Let $\left( \Omega, \mathcal{F}, \mathbb{P} \right)$ be a probability space. I have a random variable on this space X such that $ess \ inf \ X < r < ess \ sup \ X$ where $r \in \mathbb{R}$.
Now I also have an equivalent probability measure $\tilde{\mathbb{P}}$ with Radon Nikodym derivative $\tilde{Z} := \frac{d\tilde{\mathbb{P}}}{d\mathbb{P}}$ which satisfies $\mathbb{E}\left[\tilde{Z}\right] = 1$ (this is expectation under $\mathbb{P}$) and $\tilde{Z} \in \left(0,1/\alpha\right) \mathbb{P}-$a.s. where $\alpha \in \left(0,1\right)$.

This is my problem. I need to find a random variable $Z$ ($Z$ will be the Radon-Nikodym derivative of another measure - but you don't really need that information) such that

  1. $\mathbb{E}\left[Z\right]=1$,
  2. $Z \in \left[0,1/\alpha\right] \mathbb{P}-$a.s. and
  3. $\mathbb{E}\left[ZX\right] < \mathbb{E}\left[\tilde{Z}X\right]$.

My idea: Let $A=\{X<r\}$. I will try to "reweight" $\tilde{Z}$ so that it puts more mass on the set $A$ and less mass on $A^{c}$ so that condition 3 above holds whilst also trying to satisfy 1 and 2.

We have $\mathbb{E}\left[\tilde{Z}X\right] = \mathbb{E}\left[\tilde{Z}X \mathbb{1}_{A} \right] + \mathbb{E}\left[\tilde{Z}X \mathbb{1}_{A^{c}} \right] > \mathbb{E}\left[\left((1+\epsilon)\tilde{Z}\wedge\frac{1}{\alpha}\right)X \mathbb{1}_{A} \right] + \mathbb{E}\left[(1-\epsilon^{\prime})\tilde{Z}X \mathbb{1}_{A^{c}} \right]$ for all $\epsilon >0$ and $1 > \epsilon^{\prime} >0 $. Can we then let $Z$ be defined by

  • $Z= \left((1+\epsilon)\tilde{Z}\wedge\frac{1}{\alpha}\right)\mathbb{1}_{A} + (1-\epsilon^{\prime})\tilde{Z}\mathbb{1}_{A^{c}}$

where $\epsilon$ and $\epsilon^{\prime}$ are chosen so that $\mathbb{E}\left[Z\right]=1$? IF so, then we have conditions 1,2 and 3 all hold true and we are done.


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