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Consider a diffusion given by, $d X_t = \mu(X_t) dt + \sigma(X_t) dB_t$

$X_0 = x$.

Suppose the functions $\mu$ and $\sigma$ are as follows -

$f(x) = \mu(x) = \sigma(x) = \begin{cases} 2 & \text{ if } x \ge 0 \\ 1 & \text{ if } x < 0 \end{cases} $

The purpose of $f(x)$ will be clear in a moment.

By Nakao(1972) we know that there exists a strong solution. Now, suppose I am interested in computing the following -

$v(x) = \mathbb{E}^x \int_0^\infty e^{-t} f(X_t) d t $

I know how to do it mechanically. We have the following two DEs: \begin{align} v(x) - 2 v'(x) -2v"(x) -2 = 0 & \text{ if } x >0 \\ v(x) - v'(x) - \frac{1}{2} v"(x) -1 =0 & \text{ if } x < 0 \end{align}

Now, I will solve these 2 simple DEs. Each solution will have 2 constants to be determined. I will use the fact that $v(\infty) = 2$ and $v(-\infty) = 1$ to kill one constant on either side. Then, I will use continuity and differentiability (smooth-pasting) at $0$ to obtain the other 2 constants.

In doing so, however, I have assumed that $v$ is differentiable at $0$. I can prove that $v$ is continuous at $0$. But I do not know how to make the argument for differentiability. This sort of a question comes up often for applied people working with stochastic control and the "standard" method is to assume that it is smooth and then use a "verification theorem". Assuming I want to avoid that, what could be a direct way to prove differentiability?

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  • $\begingroup$ I think you should add an expectation $\mathbb E^x$ in the definition of $v(x)$, and maybe $X_0=x$ in that of $X_t$. $\endgroup$ – Jean Duchon Jul 10 '16 at 10:34
  • $\begingroup$ Absolutely! Damn, how did I miss this!! Many thanks for pointing it out. $\endgroup$ – avk255 Jul 10 '16 at 20:35
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Here is a rough probabilistic argument for differentiability of $\nu(x)$. At least formally, we have: $$ \nu^{\prime}(x) = \mathbb{E} \int_0^{\infty} e^{-t} \delta(X_t(x)) X_t^{\prime}(x) dt \tag{$\star$} $$ since the derivative of a step function, in the distributional sense, is a Dirac delta function $\delta(\cdot)$. Note that the notation $X_t(x)$ expresses the dependence of the SDE solution on its initial data $x$. In terms of the symmetric local time $L_t$ accumulated by the process $X(t)$ at the origin, we can write ($\star$) as: $$ v^{\prime}(x) = \mathbb{E} \int_0^{\infty} e^{-t} X_t^{\prime}(x) dL_t \tag{$\star\star$} $$ This local time is continuous with respect to $x$, since roughly speaking, the drift of the SDE for $X(t)$ does not involve the local time of the process. Moreover, at least formally, the process $X_t^{\prime}$ satisfies the SDE: $$ d X_t^{\prime} = X_t^{\prime} \left( d L_t + \delta(X_t) dB_t \right) \quad X_0^{\prime}(x) = 1 $$ which depends on the initial data only through $X_t$. So if $X_t$ is a.s. pathwise differentiable, then it seems from ($\star\star$) like $\nu^{\prime}(x)$ is continuous. This motivates revisiting that paper you reference by Nakao, and understanding better the regularity of $X_t$ with respect to its initial data.

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