A question about continuity of a specific function with probability measure











up vote
1
down vote

favorite












Let $X$ be a compact metric space, and $Theta$ be a finite space, endowed with their own $sigma$-algebra.



Let $f colon X times Theta to mathbb{R}$ be a Caratheodory function such that
(1) for each $x in X$, the function $f(x, cdot) colon Theta to mathbb{R}$
is measurable; and (2) for each $theta in Theta$, the function $f( cdot, theta) colon X to mathbb{R}$ is continuous.



Given each $x in X$, we have a probability distribution $pi( cdot ,| , x) colon 2^{Theta} to [0,1]$. In particular, given any fixed $x in X$,
it will generate a corresponding probability distribution $pi$ on $2^Theta$.



I am curious that




Under what kind of conditions (assumptions) imposed on this probability distribution $pi$ , the map $$X ni x mapsto int_Theta f(x,theta) , pi( mathrm{d} theta ,| ,x) in mathbb{R}$$
will be continuous on $X$?




Any idea or suggestions are most welcome!



Thank you so much!










share|cite|improve this question
























  • Thanks @Michael, you’re right. Actually, the space $Theta$ I considered was a compact metric space, but now I just want to simplify the question and restrict it to be finite.
    – Paradiesvogel
    Nov 18 at 10:50










  • Thanks @Michael . I totally agree with you. In fact, I really need the probability distribution $pi(cdot | x)$ depending on $x$. This means for each $x in X$, I have a different probability distribution defined on $2^Theta$. Also, the space $Theta$ is at least not trivial. In such a setting, what can I do to ensure the map $h(x)$ is continuous on $X$? Is it possible to do that? Thanks a million again :-)
    – Paradiesvogel
    Nov 18 at 11:02

















up vote
1
down vote

favorite












Let $X$ be a compact metric space, and $Theta$ be a finite space, endowed with their own $sigma$-algebra.



Let $f colon X times Theta to mathbb{R}$ be a Caratheodory function such that
(1) for each $x in X$, the function $f(x, cdot) colon Theta to mathbb{R}$
is measurable; and (2) for each $theta in Theta$, the function $f( cdot, theta) colon X to mathbb{R}$ is continuous.



Given each $x in X$, we have a probability distribution $pi( cdot ,| , x) colon 2^{Theta} to [0,1]$. In particular, given any fixed $x in X$,
it will generate a corresponding probability distribution $pi$ on $2^Theta$.



I am curious that




Under what kind of conditions (assumptions) imposed on this probability distribution $pi$ , the map $$X ni x mapsto int_Theta f(x,theta) , pi( mathrm{d} theta ,| ,x) in mathbb{R}$$
will be continuous on $X$?




Any idea or suggestions are most welcome!



Thank you so much!










share|cite|improve this question
























  • Thanks @Michael, you’re right. Actually, the space $Theta$ I considered was a compact metric space, but now I just want to simplify the question and restrict it to be finite.
    – Paradiesvogel
    Nov 18 at 10:50










  • Thanks @Michael . I totally agree with you. In fact, I really need the probability distribution $pi(cdot | x)$ depending on $x$. This means for each $x in X$, I have a different probability distribution defined on $2^Theta$. Also, the space $Theta$ is at least not trivial. In such a setting, what can I do to ensure the map $h(x)$ is continuous on $X$? Is it possible to do that? Thanks a million again :-)
    – Paradiesvogel
    Nov 18 at 11:02















up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $X$ be a compact metric space, and $Theta$ be a finite space, endowed with their own $sigma$-algebra.



Let $f colon X times Theta to mathbb{R}$ be a Caratheodory function such that
(1) for each $x in X$, the function $f(x, cdot) colon Theta to mathbb{R}$
is measurable; and (2) for each $theta in Theta$, the function $f( cdot, theta) colon X to mathbb{R}$ is continuous.



Given each $x in X$, we have a probability distribution $pi( cdot ,| , x) colon 2^{Theta} to [0,1]$. In particular, given any fixed $x in X$,
it will generate a corresponding probability distribution $pi$ on $2^Theta$.



