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!

  • $\begingroup$ 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. $\endgroup$ – Paradiesvogel Nov 18 '18 at 10:50
  • $\begingroup$ 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 :-) $\endgroup$ – Paradiesvogel Nov 18 '18 at 11:02

I think your integral is $$ h(x) = \sum_{\theta\in \Theta} f(x,\theta) \pi(\{\theta\}|x) \quad \forall x \in X $$ 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 $x$.

  • $\begingroup$ 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! $\endgroup$ – Paradiesvogel Nov 18 '18 at 11:16
  • $\begingroup$ 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). $\endgroup$ – Michael Nov 18 '18 at 11:20
  • $\begingroup$ 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? $\endgroup$ – Paradiesvogel Nov 18 '18 at 11:31
  • $\begingroup$ 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. $\endgroup$ – Michael Nov 18 '18 at 11:32
  • $\begingroup$ 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 :-) $\endgroup$ – Paradiesvogel Nov 18 '18 at 11:41

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