Brownian Motion generator applied on indicator function.

Question

We know that for a vanishing real-valued $C^2$-function $f$ and a brownian motion $B$ we have $$\frac{d}{dt}\mathbb{E}_x[f(B_t)]=\frac{1}{2}\frac{d^2}{dx^2}\mathbb{E}_x[f(B_t)]$$ Now, i have seen the following. Let $$\Phi_t(x)=\mathbb{P}(B_t\leq x)$$ which is the distribution function of a normal distribution with parameters $(0,t)$. Then it holds $$\frac{d}{dt}\Phi_t(x)=\frac{1}{2}\frac{d^2}{dx^2}\Phi_t(x)$$ I don't understand why this is true or where this comes from. How to rewrite $$\Phi_t(x)=\mathbb{E}_x[f(B_t)]$$ with suitable $f$ such that $f$ is in the domain of the generator of brownian motion?

Answer 1

Denote by $$p_t(y) := \frac{1}{\sqrt{2\pi t}} \exp \left(-\frac{y^2}{2t} \right)$$ the density of $B_t$. Then $$\Phi_t(x) = \mathbb{P}(B_t \leq x) = \int_{(-\infty,x]} p_t(y) \, dy.$$ Differentiation with respect to $t$ yields

$$\frac{d}{dt} \Phi_t(x) = \int_{(-\infty,x]} \frac{d}{dt} p_t(y) \, dy.$$ Since $$\frac{d}{dt} p_t(y) = \frac{1}{2} \frac{d^2}{dy^2} p_t(y)$$ we get $$\frac{d}{dt} \Phi_t(x) = \frac{1}{2} \int_{(-\infty,x]} \frac{d^2}{dy^2}p_t(y) \, dy = \frac{1}{2} \frac{d}{dx} p_t(x).\tag{1}$$

On the other hand, the fundamental theorem of calculus gives

$$\frac{d}{dx} \Phi_t(x) = p_t(x)$$

and so

$$\frac{d^2}{dx^2} \Phi_t(x) = \frac{d}{dx} p_t(x).\tag{2}$$

Combining $(1)$ and $(2)$ proves the assertion.

Answer 2

\begin{align*} \frac{\partial}{\partial t} E^x[f(B_t)] &=\frac{1}{2}\frac{\partial ^2}{\partial x^2} E^x[f(B_t)] \quad\quad (*) \\ \frac{\partial}{\partial t}\Phi_t(x) &=\frac{1}{2}\frac{\partial ^2}{\partial x^2}\Phi_t(x) \quad \quad \quad (**) \end{align*} where $\Phi_t(x)= P(B_t\leq x)$.

The answer given by @saz already explained why $(*)$ is true.

For your other question

...How to rewrite $\Phi_t(x)$ with suitable $f$ such that $f$ is in the domain of the generator of Brownian motion [so that $(*)$ can be viewed as an instance of $(**)$]?

the answer is no, $(*)$ should not be viewed as an instance of $(**)$. Not even formally with $f$ being an indicator function.

Equation $(*)$ is a special case of the Kolmogorov backward equation, and $(**)$ is related to the Kolmogorov forward equation. In general they are dual to each other. In the special case of Brownian motion, they turn out to be similar because the infinitesmal generator $$ A = \frac{1}{2} \frac{\partial^2}{\partial x^2} $$ is self-adjoint and the transition density $$ p_t(x,z) = P^x( B_t \in dz ) $$ is symmetric in the backward variable $x$ and forward variable $z$.

In general, given Ito diffusion $(X_t)$ with $X_0 = x$, the Kolmogorov backward equation is $$ \frac{\partial}{\partial t} E^x[ f(X_t) ] = A E^x[ f (X_t) ]. $$ where $f \in C^2$ and $A$ operates on the backward variable $x$. The special case when $X = B$ is precisely $(*)$.

On the other hand, let $p_t(x,z) = P^x( X_t \in dz )$ be the transition density of $(X_t)$, then the Kolmogorov forward equation is $$ \frac{\partial}{\partial t} p_t(x,z) = A^* p_t(x,z) $$ where $A^*$ is the adjoint operator of $A$ operating on the forward variable $z$.

For Brownian motion starting a $x$, the infinitesmal generator $A$ is self-adjoint, i.e. $A = A^*$, with appropriate domains. Moreover $ p_t(x,z) = \frac{1}{ \sqrt{2 \pi t} }e^{-\frac12 \frac{(x-z)^2}{t}} $ implies that $$ A^* p_t(x,z) = A p_t(x,z) $$ where $A^*$ operates on $z$ and $A$ operates on $x$. So $$ \frac{\partial}{\partial t} p_t(x,z) = A p_t(x,z), $$ which leads to $(**)$.