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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

$$\operatorname{esssup}{}_P(Y)\le \alpha \operatorname{esssup}_{P_1}(Y_1)+(1-\alpha)\operatorname{esssup}_{P_2}(Y_2)?$$

Here the definition of $\operatorname{esssup}_{Q}(\cdot)$ w.r.t some probability $Q$ can be found here:

http://en.wikipedia.org/wiki/Essential_supremum_and_essential_infimum

Thanks a lot for the help!

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Unless I am missing something, this is false, and the conditional expectation is a red herring.

Let $\Omega = \{a,b\}\newcommand{\esssup}{\operatorname{ess\,sup}}$ be a set with two points and $\mathcal{G} = \mathcal{F} = 2^\Omega$. Set $P_1 = \delta_a$ and $P_2 = \delta_b$, and $X = 1_{\{a\}}$. Since $\mathcal{G} = \mathcal{F}$ the conditional expectation doesn't do anything, and we have $Y_1 = Y_2 = Y = X$. (Technically $Y_1(b)$ and $Y_2(a)$ are undefined but it won't actually matter.) Then it is easy to see that $$\begin{align*} \esssup{}_{P_1} Y_1 &= 1 \\ \esssup{}_{P_2} Y_2 &= 0 \\ \esssup_P Y &= 1 \end{align*}$$ so your proposed inequality reads $1 \le \alpha$.

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  • $\begingroup$ Thanks so much for your suggestion. But could we prove the following inequality? $\endgroup$ Commented Apr 1, 2014 at 8:50
  • $\begingroup$ $$\operatorname{esssup}{}_P(Y)\le \max\Big( \operatorname{esssup}_{P_1}(Y_1), \operatorname{esssup}_{P_2}(Y_2)\Big)$$ $\endgroup$ Commented Apr 1, 2014 at 8:51
  • $\begingroup$ @codegolf: This should be asked as a new question, and I think it would be better for math.stackexchange.com than for MathOverflow. $\endgroup$ Commented Apr 1, 2014 at 13:09
  • $\begingroup$ It is better, thx! $\endgroup$ Commented Apr 1, 2014 at 13:54

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