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Linear stochastic differential equation

Consider linear SDE process.

dX_t=( (-\mu +\frac{\sigma^2}{2})X_t+\kappa)dt+\sigma (X_t+\eta)dW_t

Here, \kappa, \eta are constant.

The ODE for this process, 

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

Here, x represents  x+\eta, and \xi represents \xi+\eta.

\tau=-\kappa-(\mu-\frac{\sigma^2}{2})\eta

 

We write the independent solutions for this ODE after geometric Brownian motion.

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

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

 

\alpha, \beta are the solution of quadratic equation.

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

 

We define the result of differentiating y_1, y_2 as

y_1'(x)=(\alpha-1)x^{\alpha-2}w_d(x)

y_2'(x)=(\beta-1) x^{\beta-2}z_d(x) 

 

Eq(2) has fixed singular point at x=\infty, so we can apply Frobenius method.

Therefore w, w_d, z, z_d are polynomial of x^{-1}, and constant part is 1.

The coefficients of x^{-l} only depend to "k", the differentiating result by k is O(1/k). 

 

We transform the equation like Sturm-Liouville equation in the same way as geometric Brownian motion, 

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

Here, 

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

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

The independent solutions which satisfy boundary condition are

In x<\xi

y_1(x)+cy_2(x)

w(b+\eta)(b+\eta)^{\alpha}+cz(b+\eta)(b+\eta)^{\beta}=0

In x>\xi

y_1(x)

Solve the Green function in the same way as geometric Brownian motion

In x<\xi

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

In x>\xi

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

 

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

Here,

\Psi_k(x)=(\alpha-1)w_d(x)z(x)-(\beta-1)w(x)z_d(x)

 in x<\xi g(x) is

g(x)=-\frac{2w(\xi+\eta){(\xi+\eta)}^{-\beta}({(x+\eta)}^{\beta-1}z(x+\eta)w(b+\eta)-{(x+\eta)}^{\alpha-1}w(x+\eta)z(b+\eta){(b+\eta)}^{\beta-\alpha})}{\sigma^2\Psi_k(\xi+\eta)w(b+\eta)}

 in x>\xi g(x) is

g(x)=-{w(x+\eta)(x+\eta)}^{\alpha-1}\frac{2({(\xi+\eta)}^{-\alpha}z(\xi+\eta)w(b+\eta)-{(\xi+\eta)}^{-\beta}w(\xi+\eta)z(b+\eta){(b+\eta)}^{\beta-\alpha}}{\sigma^2\Psi_k(\xi+\eta)w(b+\eta)})

 

The characteristi function of first passage time is 

\int_{0}^{\infty}\exp(ikt)f(t)dt=-w(\xi+\eta)(\xi+\eta)^{-\beta}\frac{(\alpha-1)(b+\eta)^{\alpha}w_d(b+\eta)+c(\beta-1)(1+\eta)^{\beta}z_d(b+\eta)}{c( (\alpha-1) w_d(\xi+\eta)z(\xi+\eta)-(\beta-1)w(\xi+\eta)z_d(\xi+\eta) )}

Substitute  (b+\eta)^{\alpha}w(b+\eta)+c(b+\eta)^{\beta}z(b+\eta)=0

\int_{0}^{\infty}\exp(ikt)f(t)dt=(\frac{\xi+\eta}{b+\eta})^{-\beta}\frac{w(\xi+\eta)\Psi_k(b+\eta)}{w(b+\eta)\Psi_k(\xi+\eta)}

Here, 

\Psi_k(x)=(\alpha-1)w_d(x)z(x)-(\beta-1)w(x)z_d(x)

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

\alpha=\frac{-\mu-\sqrt{\mu^2-2\sigma^2ik}}{\sigma^2}

 

When \xi is large,

\Psi_k(\xi)\sim\alpha-\beta