# Differentiability of value function

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.

• 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, $V(0) = \frac{1/3 + 1/7}{2}$. – avk255 Dec 2 '15 at 2:51
• Not sure $V''$ exists – Bjørn Kjos-Hanssen Dec 2 '15 at 3:03
• 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 $0$. – avk255 Dec 2 '15 at 3:05
• Also, in going from line 1 to 2, should it not be $W_s < -x$ instead of $x$? – avk255 Dec 2 '15 at 3:37