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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 $X_t = 1+\int_0^t Z_s dB_s$.

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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: $$X_t = \exp\left(\int_0^t \frac{Z_s}{X_s}dB_s - \frac 12 \int_0^t\left(\frac{Z_s}{X_s}\right)^2ds\right)$$

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  • $\begingroup$ How is taking the logarithm of $x$ and taking the exponential of that result any kind of proof? $\endgroup$
    – CCZ23
    Jun 15, 2020 at 17:26
  • $\begingroup$ Oh, sorry, I thought you found $dY_t$ differently. I will edit the answer in a moment. $\endgroup$ Jun 15, 2020 at 17:27

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