conditional expectation under convex combinaison of probability measures(II)

Question

Let $(\Omega,\mathcal{F})$ denote some measurable space. Let $P_1$ and $P_2$ denote respectively two probability measures. Now let $\mathcal{G}$ be some sub sigma-algebra of $\mathcal{F}$. Given a positive integrable random variable $X$, we can define respectively the conditional expectation

$$Y_1=E^{P_1}[X|\mathcal{G}],~ Y_2=E^{P_2}[X|\mathcal{G}]$$

Now for some $0<\alpha<1$, we can define a new probability measure $P=\alpha P_1+(1-\alpha)P_2$, then we get

$$Y=E^{P}[X|\mathcal{G}]$$

Now my question is whether we can prove

$$\operatorname{esssup}{}_P(Y)\le \max\Big( \operatorname{esssup}_{P_1}(Y_1), \operatorname{esssup}_{P_2}(Y_2)\Big)?$$

Thanks a lot for the help!

Answer 1

Let me denote (for consistency) by $\mathbb{P}_0$, $\mathbb{P}_1$ the given probabilities (rather than $\mathbb{P}_1$, $\mathbb{P}_2$), and set $\mathbb{P}_\alpha=(1-\alpha)\mathbb{P}_0 + \alpha\mathbb{P}_1$. Let $\mathbb{Q}$ be any probability such that $\mathbb{P}_0\ll\mathbb{Q}$ and $\mathbb{P}_1\ll\mathbb{Q}$ (any $\mathbb{P}_\alpha$ would be fine in fact, but it does not matter). Then $\mathbb{P}_\alpha\ll\mathbb{Q}$ and $G_\alpha=(1-\alpha)G_0 + \alpha G_1$, where $G_\alpha=\frac{d\mathbb{P}_\alpha}{d\mathbb{Q}}$.

It is easy to check (see below) that $$ Y_\alpha = \frac{\mathbb{E}^\mathbb{Q}[XG_\alpha|\mathcal{G}]}{\mathbb{E}^\mathbb{Q}[G_\alpha|\mathcal{G}]} = \frac{(1-\alpha)\mathbb{E}^\mathbb{Q}[XG_0|\mathcal{G}]+\alpha\mathbb{E}^\mathbb{Q}[XG_1|\mathcal{G}]} {(1-\alpha)\mathbb{E}^\mathbb{Q}[G_0|\mathcal{G}]+\alpha\mathbb{E}^\mathbb{Q}[G_1|\mathcal{G}]}. $$ To prove the claimed inequality, it is sufficient to notice that the real function $$ \alpha \mapsto\frac{(1-\alpha)x_0 + \alpha x_1}{(1-\alpha)p_0 + \alpha p_1} $$ is either non--decreasing or non--increasing.

To conclude, let me prove that, if $\mathbb{P}\ll\mathbb{Q}$ and $G=\frac{d\mathbb{P}}{d\mathbb{Q}}$, then $$ \mathbb{E}^\mathbb{P}[X|\mathcal{G}] = \frac{\mathbb{E}^\mathbb{Q}[XG|\mathcal{G}]}{\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]}. $$ First, if $A=\{\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]=0\}$, then $$ \mathbb{P}[A] = \mathbb{E}^\mathbb{Q}[G \mathbf{1}_A] = \mathbb{E}^\mathbb{Q}\bigl[\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]\mathbf{1}_A\bigr] = 0 $$ and the formula makes sense. For every $A\in\mathcal{G}$, $$ \mathbb{E}^\mathbb{Q}\bigl[\mathbb{E}^\mathbb{Q}[XG|\mathcal{G}]\mathbf{1}_A\bigr] = \mathbb{E}^\mathbb{Q}[XG\mathbf{1}_A] = \mathbb{E}^\mathbb{P}[X\mathbf{1}_A] = \mathbb{E}^\mathbb{P}\bigl[\mathbb{E}^\mathbb{P}[X|\mathcal{G}]\mathbf{1}_A\bigr] $$ $$ = \mathbb{E}^\mathbb{Q}\bigl[\mathbb{E}^\mathbb{P}[X|\mathcal{G}]G\mathbf{1}_A\bigr] = \mathbb{E}^\mathbb{Q}\bigl[\mathbb{E}^\mathbb{P}[X|\mathcal{G}]\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]\mathbf{1}_A\bigr], $$ therefore $\mathbb{E}^\mathbb{Q}[XG|\mathcal{G}] = \mathbb{E}^\mathbb{P}[X|\mathcal{G}]\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]$.

Answer 2

Yes. Using $E_1,E_2$ for the two expectations, if for any $A\in\mathcal{G}$ we have $E_i(X 1_A) \le C P_i(A)$ then the same holds for the convex combination.