3
$\begingroup$

I have the following SDE and I would like to check if the solution , i.e. the explicit form for $X_t$ that I gave is either wrong or false. $$dX_t = \frac{e^t-X_t}{t-2}dt + tdW_t$$

with $t \in [0,2] $ and $X_0 = 0$ and $W_t$ is a brownian motion.

I supppose it is a modified SDE of Hull-White but with a coefficient on $X_t$ depending on time.

For my own readability I wrote the equation as follow, hoping it would be easy to plug the coefficients of each term in the solution for the general Linear SDE : $$ dX_t =\bigg( \frac{-X_t}{t-2} + \frac{e^t}{t-2} \bigg)dt + tdW_t$$

I tried to start with the process $Y_t = e^{\int_0^t \frac{-1}{s-2}ds}X_t$ and apply Ito to it. Solving the integral in the exponential I get: $$Y_t =e^{-log(|2-t|)+log(|2|)}X_t = \frac{2}{2-t}X_t \tag{1}$$

I calculate its differential by applying Ito's lemma: $dY_t = \bigg( \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} \mu_X+ \frac{1}{2}\frac{\partial ^2f}{\partial x^2}\sigma^2_X \bigg)dt + \bigg( \frac{\partial f}{\partial x}\sigma_X \bigg)$

with $ \mu_X$ and $ \sigma_X$ defined in : $dX_t = \mu_X dt + \sigma_X dW_t $

From the relation $(1)$ the partial derivatives are : $\frac{\partial f}{\partial t} = \frac{-2}{(2-t)^2}, \frac{\partial f}{\partial x} = \frac{2}{2-t} , \frac{\partial ^2f}{\partial x^2} = 0$

Finally :

$$dY_t = \frac{-2}{(2-t)^2} X_tdt + \frac{2}{2-t} \bigg[ \big( \frac{-X_t}{t-2} + \frac{e^{t}}{t-2} \big)dt \bigg]+ \frac{2}{2-t}tdW_t$$

The terms in $X_t$ simplify each other and so integrating both LHS and RHS is feasible : $$Y_t - Y_0 = \int_0^t \frac{2}{2-s} \frac{e^{s}}{s-2}ds + \int_0^t \frac{2}{2-s}sdW_s$$

From $ (1) $ and the initial $X_0 = 0$ we know that $Y_0 = 0$ and replacing $Y_t$ by its value in $(1)$ : $$X_t = \int_0^t \frac{2}{2-s} \frac{e^{s}}{s-2}ds + \int_0^t \frac{2}{2-s}sdW_s$$

And now I don't know what I can do to see if it is a correct solution. N.B. : Also I feel like I miss some perspective here, so any comments on form or context is still appreciated.

$\endgroup$
1
  • $\begingroup$ I think that your solution is not correct; note that if $(X_t)_t$ is a solution to the SDE, then $t \mapsto X_t-\int_0^t s \, dW_s$ is differentiable... I don't see how this could be true for the process you obtained. $\endgroup$
    – saz
    Jan 25, 2018 at 13:15

1 Answer 1

2
$\begingroup$

Suppose that $(X_t)_{t \geq 0}$ is a solution to the SDE

$$dX_t = \frac{e^t-X_t}{t-2} \, dt+ t \, dW_t.$$

then

$$Y_t := X_t - \int_0^t s \, dW_s = \int_0^t \frac{e^s-X_s}{s-2} \, ds$$

is a solution to the ordinary differential equation

$$dY_t = - \frac{Y_t}{t-2} \, dt + \underbrace{\frac{e^t-\int_0^t s \, dW_s}{t-2}}_{=:f(t)} \, dt \tag{1}$$

The solution to the associated homogeneous equation

$$dz_t = - \frac{z_t}{t-2}$$

is given by

$$z_t = \frac{c}{t-2}.$$

To obtain a solution to the inhomogeneous equation we use the variation of constants approach, i.e. we make the ansatz

$$Y_t = \frac{c(t)}{t-2} \tag{2}$$

and find

$$Y'(t) = - \frac{Y_t}{t-2}+ \frac{c'(t)}{t-2}.$$

Comparing this with $(1)$ we find that

$$c'(t) = f(t) = \frac{1}{t-2} \left( e^t- \int_0^t s \, dW_s \right),$$

and so

$$c(t) = \int_0^t \frac{1}{s-2} \left( e^s- \int_0^s r \, dW_r \right) \, ds.$$

Thus, by $(2)$,

$$Y_t = \frac{1}{t-2} \int_0^t \frac{1}{s-2} \left( e^s- \int_0^s r \, dW_r \right) \, ds$$

implying

$$X_t = Y_t + \int_0^t s \, dW_s = \frac{1}{t-2} \int_0^t \frac{1}{s-2} \left( e^s- \int_0^s r \, dW_r \right) \, ds + \int_0^t s \, dW_s. \tag{3}$$

Remark: By Itô's formula, we have

$$tW_t = \int_0^t s \, dW_s + \int_0^t W_s \, ds,$$

i.e.

$$\int_0^t s \, dW_s = t W_t - \int_0^t W_s \, ds.$$

This means that we can actually get rid of the stochastic integral in $(3)$.

$\endgroup$

You must log in to answer this question.