A question about continuity of a specific function with probability measure











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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!










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  • 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






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share|cite|improve this question













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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

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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











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1 Answer
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1 Answer
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active

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active

oldest

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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


















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