[Math] Solving the SDE $dX_t=bdt+cX_t dW_t$

stochastic-calculusstochastic-differential-equationsstochastic-processes

I want to solve the SDE $dX_t=bdt+cX_t dW_t$, $X_0=0$ for $b,c\in\mathbb R$. I start by rewriting this as

$$dX_t=(\mu_1+\mu_2 X_t )dt+(\sigma_1+\sigma_2 X_t )dW_t$$

where $\mu_1=b, \mu_2=0, \sigma_1=0, \sigma_2=c$. And the general solution for linear SDE is known to be$X_t=Y_t Z_t$ where

$$dY_t=\mu_2 Y_t dt+\sigma_2 Y_t dW_t, \quad Y_0=1$$
$$dZ_t = \frac{\mu_1-\sigma_1 \sigma_2}{Y_t } dt + \frac{\sigma_1}{Y_t} dW_t, \quad Z_0=X_0=0$$

So
$$dY_t=cY_t dW_t \quad \Rightarrow \quad Y_t=\exp\left(-\frac{c^2}{2}t+cW_t\right)$$
and

$$dZ_t=\frac{b}{Y_t}dt = b \exp\left(\frac{c^2}{2}t – cW_t\right)dt$$

But now I am stuck as I am not sure how to find $Z_t$.

Best Answer

Actually, there is nothing left to do. From

$$dZ_t = b \exp \left( \frac{c^2}{2} t - c W_t \right) \, dt$$

it follows that

$$Z_t = \underbrace{Z_0}_{0} + b \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) \, ds.$$

Hence,

$$X_t = Y_t \cdot Z_t = b \exp \left(- \frac{c^2}{2} t+c W_t \right) \cdot \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) \, ds.$$

CASE 1. $g(t) \neq 0$, $h(t)=0$

Then $Z_t$ would satisfy $$\mathrm{d}Z_t = Z_t(-a(t)+g^2(t))\mathrm{d}t -Z_t g(t)\mathrm{d}B_t , \ \ Z_0 = 1$$ (We solve the above SDE by a standard trick involving Ito's formula, that is, first we divide by $Z_t$ to obtain the expression for $Z_t^{-1}\mathrm{d}Z_t$ and then derive the expression $\mathrm{d}(\ln(Z_t))$)

and so $$Z_t = \exp\left( \int_0^t\left( \frac{1}{2}g^2(s) - a(s)\right) \mathrm{d}s - \int_0^t g(s)\mathrm{d}B_s\right).$$ This is the integrating factor which you obtained, and it is not the correct one if $h(t) \neq 0$.

If we continue then we obtain \begin{align*}\mathrm{d}(Z_tX_t) =& Z_tb(t)\mathrm{d}t \\ Z_tX_t =& X_0 + \int_0^t Z_sb(s) \mathrm{d}s\\ X_t =& X_0Z_t^{-1}+ Z_t^{-1}\int_0^t Z_sb(s) \mathrm{d}s\\ X_t =& \exp\left( \int_0^t\left( a(s) - \frac{1}{2}g^2(s) \right) \mathrm{d}s + \int_0^t g(s)\mathrm{d}B_s\right) \\ &\cdot \left(X_0+ \int_0^{t} b(s)\exp\left( \int_0^s\left( \frac{1}{2}g^2(r) - a(r)\right) \mathrm{d}r - \int_0^s g(r)\mathrm{d}B_r\right)\mathrm{d}s\right). \end{align*}

CASE 2. $g(t) = 0$, $h(t) \neq 0$

Then $(Z_t)$ satisfies $$\mathrm{d}Z_t = -a(t)Z_t\mathrm{d}t, \ \quad Z_0 =1 $$ and so $$Z_t = \exp\left( -\int_0^t a(s)\mathrm{d}s\right).$$

Therefore, \begin{align*}\mathrm{d}(X_tZ_t) =& Z_tb(t)\mathrm{d}t+ Z_t h(t)\mathrm{d}B_t \\ X_tZ_t =& X_0 + \int_0^t Z_sb(s) \mathrm{d}s + \int_0^t Z_sb(s) \mathrm{d}B_s\\ X_t = & X_0Z_t^{-1}+ Z_t^{-1}\left( \int_0^t Z_sb(s) \mathrm{d}s + \int_0^t Z_sb(s) \mathrm{d}B_s\right) \\ X_t = & X_0 e^{ \int_0^t a(s)\mathrm{d}s}\\ &+ e^{ \int_0^t a(s)\mathrm{d}s}\left( \int_0^t e^{ -\int_0^s a(r)\mathrm{d}r}b(s) \mathrm{d}s + \int_0^t e^{ -\int_0^s a(r)\mathrm{d}r}b(s) \mathrm{d}B_s\right). \end{align*}

