Mixing
The expected value of a random function
$begingroup$
Suppose we have a 'random function' $f(x)$ that will be one of $n$ functions $f_1(x), ..., f_n(x)$ with probabilities $p_1, ..., p_n$. Intuitively, it might seem sensible to define
$$E[f(x)] = sum_{i=1}^{n}f_i(x)p_i$$
Now suppose $f(x)$ can be any one of a continuum of functions, each indexed by a unique real number. Again, it might seem sensible to define
$$E[f(x)] = int_{-infty}^{infty}f_i(x)g(i)di$$
where $g(i)$ describes the probability densities of the different possible functions.
However, I cannot find such expressions anywhere; and when I Google 'functional integration' I find rather more complicated notions than those expressed here.
Is there some reason why the definitions above are invalid/problematic? To be clear, I understand that the set of real valued functions is larger than the set of real numbers. So if all real valued functions had positive probability, it would not be possible to index them with the real numbers as I have done here. However, suppose we somehow knew that there were only a continuum of functions with positive probability: could we then define the expected value of the function as above?
Many thanks in advance!
probability-theory infinity expected-value
asked Jan 11 at 18:20
afreelunchafreelunch
235
$endgroup$
add a comment |
$begingroup$
Suppose we have a 'random function' $f(x)$ that will be one of $n$ functions $f_1(x), ..., f_n(x)$ with probabilities $p_1, ..., p_n$. Intuitively, it might seem sensible to define
$$E[f(x)] = sum_{i=1}^{n}f_i(x)p_i$$
Now suppose $f(x)$ can be any one of a continuum of functions, each indexed by a unique real number. Again, it might seem sensible to define
$$E[f(x)] = int_{-infty}^{infty}f_i(x)g(i)di$$
where $g(i)$ describes the probability densities of the different possible functions.
However, I cannot find such expressions anywhere; and when I Google 'functional integration' I find rather more complicated notions than those expressed here.
Is there some reason why the definitions above are invalid/problematic? To be clear, I understand that the set of real valued functions is larger than the set of real numbers. So if all real valued functions had positive probability, it would not be possible to index them with the real numbers as I have done here. However, suppose we somehow knew that there were only a continuum of functions with positive probability: could we then define the expected value of the function as above?
Many thanks in advance!
probability-theory infinity expected-value
asked Jan 11 at 18:20
afreelunchafreelunch
235
$endgroup$
1
$begingroup$
I don't know why there would be a problem. You're just defining a real number $f(x)$ for each $x$ by taking the expected value of some random variable
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Also, you should write $E[f(x)] = int_{-infty}^infty f_t(x)g(t)dt$
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Thanks, I have modified this.
$endgroup$
– afreelunch
Jan 11 at 18:25
$begingroup$
I mean, of course for the integral defn of $E[f(x)]$, you must make sure $t mapsto f_t(x)g(t)$ is measurable or whatever
$endgroup$
– mathworker21
Jan 11 at 18:39
$begingroup$
Another way to see this: You have a real-valued function $f(x, omega)$ for $x in mathbb{R}$ and $omega in S$ (where $S$ is the sample space). So for each $x in mathbb{R}$ you can define random variable $Y_x:Srightarrowmathbb{R}$ by $Y_x(omega)=f(x,omega)$. Its expectation $E[Y_x]$ is defined as usual for random variables. Since the $x$ parameter determines properties of $Y_x$, the expectation $E[Y_x]$ may depend on $x$. We would also need to assume the standard things for random variables, that is for each $x$ assume $Y_x$ has a CDF $P[Y_xleq y]$ for all $y in mathbb{R}$.
$endgroup$
– Michael
Jan 11 at 21:19
add a comment |
$begingroup$
Suppose we have a 'random function' $f(x)$ that will be one of $n$ functions $f_1(x), ..., f_n(x)$ with probabilities $p_1, ..., p_n$. Intuitively, it might seem sensible to define
$$E[f(x)] = sum_{i=1}^{n}f_i(x)p_i$$
Now suppose $f(x)$ can be any one of a continuum of functions, each indexed by a unique real number. Again, it might seem sensible to define
$$E[f(x)] = int_{-infty}^{infty}f_i(x)g(i)di$$
where $g(i)$ describes the probability densities of the different possible functions.
