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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!

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  • $\begingroup$ Doesn't this follow since $Y$ is a convex combination of $Y_1$ and $Y_2$? $\endgroup$ – Steve Huntsman Apr 1 '14 at 14:08
  • $\begingroup$ It is not so evident since the $esssup$ are taken under different probability measures... $\endgroup$ – CodeGolf Apr 1 '14 at 14:29
  • $\begingroup$ I think you can say there's only one measure, namely $P$, and say that everything ($P_1$ and $P_2$) are abs. cts. with respect to it. $\endgroup$ – Anthony Quas Apr 1 '14 at 15:15
  • $\begingroup$ thx for your reply, could you please give more details for the proof? $\endgroup$ – CodeGolf Apr 1 '14 at 15:16
  • $\begingroup$ Lemma: $E^P Z = \alpha E^{P_1} Z + (1-\alpha) E^{P_2} Z$. Corollary: $Y = \alpha Y_1 + (1-\alpha) Y_2$. Also useful: since $P_1 \ll P$, $\operatorname{esssup}_{P_1} Z \le \operatorname{esssup}_{P} Z$. $\endgroup$ – Nate Eldredge Apr 1 '14 at 20:11
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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 $\mathbb{E}^\mathbb{Q}[XG|\mathcal{G}] = \mathbb{E}^\mathbb{P}[X|\mathcal{G}]\mathbb{E}^\mathbb{Q}[G|\mathcal{G}]$.

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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.

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