The general case

So far we were not able to find the solution for the general case, to find it we need to modify and rearrange our initial equation in a slightly different way. Let write the SDE for $(X_t)$ as follows \begin{align*}\mathrm{d}X_t = (a(t)X_t + b(t)+g(t)h(t) - g(t)h(t))\mathrm{d}t + (g(t)X_t+h(t))\mathrm{d}B_t \end{align*} after rearranging we have \begin{align*}\mathrm{d}X_t - \left((a(t)X_t+g(t)h(t))\mathrm{d}t+ g(t)X_t\mathrm{d}B_t\right)= (b(t)- g(t)h(t))\mathrm{d}t + h(t)\mathrm{d}B_t \end{align*} Now let ($Z_t$) be an Ito process that satisfies $$\mathrm{d}Z_t = f_1(t, Z_t)\mathrm{d}t + f_2(t, Z_t)\mathrm{d}B_t , \ \ Z_0 = 1.$$ After multiplying our SDE by $Z_t$ we obtain \begin{align*}Z_t\mathrm{d}X_t - Z_t\left((a(t)X_t+g(t)h(t))\mathrm{d}t+ g(t)X_t\mathrm{d}B_t\right)= Z_t \left((b(t)- g(t)h(t))\mathrm{d}t + h(t)\mathrm{d}B_t\right) \end{align*} Applying Ito product formula to $X_tZ_t$ we get $$ \mathrm{d}(X_tZ_t) = Z_t\mathrm{d}X_t + X_t\mathrm{d}Z_t+ \mathrm{d}X_t\mathrm{d}Z_t,$$

we want to have \begin{align*}X_t\mathrm{d}Z_t+ \mathrm{d}X_t\mathrm{d}Z_t = - Z_t\left((a(t)X_t+g(t)h(t))\mathrm{d}t+ g(t)X_t\mathrm{d}B_t\right)\end{align*}

The LHS equals to \begin{align*}X_t\left(f_1(t, Z_t)\mathrm{d}t + f_2(t, Z_t)\mathrm{d}B_t \right)+ (g(t)X_t+h(t))f_2(t, Z_t)\mathrm{d}t.\end{align*}

We compare with the RHS \begin{align*}X_t\left(f_1(t, Z_t)\mathrm{d}t + f_2(t, Z_t)\mathrm{d}B_t \right)+ (g(t)X_t+h(t))f_2(t, Z_t)\mathrm{d}t = - Z_t\left((a(t)X_t+g(t)h(t))\mathrm{d}t+ g(t)X_t\mathrm{d}B_t\right)\end{align*}

and conclude that $f_2(t, Z_t) = -g(t)Z_t$. Now

\begin{align*}X_t\left(f_1(t, Z_t)\mathrm{d}t -g(t)Z_t\mathrm{d}B_t \right)- (g(t)X_t+h(t))g(t)Z_t\mathrm{d}t = - Z_t\left((a(t)X_t+g(t)h(t))\mathrm{d}t+ g(t)X_t\mathrm{d}B_t\right)\end{align*} simplifies to \begin{align*}X_tf_1(t, Z_t)\mathrm{d}t -g(t)^2Z_tX_t\mathrm{d}t = - a(t)Z_tX_t\mathrm{d}t\end{align*} and so \begin{align*}X_tf_1(t, Z_t)\mathrm{d}t = X_t\left( (g(t)^2- a(t))Z_t\right)\mathrm{d}t\end{align*} let us conclude that $$ f_1(Z_t, t) = (-a(t)+g(t)^2)Z_t.$$ Thus $$\mathrm{d}Z_t = (-a(t)+g^2(t))Z_t \mathrm{d}t - g(t)Z_t\mathrm{d}B_t,$$ you can see the explicit form of $Z_t$ in Case 1 and $$\mathrm{d}(Z_tX_t) = (b(t)-h(t)g(t))Z_t\mathrm{d}t + h(t)Z_t\mathrm{d}B_t.$$

Using Ito's lemma we can show that $(Y_t):= (Z_t^{-1})$ is an Ito process such that it satisfies $$ \mathrm{d}Y_t = a(t)Y_t \mathrm{d}t + g(t)Y_t\mathrm{d}B_t,\ \ Y_0 = 1$$ it has an explicit form $$ Y_t = \exp\left(\int_0^t (a(s)-\frac{1}{2}g^2(s)) \mathrm{d}s +\int_0^t g(s)\mathrm{d}B_s\right).$$

Finally, we see that \begin{align*}\mathrm{d}(Z_tX_t) =& (b(t)-h(t)g(t))Z_t\mathrm{d}t + h(t)Z_t\mathrm{d}B_t \end{align*} yields \begin{align*}Z_tX_t =& Z_0X_0 + \int_0^t (b(s)-h(s)g(s))Z_s\mathrm{d}s + \int_0^t h(s)Z_s\mathrm{d}B_s \end{align*} Hence, we have the explicit solution \begin{align*}X_t =& Z_0X_0Y_t + Y_t\left(\int_0^t (b(s)-h(s)g(s))Z_s\mathrm{d}s + \int_0^t h(s)Z_s\mathrm{d}B_s\right). \end{align*}

Solution of SDE – Solving dX? = ?(t)X?dt + ?X?dW?