I am curious that




Under what kind of conditions (assumptions) imposed on this probability distribution $pi$ , the map $$X ni x mapsto int_Theta f(x,theta) , pi( mathrm{d} theta ,| ,x) in mathbb{R}$$
will be continuous on $X$?




Any idea or suggestions are most welcome!



Thank you so much!










share|cite|improve this question















Let $X$ be a compact metric space, and $Theta$ be a finite space, endowed with their own $sigma$-algebra.



Let $f colon X times Theta to mathbb{R}$ be a Caratheodory function such that
(1) for each $x in X$, the function $f(x, cdot) colon Theta to mathbb{R}$
is measurable; and (2) for each $theta in Theta$, the function $f( cdot, theta) colon X to mathbb{R}$ is continuous.



Given each $x in X$, we have a probability distribution $pi( cdot ,| , x) colon 2^{Theta} to [0,1]$. In particular, given any fixed $x in X$,
it will generate a corresponding probability distribution $pi$ on $2^Theta$.



I am curious that




Under what kind of conditions (assumptions) imposed on this probability distribution $pi$ , the map $$X ni x mapsto int_Theta f(x,theta) , pi( mathrm{d} theta ,| ,x) in mathbb{R}$$
will be continuous on $X$?




Any idea or suggestions are most welcome!



Thank you so much!







real-analysis probability functional-analysis probability-distributions conditional-probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 18 at 11:22

























asked Nov 18 at 9:36









Paradiesvogel

46139




46139












  • Thanks @Michael, you’re right. Actually, the space $Theta$ I considered was a compact metric space, but now I just want to simplify the question and restrict it to be finite.
    – Paradiesvogel
    Nov 18 at 10:50










  • Thanks @Michael . I totally agree with you. In fact, I really need the probability distribution $pi(cdot | x)$ depending on $x$. This means for each $x in X$, I have a different probability distribution defined on $2^Theta$. Also, the space $Theta$ is at least not trivial. In such a setting, what can I do to ensure the map $h(x)$ is continuous on $X$? Is it possible to do that? Thanks a million again :-)
    – Paradiesvogel
    Nov 18 at 11:02




















  • Thanks @Michael, you’re right. Actually, the space $Theta$ I considered was a compact metric space, but now I just want to simplify the question and restrict it to be finite.
    – Paradiesvogel
    Nov 18 at 10:50










  • Thanks @Michael . I totally agree with you. In fact, I really need the probability distribution $pi(cdot | x)$ depending on $x$. This means for each $x in X$, I have a different probability distribution defined on $2^Theta$. Also, the space $Theta$ is at least not trivial. In such a setting, what can I do to ensure the map $h(x)$ is continuous on $X$? Is it possible to do that? Thanks a million again :-)
    – Paradiesvogel
    Nov 18 at 11:02


















Thanks @Michael, you’re right. Actually, the space $Theta$ I considered was a compact metric space, but now I just want to simplify the question and restrict it to be finite.
– Paradiesvogel
Nov 18 at 10:50




Thanks @Michael, you’re right. Actually, the space $Theta$ I considered was a compact metric space, but now I just want to simplify the question and restrict it to be finite.
– Paradiesvogel
Nov 18 at 10:50












Thanks @Michael . I totally agree with you. In fact, I really need the probability distribution $pi(cdot | x)$ depending on $x$. This means for each $x in X$, I have a different probability distribution defined on $2^Theta$. Also, the space $Theta$ is at least not trivial. In such a setting, what can I do to ensure the map $h(x)$ is continuous on $X$? Is it possible to do that? Thanks a million again :-)
– Paradiesvogel
Nov 18 at 11:02






Thanks @Michael . I totally agree with you. In fact, I really need the probability distribution $pi(cdot | x)$ depending on $x$. This means for each $x in X$, I have a different probability distribution defined on $2^Theta$. Also, the space $Theta$ is at least not trivial. In such a setting, what can I do to ensure the map $h(x)$ is continuous on $X$? Is it possible to do that? Thanks a million again :-)
– Paradiesvogel
Nov 18 at 11:02