However, I cannot find such expressions anywhere; and when I Google 'functional integration' I find rather more complicated notions than those expressed here.
Is there some reason why the definitions above are invalid/problematic? To be clear, I understand that the set of real valued functions is larger than the set of real numbers. So if all real valued functions had positive probability, it would not be possible to index them with the real numbers as I have done here. However, suppose we somehow knew that there were only a continuum of functions with positive probability: could we then define the expected value of the function as above?
Many thanks in advance!
probability-theory infinity expected-value
asked Jan 11 at 18:20
afreelunchafreelunch
235
$endgroup$
Suppose we have a 'random function' $f(x)$ that will be one of $n$ functions $f_1(x), ..., f_n(x)$ with probabilities $p_1, ..., p_n$. Intuitively, it might seem sensible to define
$$E[f(x)] = sum_{i=1}^{n}f_i(x)p_i$$
Now suppose $f(x)$ can be any one of a continuum of functions, each indexed by a unique real number. Again, it might seem sensible to define
$$E[f(x)] = int_{-infty}^{infty}f_i(x)g(i)di$$
where $g(i)$ describes the probability densities of the different possible functions.
However, I cannot find such expressions anywhere; and when I Google 'functional integration' I find rather more complicated notions than those expressed here.
Is there some reason why the definitions above are invalid/problematic? To be clear, I understand that the set of real valued functions is larger than the set of real numbers. So if all real valued functions had positive probability, it would not be possible to index them with the real numbers as I have done here. However, suppose we somehow knew that there were only a continuum of functions with positive probability: could we then define the expected value of the function as above?
Many thanks in advance!
probability-theory infinity expected-value
probability-theory infinity expected-value
asked Jan 11 at 18:20
afreelunchafreelunch
235
asked Jan 11 at 18:20
afreelunchafreelunch
235
edited Jan 11 at 18:31
afreelunch
asked Jan 11 at 18:20
afreelunchafreelunch
235
asked Jan 11 at 18:20
afreelunchafreelunch
235
asked Jan 11 at 18:20
afreelunchafreelunch
235
235
1
$begingroup$
I don't know why there would be a problem. You're just defining a real number $f(x)$ for each $x$ by taking the expected value of some random variable
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Also, you should write $E[f(x)] = int_{-infty}^infty f_t(x)g(t)dt$
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Thanks, I have modified this.
$endgroup$
– afreelunch
Jan 11 at 18:25
$begingroup$
I mean, of course for the integral defn of $E[f(x)]$, you must make sure $t mapsto f_t(x)g(t)$ is measurable or whatever
$endgroup$
– mathworker21
Jan 11 at 18:39
$begingroup$
Another way to see this: You have a real-valued function $f(x, omega)$ for $x in mathbb{R}$ and $omega in S$ (where $S$ is the sample space). So for each $x in mathbb{R}$ you can define random variable $Y_x:Srightarrowmathbb{R}$ by $Y_x(omega)=f(x,omega)$. Its expectation $E[Y_x]$ is defined as usual for random variables. Since the $x$ parameter determines properties of $Y_x$, the expectation $E[Y_x]$ may depend on $x$. We would also need to assume the standard things for random variables, that is for each $x$ assume $Y_x$ has a CDF $P[Y_xleq y]$ for all $y in mathbb{R}$.
$endgroup$
– Michael
Jan 11 at 21:19
add a comment |
1
$begingroup$
I don't know why there would be a problem. You're just defining a real number $f(x)$ for each $x$ by taking the expected value of some random variable
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Also, you should write $E[f(x)] = int_{-infty}^infty f_t(x)g(t)dt$
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Thanks, I have modified this.
$endgroup$
– afreelunch
Jan 11 at 18:25
$begingroup$
I mean, of course for the integral defn of $E[f(x)]$, you must make sure $t mapsto f_t(x)g(t)$ is measurable or whatever
$endgroup$
– mathworker21
Jan 11 at 18:39
$begingroup$
Another way to see this: You have a real-valued function $f(x, omega)$ for $x in mathbb{R}$ and $omega in S$ (where $S$ is the sample space). So for each $x in mathbb{R}$ you can define random variable $Y_x:Srightarrowmathbb{R}$ by $Y_x(omega)=f(x,omega)$. Its expectation $E[Y_x]$ is defined as usual for random variables. Since the $x$ parameter determines properties of $Y_x$, the expectation $E[Y_x]$ may depend on $x$. We would also need to assume the standard things for random variables, that is for each $x$ assume $Y_x$ has a CDF $P[Y_xleq y]$ for all $y in mathbb{R}$.
