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Let $z$ be a brownian motion, let $\mathcal{F}$ be the filtration it generates. For $k>0$ and $m\in\mathbb{R}$, I define the process $Y$ as

$$Y_t=E\Big[\Big(\int_0^\infty e^{-ks+mz_s}ds\Big)^\eta\vert\mathcal{F}_t\Big].$$

where $\eta\in\mathbb{R}^\star$. Then, $Y$ is a martingale (this seems obvious). I assume all conditions are met for the above integral to exist (c.f. Yor, 2002 for example). Therefore there exists a process $\{\sigma_t\}_{t\geq 0}$ such that

$$dY_t=Y_t\sigma_tdz_t.$$

I want to show that $\sigma_t$ is decreasing (in expectation, perhaps?). The intuition is that the discounting $e^{-ks}$ becomes stronger and stronger so the unknown part of the integral becomes less and less volatile... Sorry for the bad phraseology, I can't think of a better way of explaining it. Thanks!

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