Finishing the solution of this SDE

Question

It is given the following SDE:

$\begin{cases}\mathrm{d}Y_t = (Y_t + B_t) \mathrm{d}B_t + Y_t\mathrm{d}t\\ Y_0 = 1 \end{cases}$

where $(Y_t)$ is the unknown Ito process, while $(B_t)$ is a (known) Brownian motion.

This is an inhomogeneous equation, so I first looked at its homogeneous counterpart: $\mathrm{d}Y_t = Y_t\mathrm{d}t$. This is an ODE, which has the solution: $Y_t=ce^t$.

Now I proceed with the method of variation of constants: I'm looking for the solution of the original equation in the form $V_t = C_te^t$ where $(C_t)$ is a yet unknown Ito process. So what I want is:

$$ \mathrm{d}V_t = (V_t + B_t)\mathrm{d}B_t + V_t\mathrm{d}t $$

What I know is:

$$ \mathrm{d}V_t = \mathrm{d}(C_te^t) = e^t\mathrm{d}C_t + C_t e^t\mathrm{d}t + (\mathrm{d}C_t)(\mathrm{d}(e^t))$$

where of course the last term vanishes, as the function $t\mapsto e^t$ is smooth. So I have:

$$\mathrm{d}C_t = e^t\mathrm{d}C_t + V_t\mathrm{d}t$$

which then means I want the following:

$$ e^t\mathrm{d}C_t = (V_t + B_t)\mathrm{d}B_t$$ $$ \mathrm{d}C_t = C_t\mathrm{d}B_t + e^{-t}B_t\mathrm{d}B_t $$ The problem is, integrating the last equation gives $$ C_t - C_0 = \int_0^tC_s\mathrm{d}B_s + \int_0^t e^{-s}B_s\mathrm{d}B_s $$ and I can't get rid of the Ito integral involving $C$ itself.

Is this approach correct/can it be finished?

Answer 1

Here is the next steps. We begin with $$dC_t = C_tdB_t +e^{-t}B_tdB_t \tag{1}$$

Let denote $Z_t$ the solution of $$dZ_t = Z_t dB_t$$ It's easy to find that $$Z_t = Z_0\cdot e^{-\frac{1}{2}t+B_t} \tag{2}$$

Denote $C_t = Z_t U_t$ and come back to $(1)$, we have: $$\begin{align} dC_t &=U_tdZ_t + Z_tdU_t +\frac{1}{2} dZ_tdU_t =\underbrace{U_tZ_t}_{=C_t} dB_t+ Z_tdU_t +\frac{1}{2} dZ_tdU_t \tag{3} \end{align}$$

We need advance technique here.

First, we suppose that $$dU_t = \alpha(U_t,t) dt + \beta(U_t,t) dW_t \tag{4}$$ then $$dZ_tdU_t = Z_t \beta(U_t,t) dt$$ From $(1),(3), (4)$ we deduce that $$Z_t (\alpha(U_t,t) dt + \beta(U_t,t) dW_t) +\frac{1}{2} Z_t \beta(U_t,t) dt = e^{-t}B_t dB_t$$ $$\Longrightarrow \begin{cases} &Z_t ( \alpha(U_t,t) + \frac{1}{2}\beta(U_t,t)) = 0 \\ &Z_t \beta(U_t,t) = e^{-t}B_t \end{cases} $$ $$ \Longrightarrow \begin{cases} &\alpha(U_t,t) = - \frac{1}{2}\beta(U_t,t) \\ & \beta(U_t,t) = \frac{e^{-t}B_t}{Z_t} \end{cases} $$ $$ \stackrel{(2)(4)}{\Longrightarrow} dU_t=\frac{e^{-t}B_t}{Z_0e^{-\frac{1}{2}t+B_t}}\left(-\frac{1}{2}dt + dB_t\right) \tag{5}$$

From $(5)$, it is easy to deduce the solution of $U_t$, then you deduce $C_t = U_tZ_t$ and the solution $Y_t$.

But how could we suppose $(5)$? Perhaps there are other solutions?

It suffices to revert to the initial equation. This equation satisfies the Lipschitz](as in this following question). Then the solution is unique.

Finally, we can conclude that the solution we found is the only the solution of the SDE.

