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!
real-analysis probability functional-analysis probability-distributions conditional-probability
add a comment |
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!
real-analysis probability functional-analysis probability-distributions conditional-probability
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
add a comment |
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!
real-analysis probability functional-analysis probability-distributions conditional-probability
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
real-analysis probability functional-analysis probability-distributions conditional-probability
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
add a comment |
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
add a comment |
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 $x$.
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
|
show 5 more comments
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 $x$.
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
|
show 5 more comments
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 $x$.
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
|
show 5 more comments
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 $x$.
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 $x$.
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
|
show 5 more comments
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
|
show 5 more comments
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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