# Find an explicit solution to the SDE $dX_t = \frac{1}{2} X_t B_t^4dt + X_t B_t^2 dB_t$

Problem statement:

Let $$(B_t)_{t\geq0}$$ be a one-dimensional standard Brownian motion. Find an explicit solution to the SDE $$dX_t = \frac{1}{2} X_t B_t^4dt + X_t B_t^2 dB_t$$

with initial condition $$X_0 = 1$$. Hint: Apply the Itô formula to $$Y_t :=\text{log}X_t$$.

Attempted solution:

Let $$Y_t = g(t,X_t) = \text{log}X_t$$ where $$g(t,x) = \text{log} x$$. The partial derivatives are $$\partial_t g = 0, \partial_x g= \frac{1}{x}, \partial_x^2 g = -\frac{1}{x^2}.$$

Itô's formula in differential form then yields:

$$$$\begin{split} dg(t,X_t) &= \partial_x g(t,X_t)dX_t + (\partial_tg(t,X_t) + \frac{1}{2}\partial_x^2g(t,X_t))dt \\ dY_t &= \frac{1}{X_t}(\frac{1}{2} X_t B_t^4dt + X_t B_t^2 dB_t) + (0 - \frac{1}{2} \frac{1}{X_t^2})dt \\ &= \frac{1}{2}(B_t^4-\frac{1}{X_t^2})dt + X_tB_t^2dB_t \end{split}$$$$

I don't know how to continue. It's an old exam question, so I have the solution:

I don't know why $$d\langle X \rangle_t = dt = X_t^2Bt^4dt$$.

Some mathematicians don't like the following abuse of notation, but it has the advantage to be limpid; it goes as follows : $$\mathrm{d}\langle X\rangle_t = \mathrm{d}X_t \cdot \mathrm{d}X_t = \left(\frac{1}{2} X_t B_t^4\mathrm{d}t + X_t B_t^2 \mathrm{d}B_t\right)^2 = X_t^2B_t^4\,\mathrm{d}t + \mathcal{O}(\mathrm{d}t)$$ since $$\mathrm{d}B_t^2 \equiv \mathrm{d}t$$.

• Could you please elaborate $\text{d}B_t^2 \equiv \text{d}t$? Mar 11, 2023 at 11:41

The function $$g$$ depends only on $$X_t$$, we have : $$$$\begin{split} dg(X_t) &= \partial_x g(X_t)dX_t + \frac{1}{2}\partial_x^2g(X_t))dt \\ dY_t &= \frac{1}{X_t}(\frac{1}{2} X_t B_t^4dt + X_t B_t^2 dB_t) -\frac{1}{2} \frac{1}{X_t^2}d\langle X\rangle_t \\ \end{split}$$$$ Since $$dX_t = \underbrace{\frac{1}{2} X_t B_t^4}_{b(X_t)}dt + \underbrace{X_t B_t^2}_{\sigma(X_t)}dB_t$$ then $$d\langle X \rangle_t$$ is given by : $$\implies d\langle X\rangle_t = \sigma^2(X_t)dt=(X_tB_t^2)^2dt=X_t^2B_t^4dt$$ You get $$dY_t=B_t^2dB_t\implies Y_t=\underbrace{Y_0}_{=0}+\int_0^tB_s^2dB_s=\int_0^tB_s^2dB_s$$ Finally, $$X_t=\exp\left(\int_0^tB_s^2dB_s\right)$$

Note : Itô processes of the form $$X_t=X_0+\int_0^t b(s,X_s)ds+\int_0^t \sigma(s,X_s)dB_s$$ have the quadratic variation $$\langle X\rangle_t=\int_0^t \sigma^2(s,X_s)ds$$ For a detailed proof on why it is equal to the previous quantity, check this answer here.

EDIT

A quick explanation on why $$dB^2_t=d\langle B\rangle_t = dt$$:

Since $$B_t=\int_0^tdB_s=\int_0^t \textbf{1}\times dB_s\implies \langle B\rangle_t=\int_0^t \textbf{1}^2 ds=t \implies d\langle B\rangle_t = dt$$

As for your formula : Suppose $$X_t=X_0+\int_0^t b(s,X_s)ds +\int_0^t \sigma(s,X_s)dB_s$$. Then for any twice-differentiable scalar function $$g$$ :

$$$$\begin{split} dg(t,X_t)&=\frac{\partial g}{\partial t}(t,X_t)dt+\frac{\partial g}{\partial x}(t,X_t)dX_t+\frac{1}{2}\frac{\partial^2 g}{\partial^2 t}(t,X_t)d\langle X \rangle_t\\ &=\frac{\partial g}{\partial t}(t,X_t)dt+\frac{\partial g}{\partial x}(t,X_t)\left(b(t,X_t)dt + \sigma(t,X_t)dB_t\right)+\frac{1}{2}\frac{\partial^2 g}{\partial^2 x}(t,X_t)(\sigma^2(t,X_t)dt)\\ &=\left(\frac{\partial g}{\partial t}(t,X_t)+ b(t,X_t)\frac{\partial g}{\partial x}(t,X_t)+\frac{1}{2}\frac{\partial^2 g}{\partial^2 x}(t,X_t)\sigma^2(t,X_t)\right)dt\\ &+\left( \sigma(t,X_t)\frac{\partial g}{\partial x}(t,X_t) \right)dB_t \end{split}$$$$ I guess you can see the term you missed out.

• I think OP's question is exactly 'why $d\langle X,X\rangle_t=\sigma(X_t)^2dt$ holds?' Mar 10, 2023 at 17:36
• In that case, should not the substitution yield $dt = d \langle X \rangle_t / \sigma(X_t)^2$? Mar 11, 2023 at 12:03