趣味の研究

趣味の数学を公開しています。初めての方はaboutをご覧ください。

The case of geometric Brownian motion

We think about the geometric Bownian motion with negative drift.

$dX_t=(-\mu +\frac{\sigma^2}{2})X_tdt+\sigma X_t dW_t$

Here, $\mu$ is the drift of $\log X_t$

The ODE for g(x) is

$\frac{\sigma^2}{2}\frac{d^2}{dx^2}(x^2g(x) )+(\mu-\frac{1}{2}\sigma^2)\frac{d}{dx}(xg(x) )+ikg(t)=-\delta(x-\xi)$

This ODE is Euler equation, we solve the Green function in the same way of Sturm-Liouville equation.

Transform the equation like Sturm-Liouville equation .

$r(x)\frac{d}{dx}(p(x)\frac{d}{dx}g(x))+q(x)=-\delta(x-\xi)$

Here,

$r(x)=\frac{\sigma^2}{2}x^{-1-\frac{2\mu}{\sigma^2}}$

$p(x)=x^{3+\frac{2\mu}{\sigma^2}}$

When the right side equals 0, the independent solutions are $x^{\lambda-1}$ .

Here, $\lambda$ satisfies

$\frac{\sigma^2}{2}\lambda^2+\mu\lambda+ik=0$ .

We define the solutions of this quadratic equation as $\alpha, \beta$ .

α<β.

Define

$y_1(x)=x^{\alpha-1}$

$y_2(x)=x^{\beta-1}$

,

The boundary condition is 0 at x=b, and coverges to 0 first enough at x=∞.

So the solution in in x< $\xi$ is

$y_1(x)+cy_2(x)$

Here, $b^{-\alpha}+cb^{-\beta}=0$ .

The solution in x> $\xi$ is

$y_1(x)$

Next, solve the Green function like the case of Sturm-Liouville equation .

The diffrence from Sturm-Liouville equation is the factor r(x), then Green function in x< $\xi$ is

$-\frac{y_1(\xi)(y_1(x)+cy_2(x))}{A(\xi)}$

the Green function in x> $\xi$ is

$-\frac{y_1(x)(y_1(\xi)+cy_2(\xi))}{A(\xi)}$

$A(\xi)=cp(\xi)r(\xi)(y'_1(\xi)y_2(\xi)-y_1(\xi)y'_2(\xi))$

Remark $\alpha,\beta,c$ are dependent with "k".

Concretely A(x) is

$A(\xi)=\frac{c\sigma^2}{2}\xi^2(\alpha-\beta)\xi^{\alpha+\beta-3}$

$\alpha+\beta=-\frac{2\mu}{\sigma^2}$

so,

$A(\xi)=\frac{c\sigma^2}{2}(\alpha-\beta)\xi^{-\frac{2\mu}{\sigma^2}-1}$

The result is

in x< $\xi$ g(x) is

$g(x)=-\frac{2\xi^{-\beta}(x^{\beta-1}-x^{\alpha-1}b^{\beta-\alpha})}{\sigma^2(\alpha-\beta)}$

in x> $\xi$ g(x) is

$g(x)=-\frac{2x^{\alpha-1}(\xi^{-\alpha}-\xi^{-\beta}b^{\beta-\alpha})}{\sigma^2(\alpha-\beta)}$

In the case of k=0, we derive the passage frequency at x starting from $\xi$ .

When k=0 and b=1,

$\alpha=-\frac{2\mu}{\sigma^2}$ , $\beta=0$ .

In x< $\xi$ ,

$g(x)=\frac{1}{\mu}({x}^{-1}-{x}^{-\frac{2\mu}{\sigma^2}-1})$

In x> $\xi$ ,

$\frac{1}{\mu}({\frac{x}{\xi}})^{-\frac{2\mu}{\sigma^2}-1}( {\xi}^{-1}-{\xi}^{-\frac{2u}{\sigma^2}-1})$

From this result, we can derive the maximum number starting from n< $\xi$ .

By using bellow formula,

$\int_{0}^{\infty}\exp(ikt)f(t)dt=- \frac{\sigma(b)^2}{2}\frac{dg(x)}{dx}|_{x=b}$

Then, we solve the characteristic function of first passage time,

$\int_{0}^{\infty}\exp(ikt)f(t)dt=\frac{\sigma^2}{2}\frac{y_1(\xi)(\alpha-\beta)}{A(\xi)}=(\frac{\xi}{b})^{-\beta}$

Here,

$\beta=\frac{-\mu+\sqrt{\mu^2-2\sigma^2ik}}{\sigma^2}$