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 $$A=\{X. I will try to "reweight" $$\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.