# Volatility of a perpetuity $E\Big[\Big(\int_0^\infty e^{-ks+mz_s}ds\Big)^\eta\vert\mathcal{F}_t\Big]$

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!