1
$\begingroup$

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 $a\in\mathbb R$.

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.