Skip to main content
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Source Link

This question was asked herehere, but it did not get enough attention, so I'm crossposting it to MO.

Let $u: \mathbb{R}_+ \times \mathbb{R}^d$ be a bounded $C^2$ function whose first and second partial derivatives are uniformly bounded (or, more generally, have at most polynomial growth as $|x| \to \infty)$ on $[0, T] \times \mathbb{R}^d$, for any $0 \le T < \infty$. For any $t \ge 0$ and any $x \in \mathbb{R}^d$ it is not hard to see that,$$E^x u(t, W_t) = u(0, x) + E^x \int_0^y \left( {\partial\over{\partial s}} + {1\over2}\Delta_s\right) u(s, W_s)\,ds.$$$($$W_t$ is a Brownian motion process which takes the value $x$ at $t=0$ so the dependence of $x$ is implicit here, most of the references will consider $W_t$ a standard Brownian motion which means it takes the value $0$ at $t=0$ and in the above formula we will have $x+W_t$ instead. $\Delta_s$ here means the sum of all the mixed derivatives.$)$

My question is, how do we conclude that under $P^x$ the process$$u(t, W_t) - u(0, x) - \left({\partial\over{\partial t}} + {1\over2}\Delta_x\right)u(t, W_t)$$is a martingale?

This question was asked here, but it did not get enough attention, so I'm crossposting it to MO.

Let $u: \mathbb{R}_+ \times \mathbb{R}^d$ be a bounded $C^2$ function whose first and second partial derivatives are uniformly bounded (or, more generally, have at most polynomial growth as $|x| \to \infty)$ on $[0, T] \times \mathbb{R}^d$, for any $0 \le T < \infty$. For any $t \ge 0$ and any $x \in \mathbb{R}^d$ it is not hard to see that,$$E^x u(t, W_t) = u(0, x) + E^x \int_0^y \left( {\partial\over{\partial s}} + {1\over2}\Delta_s\right) u(s, W_s)\,ds.$$$($$W_t$ is a Brownian motion process which takes the value $x$ at $t=0$ so the dependence of $x$ is implicit here, most of the references will consider $W_t$ a standard Brownian motion which means it takes the value $0$ at $t=0$ and in the above formula we will have $x+W_t$ instead. $\Delta_s$ here means the sum of all the mixed derivatives.$)$

My question is, how do we conclude that under $P^x$ the process$$u(t, W_t) - u(0, x) - \left({\partial\over{\partial t}} + {1\over2}\Delta_x\right)u(t, W_t)$$is a martingale?

This question was asked here, but it did not get enough attention, so I'm crossposting it to MO.

Let $u: \mathbb{R}_+ \times \mathbb{R}^d$ be a bounded $C^2$ function whose first and second partial derivatives are uniformly bounded (or, more generally, have at most polynomial growth as $|x| \to \infty)$ on $[0, T] \times \mathbb{R}^d$, for any $0 \le T < \infty$. For any $t \ge 0$ and any $x \in \mathbb{R}^d$ it is not hard to see that,$$E^x u(t, W_t) = u(0, x) + E^x \int_0^y \left( {\partial\over{\partial s}} + {1\over2}\Delta_s\right) u(s, W_s)\,ds.$$$($$W_t$ is a Brownian motion process which takes the value $x$ at $t=0$ so the dependence of $x$ is implicit here, most of the references will consider $W_t$ a standard Brownian motion which means it takes the value $0$ at $t=0$ and in the above formula we will have $x+W_t$ instead. $\Delta_s$ here means the sum of all the mixed derivatives.$)$

My question is, how do we conclude that under $P^x$ the process$$u(t, W_t) - u(0, x) - \left({\partial\over{\partial t}} + {1\over2}\Delta_x\right)u(t, W_t)$$is a martingale?

Notice removed Canonical answer required by CommunityBot
Bounty Ended with ofer zeitouni's answer chosen by CommunityBot
Notice added Canonical answer required by user61522
Bounty Started worth 100 reputation by CommunityBot
added 4 characters in body
Source Link
user61522
user61522

This question was asked here, but it did not get enough loveattention, so I'm askingcrossposting it againto MO.

