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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.

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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 $x$.

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  • $\begingroup$ 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
    Commented Dec 2, 2015 at 2:51
  • $\begingroup$ The values computed using the two approaches don't match. What am I doing wrong? $\endgroup$
    – avk255
    Commented Dec 2, 2015 at 2:59
  • $\begingroup$ Not sure $V''$ exists $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented 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
    Commented Dec 2, 2015 at 3:05
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    $\begingroup$ Also, in going from line 1 to 2, should it not be $W_s < -x$ instead of $x$? $\endgroup$
    – avk255
    Commented Dec 2, 2015 at 3:37

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