Find SDE satisfied by transformation of solution to a different SDE

Suppose that $$X_t$$ satisfies $$dX_t = Y_t dt + Y_t dW_t$$, where $$dY_t = Y_t dW_t$$. What can we say about the SDE that $$\ln{(X_t)} + \frac{t}{2}$$ solves? I'm not sure how to use the SDE that $$X_t$$ solves to find the SDE that $$\ln{(X_t)} + \frac{t}{2}$$ solves. In general, is there a way to find the SDE that the transformation of the process solves?

I know that $$dY_t = Y_t dW_t$$ is a GBM with no drift, so that means $$Y_t = Y_0 e^{-\frac{1}{2}t + W_t}$$. Then we have $$dX_t = Y_0 e^{-\frac{1}{2}t + W_t} (dt + dW_t)$$. Is there any easy way to solve this? I'm not sure how to efficiently apply Ito's lemma in this case. Once a solution to $$X_t$$ is found, would we then apply that transformation and find the SDE it solves? Am I overcomplicating things? I'm not sure if there is a better way, and I'm still stuck trying to solve for $$X_t$$. Any help is appreciated!

Let $$Z_t=\frac{t}{2}+\ln{(X_t)}$$ and denote quadratic variation by $$\langle\cdot\rangle$$. By Ito, $$dZ_t=d\left(\frac{t}{2}+\ln{(X_t)}\right)=\frac{dt}{2}+\frac{dX_t}{X_t}-\frac{\langle X_t\rangle}{X_t^2}\,dW_t$$ Now note that $$\begin{gather*} X_t=e^{Z_t-\frac{t}{2}} \\ dX_t=Y_t(dt+dW_t) \\ \langle X_t\rangle=Y_t \end{gather*}$$ Thus \begin{align*} dZ_t&=\frac{dt}{2}+Y_te^{\frac{1}{2}t-Z_t}(dt+dW_t)-Y_te^{t-2Z_t}\,dW_t \\ &=\left(\frac{1}{2}+Y_te^{\frac{1}{2}t-Z_t}\right)dt+Y_te^{\frac{1}{2}t-Z_t}\left(1-e^{\frac{1}{2}t-Z_t}\right)\,dW_t \end{align*}
You've already noticed that $$Y_t=Y_0e^{W_t-\frac{t}{2}}$$, so we can substitute that to get a slightly nicer answer: $$dZ_t=\left(\frac{1}{2}+Y_0e^{W_t-Z_t}\right)dt+e^{W_t-Z_t}\left(1-e^{\frac{1}{2}t-Z_t}\right)\,dW_t$$
• Thank you so much for your answer! Is there now a way to solve this SDE given that $Y_t = Y_0 e^{-\frac{1}{2}t + W_t}$? Dec 15, 2021 at 3:01
• @cosmicflair: No, I don't see a way. Not every SDE has an expressible solution. I did miss that substituting $Y_t$ gives a nicer final result, though, so I edited that into my answer. Dec 15, 2021 at 5:57
• why do you say is $\langle X \rangle_t = Y_t$? Isn't it $\langle X \rangle_t = \int_0^t Y_s^2 ds$? Dec 15, 2021 at 11:43
• @Sebastian: No, that's (roughly) the $L^2$ norm. The quadratic variation is $\sup_{\{t_j\}_j\vdash[0,t]}{\sum_j{(X_{t_{j+1}}-X_{t_j})^2}}$. Dec 15, 2021 at 23:51
• @JacobManaker: In your last comment, you provided the definition of quadratic variation which by the way you don't use in your answer. There is also a characterization that says that the quadratic variation process of the process $X$ is the unique process such that $X^2 - \langle X \rangle$ is a martingale. Using this characterization and the Ito isometry is easy to show that $\langle X \rangle_t = \int_0^t Y_s^2ds.$