1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










I think your integral is
$$ h(x) = sum_{thetain Theta} f(x,theta) pi({theta}|x) quad forall x in X $$
if $pi({theta}|x) = pi({theta})$ for all $x in X$ then this is a sum of a finite number of functions that are continuous in $x$, and hence is continuous in $x$. More generally, if $pi({theta}|x)$ is continuous in $x$ for each $theta in Theta$, then this is a sum of a finite numer of functions that are continuous in $x$ (and hence is continuous in $x$).



Else, it is easy to get a discontinuous example (despite my incorrect comment from before that tried to do it with $Theta$ being only a 1-element set) by defining $pi({theta}|x)$ discontinuously. Define $X=[0,1]$, define $Theta={0,1}$, $f(x,0)=0$, $f(x,1) = 1$ for all $x in [0,1]$, and define:
$$ (pi({0}|x), pi({1}|x)) = left{ begin{array}{ll}
(1,0) &mbox{ if $x in [0,1/2)$} \
(1/2,1/2) & mbox{ if $x in [1/2,1]$}
end{array}
right.$$

Then
$$h(x)= pi({1}|x) = left{ begin{array}{ll}
0 &mbox{ if $x in [0,1/2)$} \
1/2 & mbox{ if $x in [1/2,1]$}
end{array}
right.$$

and this is discontinuous in $x$.






share|cite|improve this answer























  • Thanks so much for your answer @Michael. I should clarify initially that given any fixed $x in X$, we have a different probability distribution; that can be reviewed as for each state variable $x$ in a State space $X$, I will have a different probability distribution on $Theta$ and such a probability distribution depends on different observations of $x$. In this setting, could we still get some result for continuity of the map $h$? Would you mind rethinking about the question based on this setting please? I sincerely appreciate your kind help!
    – Paradiesvogel
    Nov 18 at 11:16










  • I think this is what my example already does. Note that in this example I refined my answer to have $Theta$ now a 2-element set rather than a 1-element set (for a 1-element set then the mass function must be 1 (in order to be a mass function) so I could not really define $pi({theta}|x)$ discontinuously in $x$ when $Theta$ is a 1-element set).
    – Michael
    Nov 18 at 11:20












  • Thanks @Michael . Did you mean that a sufficient condition for the continuity of the function $h$ is $pi({theta} | x)$ is continuous in $x$ for each $theta in Theta$? Besides, may I ask what do you think if we extend the finite space of $Theta$ to a compact metric space? Is it possible to do that?
    – Paradiesvogel
    Nov 18 at 11:31










  • Yes, my first paragraph says it is sufficient to have $pi({theta}|x)$ continuous in $x$ for each $theta in Theta$, for the case when $Theta$ is a finite set.
    – Michael
    Nov 18 at 11:32












  • Thank you very much @Michael . I understand now. But still curious about what if the space $Theta$ is allowed to be compact. In this setting, since the space $Theta$ could be infinite or countably infinite, such a sufficient condition may fail. Do you think is there any reasonable assumption imposed on $pi$ that would guarantee the continuity of $h$ with integral? Many thanks :-)
    – Paradiesvogel
    Nov 18 at 11:41











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',
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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003313%2fa-question-about-continuity-of-a-specific-function-with-probability-measure%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








up vote
1
down vote



accepted










I think your integral is
$$ h(x) = sum_{thetain Theta} f(x,theta) pi({theta}|x) quad forall x in X $$
if $pi({theta}|x) = pi({theta})$ for all $x in X$ then this is a sum of a finite number of functions that are continuous in $x$, and hence is continuous in $x$. More generally, if $pi({theta}|x)$ is continuous in $x$ for each $theta in Theta$, then this is a sum of a finite numer of functions that are continuous in $x$ (and hence is continuous in $x$).



