Compute the expectation of $X_tau$ and related probability given its SDE $dX_t = X_t,dt + sigma , dB

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Suppose there is a stochastic process $X_t$ satisfying the SDE $dX_t = X_t,dt + sigma ,dB_t$ where $B_t$ denotes the standard Brownian motion starting from $0$. Let $a<b$ being two real numbers. Define $tau = inft:X_tnotin [a,b]$ and I want to find $E[X_taumid X_0=x]$.

I know one method using Feynman-Kac theorem. Denote $u(t,X_t) = E[X_taumid X_t]$ exploiting its markovian property. Then $u$ is a martingale by Tower property. By Ito's lemma:
$$
du(t,X_t)=u_t dt+u_x dX_t+frac12u_xx d[X]_t=(u_t + u_xX_t + fracsigma^22 u_xx) dt + sigma u_x dB_t
$$
So we get the PDE:
begincases
u_t + xu_x + fracsigma^22 u_xx=0\
u(t,a)=a\
u(t,b)=b
endcases
Notice $u_t=0$, we simply swich $u(t,x)=u(x)=E[X_taumid X_t=x]$ to get
begincases
xu_x + fracsigma^22 u_xx=0\
u(a)=a\
u(b)=b
endcases
Solve this and we get
$$
u(x)=afracint_x^be^-fracx^2sigma^2 dxint_a^b e^-fracx^2sigma^2 dx + bfracint_a^x e^-fracx^2sigma^2 dxint_a^b e^-fracx^2sigma^2 dx
$$
Since $u(x)$ should have a form like
$$
u(x)=aP(X_tau=a) + bP(X_tau=b)
$$
Is it safe to guess $P(X_tau=a mid X_t=x)=fracint_x^b e^-fracx^2sigma^2 dxint_a^b e^-fracx^2sigma^2 dx$ and $P(X_tau=b mid X_t=x)=fracint_a^x e^-fracx^2sigma^2 dxint_a^b e^-fracx^2sigma^2 dx$?

My reason is that it seems very reasonable if $x$ increases from $a$ to $b$, $X_tau$ varies from $a$ and more and more likely becomes $b$ when $x$ increases. Besides, $fracint_a^x e^-fracx^2sigma^2 dxint_a^b e^-fracx^2sigma^2 dx + fracint_x^b e^-fracx^2sigma^2 dxint_a^b e^-fracx^2sigma^2 dx=1$.

I generally have three questions:

How to justify $P(X_tau = a) + P(X_tau = b) = 1$, namely $P(tau = infty) = 0$?

How to justify $P(X_tau = a)$ and $P(X_tau = b)$ happen to be the coefficient of $a$ and $b$ in the expression of $u(x)$?

I also learned a technique to compute this type of problem by constructing martingales like $B_t^2-t$ when we compute the exit time of Brownian motion. Is there similar way to solve this problem? I notice $e^-tX_t$ is a martingale but don't know how to proceed.

Thank you for any help!

edited Jul 26 at 22:33

Michael Hardy

204k23186461

asked Jul 26 at 20:55

Edward Wang

598311

add a commentÂ |Â

up vote
3
down vote

favorite

I generally have three questions:

How to justify $P(X_tau = a) + P(X_tau = b) = 1$, namely $P(tau = infty) = 0$?

How to justify $P(X_tau = a)$ and $P(X_tau = b)$ happen to be the coefficient of $a$ and $b$ in the expression of $u(x)$?

I also learned a technique to compute this type of problem by constructing martingales like $B_t^2-t$ when we compute the exit time of Brownian motion. Is there similar way to solve this problem? I notice $e^-tX_t$ is a martingale but don't know how to proceed.

Thank you for any help!

edited Jul 26 at 22:33

Michael Hardy

204k23186461

asked Jul 26 at 20:55

Edward Wang

598311

add a commentÂ |Â

up vote
3
down vote

favorite

I generally have three questions:

How to justify $P(X_tau = a) + P(X_tau = b) = 1$, namely $P(tau = infty) = 0$?

How to justify $P(X_tau = a)$ and $P(X_tau = b)$ happen to be the coefficient of $a$ and $b$ in the expression of $u(x)$?

