Let $B$ and $\tilde{B}$ be independent standard Brownian motions defined on the same probability space with $B_0=\tilde{B}_0=0$. Let $$X_t=e^{B_t}\int_0^te^{-B_s}d\tilde{B}_s,\hspace{1cm}Y_t=\sinh(B_t).$$ I am required to show that $X$ and $Y$ have the same law, with the question giving that I can use that SDEs with Lipschitz coefficients satisfy the uniqueness in law property.

With the implication that the SDE hint gives, I used Ito's fomula (hopefully without making any computational errors) with $f(x,y)=e^xy$ to derive that $X_t$ satisfies the SDE $$dX_t=\frac{1}{2}X_tdt+X_tdB_t+d\tilde{B}_t.$$ Then the coefficients $b(x)=x/2$ and $\sigma(x)=(x,1)$ are Lipschitz and so this SDE will have uniqueness in law. I think then that I am supposed to show that $Y$ satisfies the SDE also. Using Ito's formula with $g(x)=\sinh(x)$ and the fact that $\cosh(B_t)=\sinh(B_t)+e^{-B_t}$, I obtained that $$dY_t=\frac{1}{2}Y_tdt+(Y_t+e^{-B_t})dB_t.$$ But, defining $Z_t$ by $dZ_t=e^{-B_t}dB_t$, we do not have that $Z$ is a BM independent of $B$, so it does not appear that these SDEs are the same.

I would greatly appreciate if someone could point me to where I've gone wrong with my efforts.


1 Answer 1


Recall the identity $\cosh^2x - \sinh^2x =1$. Then, after applying Itô's lemma, the SDE for $Y$ is: $$ \begin{align} dY_t &= \cosh B_t dB_t + \frac{1}{2}\sinh B_t dt \\ &= \sqrt{ 1 + \sinh^2 B_t }dB_t + \frac{1}{2}\sinh B_t dt \\ &= \sqrt{1 + Y_t^2 }dB_t + \frac{1}{2}Y_t dt \end{align}$$

Now, in the SDE for $X$ notice that the term $d\tilde{B}_t + X_t dB_t$ defines a martingale whose quadratic variation is $(1 + X_t^2)dt$; thus, by the martingale representation theorem, we can construct a Brownian motion $W$ such that $$d\tilde{B}_t + X_t dB_t = \sqrt{1 + X_t^2} dW_t$$ Thus, the SDE for $X$ becomes $$dX_t = \sqrt{1+X_t^2} dW_t + \frac{1}{2}X_t dt$$ Since $X$ and $Y$ satisfy the same SDE, if we can prove that the solution to the SDE is unique we may conclude that $X$ and $Y$ have the same law. Uniqueness for the SDE follows by observing that both coefficients $x \mapsto \frac{x}{2}$ and $x \mapsto \sqrt{1 + x^2}$ are Lipschitz (for the second one, observe it is $C^1$ with bounded derivative), so we conclude the processes agree in law.


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