Else, it is easy to get a discontinuous example (despite my incorrect comment from before that tried to do it with $Theta$ being only a 1-element set) by defining $pi({theta}|x)$ discontinuously. Define $X=[0,1]$, define $Theta={0,1}$, $f(x,0)=0$, $f(x,1) = 1$ for all $x in [0,1]$, and define:
$$ (pi({0}|x), pi({1}|x)) = left{ begin{array}{ll}
(1,0) &mbox{ if $x in [0,1/2)$} \
(1/2,1/2) & mbox{ if $x in [1/2,1]$}
end{array}
right.$$

Then
$$h(x)= pi({1}|x) = left{ begin{array}{ll}
0 &mbox{ if $x in [0,1/2)$} \
1/2 & mbox{ if $x in [1/2,1]$}
end{array}
right.$$

and this is discontinuous in $x$.






share|cite|improve this answer























  • Thanks so much for your answer @Michael. I should clarify initially that given any fixed $x in X$, we have a different probability distribution; that can be reviewed as for each state variable $x$ in a State space $X$, I will have a different probability distribution on $Theta$ and such a probability distribution depends on different observations of $x$. In this setting, could we still get some result for continuity of the map $h$? Would you mind rethinking about the question based on this setting please? I sincerely appreciate your kind help!
    – Paradiesvogel
    Nov 18 at 11:16










  • I think this is what my example already does. Note that in this example I refined my answer to have $Theta$ now a 2-element set rather than a 1-element set (for a 1-element set then the mass function must be 1 (in order to be a mass function) so I could not really define $pi({theta}|x)$ discontinuously in $x$ when $Theta$ is a 1-element set).
    – Michael
    Nov 18 at 11:20












  • Thanks @Michael . Did you mean that a sufficient condition for the continuity of the function $h$ is $pi({theta} | x)$ is continuous in $x$ for each $theta in Theta$? Besides, may I ask what do you think if we extend the finite space of $Theta$ to a compact metric space? Is it possible to do that?
    – Paradiesvogel
    Nov 18 at 11:31










  • Yes, my first paragraph says it is sufficient to have $pi({theta}|x)$ continuous in $x$ for each $theta in Theta$, for the case when $Theta$ is a finite set.
    – Michael
    Nov 18 at 11:32












  • Thank you very much @Michael . I understand now. But still curious about what if the space $Theta$ is allowed to be compact. In this setting, since the space $Theta$ could be infinite or countably infinite, such a sufficient condition may fail. Do you think is there any reasonable assumption imposed on $pi$ that would guarantee the continuity of $h$ with integral? Many thanks :-)
    – Paradiesvogel
    Nov 18 at 11:41















up vote
1
down vote



accepted










I think your integral is
$$ h(x) = sum_{thetain Theta} f(x,theta) pi({theta}|x) quad forall x in X $$
if $pi({theta}|x) = pi({theta})$ for all $x in X$ then this is a sum of a finite number of functions that are continuous in $x$, and hence is continuous in $x$. More generally, if $pi({theta}|x)$ is continuous in $x$ for each $theta in Theta$, then this is a sum of a finite numer of functions that are continuous in $x$ (and hence is continuous in $x$).



Else, it is easy to get a discontinuous example (despite my incorrect comment from before that tried to do it with $Theta$ being only a 1-element set) by defining $pi({theta}|x)$ discontinuously. Define $X=[0,1]$, define $Theta={0,1}$, $f(x,0)=0$, $f(x,1) = 1$ for all $x in [0,1]$, and define:
$$ (pi({0}|x), pi({1}|x)) = left{ begin{array}{ll}
(1,0) &mbox{ if $x in [0,1/2)$} \
(1/2,1/2) & mbox{ if $x in [1/2,1]$}
end{array}
right.$$

Then
$$h(x)= pi({1}|x) = left{ begin{array}{ll}
0 &mbox{ if $x in [0,1/2)$} \
1/2 & mbox{ if $x in [1/2,1]$}
end{array}
right.$$

and this is discontinuous in $x$.






