# How to solve $dX_t = X_t(\sigma_t dW_t + \mu_t dt)$?

Solve the SDE $$dX_t = X_t(\sigma_t dW_t + \mu_t dt)$$ where $\sigma_t$,$\mu_t$ are deterministic.

Attempted solution

We have $$dX_t = X_t(\sigma_t dW_t + \mu_t dt)$$ Let $f(x) = \log X$, applying Ito formula we have

$$d \log(X_t) = \sigma_t dW_t X_t \frac{1}{X_t} + \mu_t dt X_t \frac{1}{X_t} + \frac{1}{2}\sigma^2 (-\frac{1}{X_t^2}$$

Cancelling terms we have $$d \log(X_t) = \sigma_t dW_t + \mu_t dt - \frac{1}{2}\sigma^2\frac{1}{X_t^2}$$

This is where I am lost I am pretty sure this is how we solve for this SDE but the $\frac{1}{X_t^2}$ term is throwing me off.

– user16651
Nov 23, 2016 at 6:01
• I downvoted this question as it a) is one of the first examples for an application of the Ito formula in any continuous time finance textbook and b) has been answered in a similar form multiple times already: quant.stackexchange.com/questions/14109, quant.stackexchange.com/questions/28272, quant.stackexchange.com/questions/1345 (there are more). Nov 23, 2016 at 9:06
• (-1) for the same reasons. May I encourage you to first search amongst existing questions before asking one @MorganWeis. Nov 23, 2016 at 9:10
• @LocalVolatility In Mathematics Stack Exchange and other communities the simple question such as this question was asked over and over again but it was not closed or donwn-voted because wolfy has shown his\her attempt. Please note the his\her mistake: $d[X_t,X_t]=\sigma^2 dt$. This mistake is prevalent. And my last question: Why Quantitative Finance belongs the beta sites?
– user16651
Nov 23, 2016 at 9:39
• I got it guys, sorry should of looked at existing questions before posting this, won't happen again Nov 23, 2016 at 14:50

let $$dY_t=\mu(t,Y_t)dt+\sigma(t,Y_t)dW_t$$ and $f\in \mathbb{C^2}$.By application of Ito's lemma, we have $$df(Y_t)=\frac{\partial f}{\partial y}dY_t+\frac{1}{2}\frac{\partial^2 f}{\partial y^2}d[Y_t,Y_t]$$ where $$\color{red}{d[Y_t,Y_t]=\sigma^2(t,Y_t)dt}$$ We have $$dX_t=\mu_t X_tdt+\sigma_t X_t dW_t$$ thus $$d \ln(X_t)=\frac{1}{X_t}(\mu_t X_tdt+\sigma_t X_t dW_t)+\frac{1}{2}\left(\frac{-1}{X_t^2}\right)(\sigma_t X_t)^2 dt$$ therefore $$d \ln(X_t)=\left(\mu_t-\frac{1}{2}\sigma_t^2 \right)dt+\sigma_t dW_t$$ by integration on $[0,t]$ we have $$\ln\left(\frac{X_t}{X_0}\right)=\int_{0}^{t}\left(\mu_s-\frac{1}{2}\sigma_s^2 \right)ds+\int_{0}^{t}\sigma_s dW_s$$ in other words $$X_t=x \exp\left(\int_{0}^{t}\left(\mu_s-\frac{1}{2}\sigma_s^2 \right)ds+\int_{0}^{t}\sigma_s dW_s\right)$$ where $X_0=x$
• Yeah we replied simultaneously, by the way, I think it should be $d[X_t,X_t]=\sigma^2 X_t^2 dt$, is it what you wanted to write in red?