# Does the convergence of drifted Brownian motion imply the convergence of expectation?

Let $$(f_{\epsilon})_{\epsilon>0}$$ be a family of non-increasing and continuous functions on $$\mathbb R_+$$ s.t. $$f_{\epsilon}(0)=1$$ and $$f_{\epsilon}(\infty)=0$$. Assume that $$\epsilon\mapsto f_\epsilon(t)$$ is non-increasing for all $$t\ge 0$$ and denote by $$f:\mathbb R_+ \to [0,1]$$ the pointwise limit of $$(f_{\epsilon})_{\epsilon>0}$$. For every $$t\ge 0$$, set

$$X^{\epsilon}_t=X_0+W_t-1+f_\epsilon(t) \quad\mbox{and} \quad X_t=X_0+W_t-1+f(t),$$

where $$X_0>0$$ is a random variable and $$(W_{t})_{t\ge 0}$$ is an independent Brownian motion. Can we prove

$$\lim_{\epsilon\to 0+} \left\{ \mathbb E \left[ \exp\left(-\frac 1 \epsilon \int_0^t \big(X_s^{\epsilon}\big)^-ds\right)\right] - \mathbb E \left[ \exp\left(-\frac 1 \epsilon \int_0^t \big(X_s\big)^-ds\right)\right]\right\} = 0$$

holds for almost every $$t \ge 0$$? Here $$a^-:=\max(-a, 0)$$ for $$a\in\mathbb R$$.

## 1 Answer

First we apply MVT to get the bound

$$e^{\xi}\frac{1}{\epsilon}\left|\int_{0}^{T}(-X^{\epsilon}(s))\vee 0-(-X(s))\vee 0 ds\right|.(1)$$

Lets truncate $$[0,T]$$ and assume $$f$$ is continuous. Then due to $$f_{\epsilon}\downarrow f$$, by Dini's theorem we have uniform convergence

$$\sup_{s\in [0,T]}|X_{\epsilon}(s)-X(s)|=\sup_{s\in [0,T]}|f_{\epsilon}(s)-f(s)|\to 0,$$

which means that the second factor is bounded. After $$t\geq t_{0}$$ and if we restrict $$X_{0}\in (0,1-\delta_{0})$$ for some $$\delta_{0}>0$$, we have

$$X_0+W_t+f_\epsilon(t) <1\Leftrightarrow (-X^{\epsilon}(t))\vee 0=-X^{\epsilon}(t).$$

Therefore,

$$\frac{1}{\epsilon}e^{\xi}\leq \frac{1}{\epsilon}\exp\left(-\frac 1 \epsilon \int_0^T \big(-X_s^{\epsilon}\big)\vee 0 ds\right)\to 0\text{ as \epsilon\to 0. }$$

So the (1) overall goes to zero.