Optimal measure

Question

I am studying conditional expectations and I have a doubt that if it is possible to maximise over a family of measures. It means, let $X$ be a random variable, $A$ be a set and $\mathcal{P}$ be a family of probability measures is it possible to compute \begin{equation} Y=\sup_{\mathbb{P}\in \mathcal{P}} \mathbb{E}^{\mathbb{P}}\left[X|A\right] \end{equation} I mean, for instance to compute the maximum of a function $f$ we can compute the first derivative $f'$. Does it exist a similar way to find Y?. Maybe in the case when each $\mathbb{P}$ has density function $f_\mathbb{P}$ is it possible to obtain using something like \begin{equation} \frac{d Y}{df_\mathbb{P}} \end{equation} I would like to know if there exist some similar result or theory about it. Any help or advice is welcome. Thank you!.

Answer 1

This answer assumes that $\ \mathcal{P}\ $ is the set of all probability measures on some $\ \sigma$-algebra $\ \mathscr{A}\ $ for which $\ A\in\mathscr{A}\ $ and with respect to which $\ X\ $ is a random variable (i.e. measurable). I made this assumption because I initially misread the description of $\ \mathcal{P}\ $ as being the family of probability measures rather than a family of probability measures. I don't believe the question is answerable for an arbitrary family of probability measures about which no further information is available.

$\def\ed{\stackrel{\text{def}}{=}}$ If $\ a\in A\ $ and $$ \mathbb{P}_a(B)\ed\cases{1&if $\ X(a)\in B$\\ 0&if $\ X(a)\not\in B$}\ , $$ then $$ \mathbb{E}^{\mathbb{P}_a}(X\,|\,A\,)=X(a)\ , $$ and since $\ \mathbb{E}^{\mathbb{P}}(X\,|\,A\,)\le\sup_\limits{a\in A}X(a)\ $ for any probability measure $\ \mathbb{P}\ ,$ it follows from this that $$ \sup_{\mathbb{P}\in\mathcal{P}}\mathbb{E}^{\mathbb{P}}(X\,|\,A\,)=\sup_{a\in A}X(a)\ .\label{e1}\tag{1} $$ Also, $\ \mathbb{E}^{\mathbb{P}}(X\,|\,A\,)\ $ attains the supremum in (\ref{e1}) at some $\ \mathbb{P}^*\in\mathcal{P}\ $ if and only if $\ X(a)\ $ attains it at some $\ a^*\in A\ .$

If $\ a^*\in A\ $ satisfies $$ X\big(a^*\big)=\sup_{a\in A}X(a) $$ then \begin{align} \mathbb{E}^{\mathbb{P}_{a^*}}(X\,|\,A\,)&=X\big(a^*\big)\\ &=\sup_{a\in A}X(a)\\ &=\sup_{\mathbb{P}\in\mathcal{P}}\mathbb{E}^{\mathbb{P}}(X\,|\,A\,)\ . \end{align} On the other hand, if $\ X(a)<\sup_\limits{a\in A}X(a)\ $ for all $\ a\in A\ ,$ and $\ \mathbb{P}\in\mathcal{P}\ ,$ then $$ A=\bigcup_{j=1}^\infty\left\{a\in A\,\left|\,X(a)\le\sup_\limits{\alpha\in A}X(\alpha)-\frac{1}{j}\,\right.\right\}\ . $$ Therefore, \begin{align} 1&=\mathbb{P}(A\,|\,A\,)\\ &=\mathbb{P}\left(\bigcup_{j=1}^\infty\left\{a\in A\,\left|\,X(a)\le\sup_\limits{\alpha\in A}X(\alpha)-\frac{1}{j}\,\right\}\right|A\right)\\ &=\lim_{n\rightarrow\infty}\mathbb{P}\left(\bigcup_{j=1}^n\left\{a\in A\,\left|\,X(a)\le\sup_\limits{\alpha\in A}X(\alpha)-\frac{1}{j}\,\right\}\right|A\right)\\ &=\lim_{n\rightarrow\infty}\mathbb{P}\left(\left\{a\in A\,\left|\,X(a)\le\sup_\limits{\alpha\in A}X(\alpha)-\frac{1}{n}\,\right\}\right|A\right)\ , \end{align} from which it follow that there exist a positive integer $\ N\ $ such that \begin{align} \mathfrak{p}&\ed\mathbb{P}\left(\left\{a\in A\,\left|\,X(a)\le\sup_\limits{\alpha\in A}X(\alpha)-\frac{1}{N}\,\right\}\right|A\right)\\ &\ge\frac{1}{2}\ . \end{align} Let $$ B\ed\left\{a\in A\,\left|\,X(a)\le\sup_\limits{\alpha\in A}X(\alpha)-\frac{1}{N}\,\right\}\right.\ . $$ Then \begin{align} \mathbb{E}^{\mathbb{P}}(X\,|\,A\,)&=\mathbb{E}^{\mathbb{P}}\big(I_BX+I_{A\setminus B}X\,\big|\,A\,\big)\\ &=\mathbb{E}^{\mathbb{P}}\big(I_BX\,\big|\,A\,\big)+\mathbb{E}^{\mathbb{P}}\big(I_{A\setminus B}X\,\big|\,A\,\big)\\ &\le\mathfrak{p}\left(\sup_\limits{\alpha\in A}X(\alpha)-\frac{1}{N}\right)+(1-\mathfrak{p})\sup_\limits{\alpha\in A}X(\alpha)\\ &=\sup_\limits{\alpha\in A}X(\alpha)-\frac{\mathfrak{p}}{N}\\ &<\sup_\limits{\alpha\in A}X(\alpha)\ . \end{align} Thus, if $\ X(a)\ $ doesn't attain the supremum in (\ref{e1}) for any $\ a\in A\ ,$ then neither does $\ \mathbb{E}^{\mathbb{P}}(X\,|\,A\,)\ $ attain it for any $\ \mathbb{P}\in\mathcal{P}\ .$