$endgroup$
– Michael
Jan 11 at 21:19
1
1
$begingroup$
I don't know why there would be a problem. You're just defining a real number $f(x)$ for each $x$ by taking the expected value of some random variable
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
I don't know why there would be a problem. You're just defining a real number $f(x)$ for each $x$ by taking the expected value of some random variable
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Also, you should write $E[f(x)] = int_{-infty}^infty f_t(x)g(t)dt$
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Also, you should write $E[f(x)] = int_{-infty}^infty f_t(x)g(t)dt$
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Thanks, I have modified this.
$endgroup$
– afreelunch
Jan 11 at 18:25
$begingroup$
Thanks, I have modified this.
$endgroup$
– afreelunch
Jan 11 at 18:25
$begingroup$
I mean, of course for the integral defn of $E[f(x)]$, you must make sure $t mapsto f_t(x)g(t)$ is measurable or whatever
$endgroup$
– mathworker21
Jan 11 at 18:39
$begingroup$
I mean, of course for the integral defn of $E[f(x)]$, you must make sure $t mapsto f_t(x)g(t)$ is measurable or whatever
$endgroup$
– mathworker21
Jan 11 at 18:39
$begingroup$
Another way to see this: You have a real-valued function $f(x, omega)$ for $x in mathbb{R}$ and $omega in S$ (where $S$ is the sample space). So for each $x in mathbb{R}$ you can define random variable $Y_x:Srightarrowmathbb{R}$ by $Y_x(omega)=f(x,omega)$. Its expectation $E[Y_x]$ is defined as usual for random variables. Since the $x$ parameter determines properties of $Y_x$, the expectation $E[Y_x]$ may depend on $x$. We would also need to assume the standard things for random variables, that is for each $x$ assume $Y_x$ has a CDF $P[Y_xleq y]$ for all $y in mathbb{R}$.
$endgroup$
– Michael
Jan 11 at 21:19
$begingroup$
Another way to see this: You have a real-valued function $f(x, omega)$ for $x in mathbb{R}$ and $omega in S$ (where $S$ is the sample space). So for each $x in mathbb{R}$ you can define random variable $Y_x:Srightarrowmathbb{R}$ by $Y_x(omega)=f(x,omega)$. Its expectation $E[Y_x]$ is defined as usual for random variables. Since the $x$ parameter determines properties of $Y_x$, the expectation $E[Y_x]$ may depend on $x$. We would also need to assume the standard things for random variables, that is for each $x$ assume $Y_x$ has a CDF $P[Y_xleq y]$ for all $y in mathbb{R}$.
$endgroup$
– Michael
Jan 11 at 21:19
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%2f3070177%2fthe-expected-value-of-a-random-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%2f3070177%2fthe-expected-value-of-a-random-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
$begingroup$
I don't know why there would be a problem. You're just defining a real number $f(x)$ for each $x$ by taking the expected value of some random variable
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Also, you should write $E[f(x)] = int_{-infty}^infty f_t(x)g(t)dt$
$endgroup$
– mathworker21
Jan 11 at 18:24
$begingroup$
Thanks, I have modified this.
$endgroup$
– afreelunch
Jan 11 at 18:25
$begingroup$
I mean, of course for the integral defn of $E[f(x)]$, you must make sure $t mapsto f_t(x)g(t)$ is measurable or whatever
$endgroup$
– mathworker21
Jan 11 at 18:39
$begingroup$
Another way to see this: You have a real-valued function $f(x, omega)$ for $x in mathbb{R}$ and $omega in S$ (where $S$ is the sample space). So for each $x in mathbb{R}$ you can define random variable $Y_x:Srightarrowmathbb{R}$ by $Y_x(omega)=f(x,omega)$. Its expectation $E[Y_x]$ is defined as usual for random variables. Since the $x$ parameter determines properties of $Y_x$, the expectation $E[Y_x]$ may depend on $x$. We would also need to assume the standard things for random variables, that is for each $x$ assume $Y_x$ has a CDF $P[Y_xleq y]$ for all $y in mathbb{R}$.
$endgroup$
– Michael
Jan 11 at 21:19