
$begingroup$
@SangchulLee Yes, my bad! I edited my question to correct the mistake! Thanks. I am not sure how to describe the "randomness." But as a "designer", say, it doesn't make a lot of sense to explicitly define the formula of function $f$, right?
$endgroup$
– johnny09
Feb 1 at 17:39
$begingroup$
@SangchulLee Yes, my bad! I edited my question to correct the mistake! Thanks. I am not sure how to describe the "randomness." But as a "designer", say, it doesn't make a lot of sense to explicitly define the formula of function $f$, right?
$endgroup$
– johnny09
Feb 1 at 17:39
$begingroup$
@SangchulLee For example, if $f$ is a social preference weight function then ideally we'd want to randomize the mapping of the elements of $A$ to the interval $[0,1]$. The output will give you a scale of preference then.
$endgroup$
– johnny09
Feb 1 at 17:41
$begingroup$
@SangchulLee For example, if $f$ is a social preference weight function then ideally we'd want to randomize the mapping of the elements of $A$ to the interval $[0,1]$. The output will give you a scale of preference then.
$endgroup$
– johnny09
Feb 1 at 17:41
2
2
$begingroup$
What I wanted to say was that there are enormously many different ways to create a random function $f : A to [0,1]$, hence the word 'randomize' is not so well-defined. One simple choice is to assign each $f(a_i)$ a value uniformly chosen from $[0, 1]$, independently of all the others. But this means that any profile is equally likely, which is very unlikely if you want to model a real situation. In full generality, you may regard $f=(f(a_1),cdots,f(a_n))$ as random vector, and in order to generate a random vector, you need to specify the joint distribution.
$endgroup$
– Sangchul Lee
Feb 1 at 18:02
$begingroup$
What I wanted to say was that there are enormously many different ways to create a random function $f : A to [0,1]$, hence the word 'randomize' is not so well-defined. One simple choice is to assign each $f(a_i)$ a value uniformly chosen from $[0, 1]$, independently of all the others. But this means that any profile is equally likely, which is very unlikely if you want to model a real situation. In full generality, you may regard $f=(f(a_1),cdots,f(a_n))$ as random vector, and in order to generate a random vector, you need to specify the joint distribution.
$endgroup$
– Sangchul Lee
Feb 1 at 18:02
add a comment |
0
active
oldest
votes
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%2f3096478%2fhow-can-we-randomize-a-mapping-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3096478%2fhow-can-we-randomize-a-mapping-function%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
1
$begingroup$
Do you mean injective functions? Since there is no bijection from a finite set to $[0,1]$. Also, what kind of randomness do you want? Even for coin tossing, where you have only two possible outcomes, you have freedom to choose the probability at which the head appears. Then in this case, you have infinitely many possible outcomes...
$endgroup$
– Sangchul Lee
Feb 1 at 17:33
$begingroup$
@SangchulLee Yes, my bad! I edited my question to correct the mistake! Thanks. I am not sure how to describe the "randomness." But as a "designer", say, it doesn't make a lot of sense to explicitly define the formula of function $f$, right?
$endgroup$
– johnny09
Feb 1 at 17:39
$begingroup$
@SangchulLee For example, if $f$ is a social preference weight function then ideally we'd want to randomize the mapping of the elements of $A$ to the interval $[0,1]$. The output will give you a scale of preference then.
$endgroup$
– johnny09
Feb 1 at 17:41
2
$begingroup$
What I wanted to say was that there are enormously many different ways to create a random function $f : A to [0,1]$, hence the word 'randomize' is not so well-defined. One simple choice is to assign each $f(a_i)$ a value uniformly chosen from $[0, 1]$, independently of all the others. But this means that any profile is equally likely, which is very unlikely if you want to model a real situation. In full generality, you may regard $f=(f(a_1),cdots,f(a_n))$ as random vector, and in order to generate a random vector, you need to specify the joint distribution.
$endgroup$
– Sangchul Lee
Feb 1 at 18:02