Let $u: \mathbb{R}_+ \times \mathbb{R}^d$ be a bounded $C^2$ function whose first and second partial derivatives are uniformly bounded (or, more generally, have at most polynomial growth as $|x| \to \infty)$ on $[0, T] \times \mathbb{R}^d$, for any $0 \le T < \infty$. For any $t \ge 0$ and any $x \in \mathbb{R}^d$ it is not hard to see that,$$E^x u(t, W_t) = u(0, x) + E^x \int_0^y \left( {\partial\over{\partial s}} + {1\over2}\Delta_s\right) u(s, W_s)\,ds.$$$($$W_t$ is a Brownian motion process which takes the value $x$ at $t=0$ so the dependence of $x$ is implicit here, most of the references will consider $W_t$ a standard Brownian motion which means it takes the value $0$ at $t=0$ and in the above formula we will have $x+W_t$ instead. $\Delta_s$ here means the sum of all the mixed derivatives.$)$

My question is, how do we conclude that under $P^x$ the process$$u(t, W_t) - u(0, x) - \left({\partial\over{\partial t}} + {1\over2}\Delta_x\right)u(t, W_t)$$is a martingale?

This question was asked here, but it did not get enough love, so I'm asking it again.

Let $u: \mathbb{R}_+ \times \mathbb{R}^d$ be a bounded $C^2$ function whose first and second partial derivatives are uniformly bounded (or, more generally, have at most polynomial growth as $|x| \to \infty)$ on $[0, T] \times \mathbb{R}^d$, for any $0 \le T < \infty$. For any $t \ge 0$ and any $x \in \mathbb{R}^d$ it is not hard to see that,$$E^x u(t, W_t) = u(0, x) + E^x \int_0^y \left( {\partial\over{\partial s}} + {1\over2}\Delta_s\right) u(s, W_s)\,ds.$$$($$W_t$ is a Brownian motion process which takes the value $x$ at $t=0$ so the dependence of $x$ is implicit here, most of the references will consider $W_t$ a standard Brownian motion which means it takes the value $0$ at $t=0$ and in the above formula we will have $x+W_t$ instead. $\Delta_s$ here means the sum of all the mixed derivatives.$)$

My question is, how do we conclude that under $P^x$ the process$$u(t, W_t) - u(0, x) - \left({\partial\over{\partial t}} + {1\over2}\Delta_x\right)u(t, W_t)$$is a martingale?

This question was asked here, but it did not get enough attention, so I'm crossposting it to MO.

Let $u: \mathbb{R}_+ \times \mathbb{R}^d$ be a bounded $C^2$ function whose first and second partial derivatives are uniformly bounded (or, more generally, have at most polynomial growth as $|x| \to \infty)$ on $[0, T] \times \mathbb{R}^d$, for any $0 \le T < \infty$. For any $t \ge 0$ and any $x \in \mathbb{R}^d$ it is not hard to see that,$$E^x u(t, W_t) = u(0, x) + E^x \int_0^y \left( {\partial\over{\partial s}} + {1\over2}\Delta_s\right) u(s, W_s)\,ds.$$$($$W_t$ is a Brownian motion process which takes the value $x$ at $t=0$ so the dependence of $x$ is implicit here, most of the references will consider $W_t$ a standard Brownian motion which means it takes the value $0$ at $t=0$ and in the above formula we will have $x+W_t$ instead. $\Delta_s$ here means the sum of all the mixed derivatives.$)$

My question is, how do we conclude that under $P^x$ the process$$u(t, W_t) - u(0, x) - \left({\partial\over{\partial t}} + {1\over2}\Delta_x\right)u(t, W_t)$$is a martingale?

Source Link
user61522
user61522

Extension of Dynkin's formula, conclude that process is a martingale

This question was asked here, but it did not get enough love, so I'm asking it again.

Let $u: \mathbb{R}_+ \times \mathbb{R}^d$ be a bounded $C^2$ function whose first and second partial derivatives are uniformly bounded (or, more generally, have at most polynomial growth as $|x| \to \infty)$ on $[0, T] \times \mathbb{R}^d$, for any $0 \le T < \infty$. For any $t \ge 0$ and any $x \in \mathbb{R}^d$ it is not hard to see that,$$E^x u(t, W_t) = u(0, x) + E^x \int_0^y \left( {\partial\over{\partial s}} + {1\over2}\Delta_s\right) u(s, W_s)\,ds.$$$($$W_t$ is a Brownian motion process which takes the value $x$ at $t=0$ so the dependence of $x$ is implicit here, most of the references will consider $W_t$ a standard Brownian motion which means it takes the value $0$ at $t=0$ and in the above formula we will have $x+W_t$ instead. $\Delta_s$ here means the sum of all the mixed derivatives.$)$

My question is, how do we conclude that under $P^x$ the process$$u(t, W_t) - u(0, x) - \left({\partial\over{\partial t}} + {1\over2}\Delta_x\right)u(t, W_t)$$is a martingale?