# Use Ito formula to solve SDE

If $$Z_t$$ is a progressive process and $$X_t$$ is defined by $$X_t = 1+\int_0^t Z_s dB_s$$, then use Ito's formula to $$Y_t=ln(X_t)$$, to show that $$X_t = \exp\Bigl(\int_0^t \frac{Z_s}{X_s} dB_s - \frac{1}{2}\int_0^t \bigl(\frac{Z_s}{X_s}\bigr)^2 ds \Bigr)$$

Consider $$Y_t = ln(X_t) = \int_0^t \frac{Z_s}{X_s} dB_s - \frac{1}{2}\int_0^t \bigl(\frac{Z_s}{X_s}\bigr)^2 ds$$.

So $$dY_t = \frac{Z_t}{X_t} dB_s - \frac{1}{2}\bigl(\frac{Z_t}{X_t}\bigr)^2$$ and $$d\langle Y \rangle_t = (\frac{Z_S}{X_s})^2.$$ Then by Ito's formula:

$$X= df(Y_t) = f'dY_t+\frac{1}{2}f''d\langle Y\rangle_t$$ $$= \frac{1}{Y_t}dY_t - \frac{1}{2Y_t^2}\left(\frac{Z_S}{X_s}\right)^2dB_s$$

This is clearly completely wrong, how should I have approached this instead...? My end goal is to get this in the form $$X_t = 1+\int_0^t Z_s dB_s$$.

From Ito's formula applied to $$Y_t := \ln(X_t)$$, we have
\begin{align*} dY_t &= \frac{1}{X_t}dX_t - \frac 12 \frac{1}{X_t^2}d\langle X,X\rangle_t \\ &= \frac{Z_t}{X_t} dB_t - \frac 12 \left( \frac{Z_t}{X_t} \right)^2dt \end{align*}
so $$\ln(X_t) = \ln(X_0) + \int_0^t \frac{Z_s}{X_s}dB_s - \frac 12 \int_0^t(\frac{Z_s}{X_s})^2ds = \int_0^t \frac{Z_s}{X_s}dB_s - \frac 12 \int_0^t(\frac{Z_s}{X_s})^2ds.$$ Now just exponentiate both sides: $$X_t = \exp\left(\int_0^t \frac{Z_s}{X_s}dB_s - \frac 12 \int_0^t\left(\frac{Z_s}{X_s}\right)^2ds\right)$$
• How is taking the logarithm of $x$ and taking the exponential of that result any kind of proof? Jun 15, 2020 at 17:26
• Oh, sorry, I thought you found $dY_t$ differently. I will edit the answer in a moment. Jun 15, 2020 at 17:27