share|cite|improve this answer























  • Thanks so much for your answer @Michael. I should clarify initially that given any fixed $x in X$, we have a different probability distribution; that can be reviewed as for each state variable $x$ in a State space $X$, I will have a different probability distribution on $Theta$ and such a probability distribution depends on different observations of $x$. In this setting, could we still get some result for continuity of the map $h$? Would you mind rethinking about the question based on this setting please? I sincerely appreciate your kind help!
    – Paradiesvogel
    Nov 18 at 11:16










  • I think this is what my example already does. Note that in this example I refined my answer to have $Theta$ now a 2-element set rather than a 1-element set (for a 1-element set then the mass function must be 1 (in order to be a mass function) so I could not really define $pi({theta}|x)$ discontinuously in $x$ when $Theta$ is a 1-element set).
    – Michael
    Nov 18 at 11:20












  • Thanks @Michael . Did you mean that a sufficient condition for the continuity of the function $h$ is $pi({theta} | x)$ is continuous in $x$ for each $theta in Theta$? Besides, may I ask what do you think if we extend the finite space of $Theta$ to a compact metric space? Is it possible to do that?
    – Paradiesvogel
    Nov 18 at 11:31










  • Yes, my first paragraph says it is sufficient to have $pi({theta}|x)$ continuous in $x$ for each $theta in Theta$, for the case when $Theta$ is a finite set.
    – Michael
    Nov 18 at 11:32












  • Thank you very much @Michael . I understand now. But still curious about what if the space $Theta$ is allowed to be compact. In this setting, since the space $Theta$ could be infinite or countably infinite, such a sufficient condition may fail. Do you think is there any reasonable assumption imposed on $pi$ that would guarantee the continuity of $h$ with integral? Many thanks :-)
    – Paradiesvogel
    Nov 18 at 11:41













up vote
1
down vote



accepted







up vote
1
down vote



accepted






I think your integral is
$$ h(x) = sum_{thetain Theta} f(x,theta) pi({theta}|x) quad forall x in X $$
if $pi({theta}|x) = pi({theta})$ for all $x in X$ then this is a sum of a finite number of functions that are continuous in $x$, and hence is continuous in $x$. More generally, if $pi({theta}|x)$ is continuous in $x$ for each $theta in Theta$, then this is a sum of a finite numer of functions that are continuous in $x$ (and hence is continuous in $x$).



Else, it is easy to get a discontinuous example (despite my incorrect comment from before that tried to do it with $Theta$ being only a 1-element set) by defining $pi({theta}|x)$ discontinuously. Define $X=[0,1]$, define $Theta={0,1}$, $f(x,0)=0$, $f(x,1) = 1$ for all $x in [0,1]$, and define:
$$ (pi({0}|x), pi({1}|x)) = left{ begin{array}{ll}
(1,0) &mbox{ if $x in [0,1/2)$} \
(1/2,1/2) & mbox{ if $x in [1/2,1]$}
end{array}
right.$$

Then
$$h(x)= pi({1}|x) = left{ begin{array}{ll}
0 &mbox{ if $x in [0,1/2)$} \
1/2 & mbox{ if $x in [1/2,1]$}
end{array}
right.$$

and this is discontinuous in $x$.






share|cite|improve this answer














I think your integral is
$$ h(x) = sum_{thetain Theta} f(x,theta) pi({theta}|x) quad forall x in X $$
if $pi({theta}|x) = pi({theta})$ for all $x in X$ then this is a sum of a finite number of functions that are continuous in $x$, and hence is continuous in $x$. More generally, if $pi({theta}|x)$ is continuous in $x$ for each $theta in Theta$, then this is a sum of a finite numer of functions that are continuous in $x$ (and hence is continuous in $x$).



