2
$\begingroup$

Suppose $X$ is a process given by -

$dX_t = db_t$ where $b_t$ is a standard Brownian motion with its filtration $(\mathcal{F}_t)$.

Suppose an agent earns a payoff given by

$V(x) = \mathbb{E} [\int_0^\infty e^{-\int_0^t r(X_s)ds} dt|X_0 =x] $

where $r(x) = \begin{cases} 3 & \text{ if } x \ge 0 \\ 7 & \text{ otherwise} \end{cases}$

I am interested in computing $V(x)$. In particular, I am interested in knowing if $V(x)$ is differentiable at $0$?

I suspect that the answer is no for differentiability. But I don't have a proof.

Thanks.

$\endgroup$

1 Answer 1

1
$\begingroup$

I think it's differentiable (everywhere). Interchanging operations freely, we have \begin{eqnarray*} \newcommand{\D}{\frac{\mathrm d}{{\mathrm d}x}} \newcommand{\E}{\mathbb E} V'(x) &=& \D \E_x \int_0^\infty \exp\left(-\left(\int_0^t 3+4[W_s<0]\mathrm ds\right)\right)\mathrm dt \\ &=& \int_0^\infty \E_0 \D\exp\left(-\left(\int_0^t 3+4[W_s<-x]\mathrm ds\right)\right)\mathrm dt \end{eqnarray*} Then we end up needing $$ L^x(t) := \D \int_0^t [W_s<x]\,\mathrm ds. $$ If we try to interchange the $\D$ with the $\int$ here it seems to not exist at $x=W_s$. However, we don't need to as, as far as I can tell, this is just the local time of $W$ at $x$.

$\endgroup$
9
  • $\begingroup$ Thanks a lot. I can sort of understand your argument. But here's a related question then. Assuming it is differentiable, I can compute the functional form explicitly as, when $x > 0$, $3 V(x) = 1 + 1/2 V''(x)$ and a similar DE when $x \le 0$. Now, I can solve these, kill one constant by conditions at $\infty$ and $-\infty$. For the remaining two constants I use continuity and differentiability at $0$. But here's an alternative way of getting $V(0)$. If $r = 3$ throughout then $V(0) = 1/3$. If $r = 7$, then $V(0) = 1/7$. Therefore, in our case, $V(0) = \frac{1/3 + 1/7}{2}$. $\endgroup$
    – avk255
    Dec 2, 2015 at 2:51
  • $\begingroup$ The values computed using the two approaches don't match. What am I doing wrong? $\endgroup$
    – avk255
    Dec 2, 2015 at 2:59
  • $\begingroup$ Not sure $V''$ exists $\endgroup$ Dec 2, 2015 at 3:03
  • $\begingroup$ Yes, $V''$ won't exist at $0$, certainly. But when $x > 0$ and when $x < 0$, it will exist and I can use the DEs I mentioned, I think. And then, I can get the two constants using continuity and differentiability at $0$. $\endgroup$
    – avk255
    Dec 2, 2015 at 3:05
  • 1
    $\begingroup$ Also, in going from line 1 to 2, should it not be $W_s < -x$ instead of $x$? $\endgroup$
    – avk255
    Dec 2, 2015 at 3:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.