The reason $\ln(x)$ is chosen is this:

(Note: where I say $B_{t}$ I mean Brownian motion, which you denoted in your question as $W_{t}$)

The SDE you provided is one of the few we can explicitly solve. I'll talk about Geometric Brownian Motion (GBM) $dX_{t} = \mu X_{t} \,dt + \sigma X_{t} \,dB_{t}$, but as you mentioned in your question, your case is the same when $\mu$ becomes a function of $t$.

You can "multiply" the stochastic differential equation (SDE) in its differential form by $\frac{1}{X_{t}}$ to get $$\frac{1}{X_{t}}dX_{t} = \frac{1}{X_{t}}\mu X_{t} \,dt + \frac{1}{X_{t}}\sigma X_{t} \,dB_{t} $$ and this simplifies to $$ \frac{1}{X_{t}}dX_{t} = \mu \,dt + \sigma \,dB_{t}.$$

Notice that the right hand side no longer depends on $X_{t}$. Now recall Ito's formula for a $C^{2,1}$ function $f(t,x)$ ($C^{2,1}$ means $f$ is twice differentiable in $x$ and once differentiable in $t$). Ito's formula tells us if $X_{t}$ satisfies the previous SDE, then $f(t,X_{t})$ will satisfy:

$$d(f(t,X_{t})) = \frac{\partial f}{\partial t}(t,X_{t}) \,dt + \frac{\partial f}{\partial x}(t,X{t}) \,dX_{t} + \frac{1}{2} \frac{\partial^{2} f}{\partial x^{2}}(t,X_{t}) \,d[X]_{t}$$ where $[X]_{t}$ is the quadratic variation process of $X_{t}$. Notice that in the SDE given to us by Ito's formula, one of the terms on the right hand size is $\frac{\partial f}{\partial x}(t,X{t}) \,dX_{t}$. This almost looks like our $\frac{1}{X_{t}} \,dX_{t}$, which was the left hand side of our original SDE. If we can choose $f(t,X_{t})$ wisely such that these are equal (i.e., such that $\frac{\partial f}{\partial x}(t,X{t})= \frac{1}{X_{t}}$, then we can solve this special SDE.

Hopefully you see that we should have $f(t,x) = \ln(x)$ for the above equality to hold. Okay, so since $f$ doesn't depend on $t$, let's call it $f(x)$ to save space. Substituting this $f$ into Ito's formula above gives:

$$d(\ln(X_{t})) = 0 \,dt + \frac{1}{X_{t}} \,dX_{t} - \frac{1}{2} \frac{1}{X_{t}^2} \,d[X]_{t}$$

Hmm... on the right hand side there is conveniently a $\frac{1}{X_{t}} \,dX_{t}$ (which was the whole point! That's why we choose $f$ as we did) and we have an SDE for this already. We have that $\frac{1}{X_{t}}dX_{t} = \mu \,dt + \sigma \,dB_{t}$.

Okay, so solving for $\frac{1}{X_{t}}\,dX_{t}$ in the Ito's formula SDE gives

$$\frac{1}{X_{t}} \,dX_{t} =d(\ln(X_{t})) + \frac{1}{2} \frac{1}{X_{t}^2} \,d[X]_{t} $$

and this allows us to set the right hand side of the above equal to $\mu \,dt + \sigma \,dB_{t}$. So we have $$d(\ln(X_{t})) + \frac{1}{2} \frac{1}{X_{t}^2} \,d[X]_{t} = \mu \,dt + \sigma \,dB_{t}. $$

What is $d[X]_{t}$? If you do the computation, you get $\sigma^{2} X_{t}^{2} \,dt$, so that our equation becomes $$d(\ln(X_{t})) + \frac{1}{2} \sigma^{2} \,dt = \mu \,dt + \sigma \,dB_{t}. $$

This simplifies to $$\ln(X_{t}) = \ln(X_{0}) + \int \limits_{0}^{t}(\mu - \frac{1}{2} \sigma^{2}) \,dt + \int \limits_{0}^{t}\sigma \,dB_{t} $$

so that $X_{t} = X_{0}e^{\int \limits_{0}^{t}(\mu - \frac{1}{2} \sigma^{2}) \,dt + \int \limits_{0}^{t}\sigma \,dB_{t}}$.

Best Answer

Related Solutions

Stochastic Differential Equations – Solution to General Linear SDE

CASE 1. $g(t) \neq 0$, $h(t)=0$

CASE 2. $g(t) = 0$, $h(t) \neq 0$

The general case

Solution of SDE – Solving dX? = ?(t)X?dt + ?X?dW?

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