Else, it is easy to get a discontinuous example (despite my incorrect comment from before that tried to do it with $Theta$ being only a 1-element set) by defining $pi({theta}|x)$ discontinuously. Define $X=[0,1]$, define $Theta={0,1}$, $f(x,0)=0$, $f(x,1) = 1$ for all $x in [0,1]$, and define:
$$ (pi({0}|x), pi({1}|x)) = left{ begin{array}{ll}
(1,0) &mbox{ if $x in [0,1/2)$} \
(1/2,1/2) & mbox{ if $x in [1/2,1]$}
end{array}
right.$$

Then
$$h(x)= pi({1}|x) = left{ begin{array}{ll}
0 &mbox{ if $x in [0,1/2)$} \
1/2 & mbox{ if $x in [1/2,1]$}
end{array}
right.$$

and this is discontinuous in $x$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 18 at 11:12

























answered Nov 18 at 11:06









Michael

13.2k11325




13.2k11325












  • Thanks so much for your answer @Michael. I should clarify initially that given any fixed $x in X$, we have a different probability distribution; that can be reviewed as for each state variable $x$ in a State space $X$, I will have a different probability distribution on $Theta$ and such a probability distribution depends on different observations of $x$. In this setting, could we still get some result for continuity of the map $h$? Would you mind rethinking about the question based on this setting please? I sincerely appreciate your kind help!
    – Paradiesvogel
    Nov 18 at 11:16










  • I think this is what my example already does. Note that in this example I refined my answer to have $Theta$ now a 2-element set rather than a 1-element set (for a 1-element set then the mass function must be 1 (in order to be a mass function) so I could not really define $pi({theta}|x)$ discontinuously in $x$ when $Theta$ is a 1-element set).
    – Michael
    Nov 18 at 11:20












  • Thanks @Michael . Did you mean that a sufficient condition for the continuity of the function $h$ is $pi({theta} | x)$ is continuous in $x$ for each $theta in Theta$? Besides, may I ask what do you think if we extend the finite space of $Theta$ to a compact metric space? Is it possible to do that?
    – Paradiesvogel
    Nov 18 at 11:31










  • Yes, my first paragraph says it is sufficient to have $pi({theta}|x)$ continuous in $x$ for each $theta in Theta$, for the case when $Theta$ is a finite set.
    – Michael
    Nov 18 at 11:32












  • Thank you very much @Michael . I understand now. But still curious about what if the space $Theta$ is allowed to be compact. In this setting, since the space $Theta$ could be infinite or countably infinite, such a sufficient condition may fail. Do you think is there any reasonable assumption imposed on $pi$ that would guarantee the continuity of $h$ with integral? Many thanks :-)
    – Paradiesvogel
    Nov 18 at 11:41


















  • Thanks so much for your answer @Michael. I should clarify initially that given any fixed $x in X$, we have a different probability distribution; that can be reviewed as for each state variable $x$ in a State space $X$, I will have a different probability distribution on $Theta$ and such a probability distribution depends on different observations of $x$. In this setting, could we still get some result for continuity of the map $h$? Would you mind rethinking about the question based on this setting please? I sincerely appreciate your kind help!
    – Paradiesvogel
    Nov 18 at 11:16










  • I think this is what my example already does. Note that in this example I refined my answer to have $Theta$ now a 2-element set rather than a 1-element set (for a 1-element set then the mass function must be 1 (in order to be a mass function) so I could not really define $pi({theta}|x)$ discontinuously in $x$ when $Theta$ is a 1-element set).
    – Michael
    Nov 18 at 11:20












  • Thanks @Michael . Did you mean that a sufficient condition for the continuity of the function $h$ is $pi({theta} | x)$ is continuous in $x$ for each $theta in Theta$? Besides, may I ask what do you think if we extend the finite space of $Theta$ to a compact metric space? Is it possible to do that?
    – Paradiesvogel
    Nov 18 at 11:31










  • Yes, my first paragraph says it is sufficient to have $pi({theta}|x)$ continuous in $x$ for each $theta in Theta$, for the case when $Theta$ is a finite set.
    – Michael
    Nov 18 at 11:32