I also learned a technique to compute this type of problem by constructing martingales like $B_t^2-t$ when we compute the exit time of Brownian motion. Is there similar way to solve this problem? I notice $e^-tX_t$ is a martingale but don't know how to proceed.

Thank you for any help!

edited Jul 26 at 22:33

Michael Hardy

204k23186461

asked Jul 26 at 20:55

Edward Wang

598311

I generally have three questions:

How to justify $P(X_tau = a) + P(X_tau = b) = 1$, namely $P(tau = infty) = 0$?

How to justify $P(X_tau = a)$ and $P(X_tau = b)$ happen to be the coefficient of $a$ and $b$ in the expression of $u(x)$?

I also learned a technique to compute this type of problem by constructing martingales like $B_t^2-t$ when we compute the exit time of Brownian motion. Is there similar way to solve this problem? I notice $e^-tX_t$ is a martingale but don't know how to proceed.

Thank you for any help!

edited Jul 26 at 22:33

Michael Hardy

204k23186461

asked Jul 26 at 20:55

Edward Wang

598311

edited Jul 26 at 22:33

Michael Hardy

204k23186461

edited Jul 26 at 22:33

Michael Hardy

204k23186461

edited Jul 26 at 22:33

Michael Hardy

204k23186461

asked Jul 26 at 20:55

Edward Wang

598311

asked Jul 26 at 20:55

Edward Wang

598311

asked Jul 26 at 20:55

Edward Wang

598311

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
0
down vote

accepted

We can prove a more general result. For a bounded region $G$, let's assume the diffusion process $X_t$ starts from $X_0=x$ and the exit time is $tau$. Suppose we can prove the following:
$$
forall xin G, P(X_1 in G^c) geq alpha>0
$$
namely $P(tau leq 1) geq alpha$, then from the Markov property, $P(tau geq 2) leq (1-alpha)^2$. Thus:
$$sum_n=1^inftyP(tau geq n) leq sum_n=1^infty(1-alpha)^n = frac1alpha-1 < infty$$
By Borel-Cantelli lemma, $P(limsup_nto infty (tau > n))=0$, so $P(tau<infty)=1$.
Since $X_t$ is Gaussian, to prove $forall x in [a,b]$, $exists alpha > 0$ such that $P(X_1notin[a,b])>alpha > 0$ is easy.

We can solve the same PDE with the boundary condition $u(a)=1$ and $u(b)=0$ to compute $P(X_tau=a)$.

The generator of the process is $mathscrL(f)=xf'(x)+fracsigma^22 f''(x)$, so $X_t-int_0^tmathscrL(f)(X_s)ds$ is a martingale where $f(x)=x$. Thus
$$
E[X_tau] = E[int_0^tauX_s ds]
$$

answered Aug 2 at 17:26

Edward Wang

598311

add a commentÂ |Â

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
0
down vote

accepted

We can prove a more general result. For a bounded region $G$, let's assume the diffusion process $X_t$ starts from $X_0=x$ and the exit time is $tau$. Suppose we can prove the following:
$$
forall xin G, P(X_1 in G^c) geq alpha>0
$$
namely $P(tau leq 1) geq alpha$, then from the Markov property, $P(tau geq 2) leq (1-alpha)^2$. Thus:
$$sum_n=1^inftyP(tau geq n) leq sum_n=1^infty(1-alpha)^n = frac1alpha-1 < infty$$
By Borel-Cantelli lemma, $P(limsup_nto infty (tau > n))=0$, so $P(tau<infty)=1$.
Since $X_t$ is Gaussian, to prove $forall x in [a,b]$, $exists alpha > 0$ such that $P(X_1notin[a,b])>alpha > 0$ is easy.

We can solve the same PDE with the boundary condition $u(a)=1$ and $u(b)=0$ to compute $P(X_tau=a)$.