Q.E.D

Answer 2

Here is another approach to solve $\mathrm{d}Y_t = Y_t\mathrm{d}t + (B_t+Y_t)\mathrm{d}B_t$. As the diffusion term involves two stochastic processes, namely $Y_t$ and $B_t$, instead of $Y_t$ and time only, we will need a generalized version of Itô's lemma for a bivariate transformation, as follows : $$ \mathrm{d}f(t,x,y) = f_t\mathrm{d}t + f_x\mathrm{d}x + f_y\mathrm{d}y + \frac{1}{2}f_{xx}\mathrm{d}x^2 + f_{xy}\mathrm{d}x\mathrm{d}y + \frac{1}{2}f_{yy}\mathrm{d}y^2 + o(\mathrm{d}t) $$ where $f_z = \partial_zf$ denotes a partial derivative. In the present case, we have $x = B_t$ and $y = Y_t$, with $\mathrm{d}B_t^2 \equiv \mathrm{d}t$, $\mathrm{d}Y_t^2 \equiv (B_t+Y_t)^2\mathrm{d}t$ and $\mathrm{d}B_t\mathrm{d}Y_t \equiv (B_t+Y_t)\mathrm{d}t$, hence : $$ \mathrm{d}f(t,B_t,Y_t) = \left(f_t + Y_tf_y + \frac{1}{2}f_{xx} + (B_t+Y_t)f_{xy} + \frac{1}{2}(B_t+Y_t)^2f_{yy}\right)\mathrm{d}t + \left(f_x + (B_t+Y_t)f_y\right)\mathrm{d}B_t $$ For the sake of simplicity, we might consider a transformation with separated variables, i.e. $f(t,x,y) = \alpha(t)\beta(x)\gamma(y)$. In order to integrate this equation explicitly, we now need to choose a transformation such that $Y_t$ disappears completely from the right-hand side. Let's design a suitable $f$.

As each derivative with respect to $y$ comes with a factor $(B_t+Y_t)$, the best choice is $\gamma(y) = y^n$, such that the powers of $Y_t$ remain the same and can cancel each others. More precisely, we must take $n = 1$ in order to avoid some remaining terms like $B_tY_t$.

Then, we ought to take $\beta(x) = e^{-x}$, so that the derivatives with respect to $x$ permit to switch the sign and allow the cancellation of the various terms. For the same reason, as well as for a purpose of ensuring the correct scaling, let's take $\alpha(t) = e^{\lambda t}$.

After inspection, we thus have $f(t,x,y) = e^{\lambda t - x}y$, hence $$ \mathrm{d}f(t,B_t,Y_t) = e^{\lambda t - B_t}\left((\lambda+\frac{1}{2})Y_t - B_t\right)\mathrm{d}t + e^{\lambda t - B_t}B_t\mathrm{d}B_t $$ By selecting $\lambda = -1/2$, we finally end up with $$ \mathrm{d}\left(e^{-t/2-B_t}Y_t\right) = -e^{-t/2-B_t}B_t\mathrm{d}t + e^{-t/2-B_t}B_t\mathrm{d}B_t $$ and, after integration, $$ Y_t = e^{t/2+B_t}\left(Y_0 - \int_0^t e^{-s/2-B_s}B_s \,\mathrm{d}s + \int_0^{B_t} e^{-s/2-B_s}B_s \,\mathrm{d}B_s\right) $$

Answer 3

It is simpler to start with a more appropriate integrating factor. In particular, consider $dA_t = -A_t dB_t$. Then

\begin{align*} d(A_t Y_t) &= A_t dY_t + Y_t dA_t + dA_t dY_t \\ &= A_t [(Y_t + B_t)dB_t + Y_t dt] + Y_t(-A_t dB_t) + (-A_t)(Y_t + B_t)dt \\ &= A_t B_t (dB_t - dt) \end{align*}

It is easy to see that $A_t = A_0 \exp(-B_t - t/2)$ so integrating both sides,

$$ A_0 \exp(-B_t - t/2) Y_t = A_0 Y_0 + \int_0^t A_0 \exp(-B_s - s/2) (dB_s - ds) $$

i.e.

$$ Y_t = Y_0 \exp(B_t + t/2) + \int_0^t \exp(B_t-B_s + (t - s)/2) (dB_s - ds) $$