  • Thank you very much @Michael . I understand now. But still curious about what if the space $Theta$ is allowed to be compact. In this setting, since the space $Theta$ could be infinite or countably infinite, such a sufficient condition may fail. Do you think is there any reasonable assumption imposed on $pi$ that would guarantee the continuity of $h$ with integral? Many thanks :-)
    – Paradiesvogel
    Nov 18 at 11:41
















Thanks so much for your answer @Michael. I should clarify initially that given any fixed $x in X$, we have a different probability distribution; that can be reviewed as for each state variable $x$ in a State space $X$, I will have a different probability distribution on $Theta$ and such a probability distribution depends on different observations of $x$. In this setting, could we still get some result for continuity of the map $h$? Would you mind rethinking about the question based on this setting please? I sincerely appreciate your kind help!
– Paradiesvogel
Nov 18 at 11:16




Thanks so much for your answer @Michael. I should clarify initially that given any fixed $x in X$, we have a different probability distribution; that can be reviewed as for each state variable $x$ in a State space $X$, I will have a different probability distribution on $Theta$ and such a probability distribution depends on different observations of $x$. In this setting, could we still get some result for continuity of the map $h$? Would you mind rethinking about the question based on this setting please? I sincerely appreciate your kind help!
– Paradiesvogel
Nov 18 at 11:16












I think this is what my example already does. Note that in this example I refined my answer to have $Theta$ now a 2-element set rather than a 1-element set (for a 1-element set then the mass function must be 1 (in order to be a mass function) so I could not really define $pi({theta}|x)$ discontinuously in $x$ when $Theta$ is a 1-element set).
– Michael
Nov 18 at 11:20






I think this is what my example already does. Note that in this example I refined my answer to have $Theta$ now a 2-element set rather than a 1-element set (for a 1-element set then the mass function must be 1 (in order to be a mass function) so I could not really define $pi({theta}|x)$ discontinuously in $x$ when $Theta$ is a 1-element set).
– Michael
Nov 18 at 11:20














Thanks @Michael . Did you mean that a sufficient condition for the continuity of the function $h$ is $pi({theta} | x)$ is continuous in $x$ for each $theta in Theta$? Besides, may I ask what do you think if we extend the finite space of $Theta$ to a compact metric space? Is it possible to do that?
– Paradiesvogel
Nov 18 at 11:31




Thanks @Michael . Did you mean that a sufficient condition for the continuity of the function $h$ is $pi({theta} | x)$ is continuous in $x$ for each $theta in Theta$? Besides, may I ask what do you think if we extend the finite space of $Theta$ to a compact metric space? Is it possible to do that?
– Paradiesvogel
Nov 18 at 11:31












Yes, my first paragraph says it is sufficient to have $pi({theta}|x)$ continuous in $x$ for each $theta in Theta$, for the case when $Theta$ is a finite set.
– Michael
Nov 18 at 11:32






Yes, my first paragraph says it is sufficient to have $pi({theta}|x)$ continuous in $x$ for each $theta in Theta$, for the case when $Theta$ is a finite set.
– Michael
Nov 18 at 11:32














Thank you very much @Michael . I understand now. But still curious about what if the space $Theta$ is allowed to be compact. In this setting, since the space $Theta$ could be infinite or countably infinite, such a sufficient condition may fail. Do you think is there any reasonable assumption imposed on $pi$ that would guarantee the continuity of $h$ with integral? Many thanks :-)
– Paradiesvogel
Nov 18 at 11:41




Thank you very much @Michael . I understand now. But still curious about what if the space $Theta$ is allowed to be compact. In this setting, since the space $Theta$ could be infinite or countably infinite, such a sufficient condition may fail. Do you think is there any reasonable assumption imposed on $pi$ that would guarantee the continuity of $h$ with integral? Many thanks :-)
– Paradiesvogel
Nov 18 at 11:41


















draft saved

draft discarded




















































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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003313%2fa-question-about-continuity-of-a-specific-function-with-probability-measure%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

AnyDesk - Fatal Program Failure

QoS: MAC-Priority for clients behind a repeater

Актюбинская область