The generator of the process is $mathscrL(f)=xf'(x)+fracsigma^22 f''(x)$, so $X_t-int_0^tmathscrL(f)(X_s)ds$ is a martingale where $f(x)=x$. Thus
$$
E[X_tau] = E[int_0^tauX_s ds]
$$

answered Aug 2 at 17:26

Edward Wang

598311

add a commentÂ |Â

up vote
0
down vote

accepted

We can prove a more general result. For a bounded region $G$, let's assume the diffusion process $X_t$ starts from $X_0=x$ and the exit time is $tau$. Suppose we can prove the following:
$$
forall xin G, P(X_1 in G^c) geq alpha>0
$$
namely $P(tau leq 1) geq alpha$, then from the Markov property, $P(tau geq 2) leq (1-alpha)^2$. Thus:
$$sum_n=1^inftyP(tau geq n) leq sum_n=1^infty(1-alpha)^n = frac1alpha-1 < infty$$
By Borel-Cantelli lemma, $P(limsup_nto infty (tau > n))=0$, so $P(tau<infty)=1$.
Since $X_t$ is Gaussian, to prove $forall x in [a,b]$, $exists alpha > 0$ such that $P(X_1notin[a,b])>alpha > 0$ is easy.

We can solve the same PDE with the boundary condition $u(a)=1$ and $u(b)=0$ to compute $P(X_tau=a)$.

The generator of the process is $mathscrL(f)=xf'(x)+fracsigma^22 f''(x)$, so $X_t-int_0^tmathscrL(f)(X_s)ds$ is a martingale where $f(x)=x$. Thus
$$
E[X_tau] = E[int_0^tauX_s ds]
$$

answered Aug 2 at 17:26

Edward Wang

598311

add a commentÂ |Â

up vote
0
down vote

accepted

We can prove a more general result. For a bounded region $G$, let's assume the diffusion process $X_t$ starts from $X_0=x$ and the exit time is $tau$. Suppose we can prove the following:
$$
forall xin G, P(X_1 in G^c) geq alpha>0
$$
namely $P(tau leq 1) geq alpha$, then from the Markov property, $P(tau geq 2) leq (1-alpha)^2$. Thus:
$$sum_n=1^inftyP(tau geq n) leq sum_n=1^infty(1-alpha)^n = frac1alpha-1 < infty$$
By Borel-Cantelli lemma, $P(limsup_nto infty (tau > n))=0$, so $P(tau<infty)=1$.
Since $X_t$ is Gaussian, to prove $forall x in [a,b]$, $exists alpha > 0$ such that $P(X_1notin[a,b])>alpha > 0$ is easy.

We can solve the same PDE with the boundary condition $u(a)=1$ and $u(b)=0$ to compute $P(X_tau=a)$.

The generator of the process is $mathscrL(f)=xf'(x)+fracsigma^22 f''(x)$, so $X_t-int_0^tmathscrL(f)(X_s)ds$ is a martingale where $f(x)=x$. Thus
$$
E[X_tau] = E[int_0^tauX_s ds]
$$

answered Aug 2 at 17:26

Edward Wang

598311

We can prove a more general result. For a bounded region $G$, let's assume the diffusion process $X_t$ starts from $X_0=x$ and the exit time is $tau$. Suppose we can prove the following:
$$
forall xin G, P(X_1 in G^c) geq alpha>0
$$
namely $P(tau leq 1) geq alpha$, then from the Markov property, $P(tau geq 2) leq (1-alpha)^2$. Thus:
$$sum_n=1^inftyP(tau geq n) leq sum_n=1^infty(1-alpha)^n = frac1alpha-1 < infty$$
By Borel-Cantelli lemma, $P(limsup_nto infty (tau > n))=0$, so $P(tau<infty)=1$.
Since $X_t$ is Gaussian, to prove $forall x in [a,b]$, $exists alpha > 0$ such that $P(X_1notin[a,b])>alpha > 0$ is easy.

We can solve the same PDE with the boundary condition $u(a)=1$ and $u(b)=0$ to compute $P(X_tau=a)$.

The generator of the process is $mathscrL(f)=xf'(x)+fracsigma^22 f''(x)$, so $X_t-int_0^tmathscrL(f)(X_s)ds$ is a martingale where $f(x)=x$. Thus
$$
E[X_tau] = E[int_0^tauX_s ds]
$$

answered Aug 2 at 17:26

Edward Wang

598311

answered Aug 2 at 17:26

Edward Wang

598311

answered Aug 2 at 17:26

Edward Wang

598311

answered Aug 2 at 17:26

Edward Wang

598311

add a commentÂ |Â

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