Let $X$ be a strictly positive continuous semimartingale with $X_0 = 1$ and define the process $Y$ by $$ Y_t = \int_0^t \frac{1}{X} dX - \frac12 \int_0^t \frac{1}{X^2} d \langle X \rangle. $$ Let the process $Z$ be given by $Z_t = e^{Y_t}$.

I now want to compute $d Z_t$ and show that $Z=X$. To compute $dZ_t$, I noticed that $Y$ follows by defining $Y_t = f(X_t) = \log(X_t)$ then by Ito's formula this gives us precisely that $$ dY_t = d\log(X_t) = f'(X_t) d X_t + \frac12 f''(X_t)d \langle X_t \rangle = \frac{1}{X_t} d X_t - \frac12 \frac{1}{X_t^2}d \langle X_t \rangle,$$ which is the same as above but in differential notation. Then to compute $dZ_t$ I defined $g(Y_t) := e^{Y_t} = Z_t$ such that \begin{align*} d Z_t &= d \exp(Y_t) = g'(Y_t)d Y_t + \frac12 g''(Y_t) d \langle Y \rangle_t \\ &= g'(f(X_t))d Y_t + \frac12 g''(f(X_t)) d \langle Y \rangle_t \\ &= f'(X_t)g(f(X_t))d Y_t + \frac12 (f'(X_t))^2g(f(X_t)) d \langle Y \rangle_t \\ &= dY_t + \frac12 \frac{1}{X_t} d \langle Y \rangle_t. \end{align*} I now wonder is this a correct derivation and how could I prove that $Z = X$? I guess I could prove that $dZ_t = dX_t$ but writing out $dY_t$ does not help me to obtain this. Thanks for any help.

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    $\begingroup$ No, $Y$ is not defined by $Y=\log X$ but by the integrated form equivalent to $dY=X^{-1}dX-\frac12X^{-2}d\langle X\rangle$, and no, one is not supposed to apply Itô to $\log X$ to solve this, but to apply Itô to $Z=e^Y$ and to deduce from that that $Z=X$. $\endgroup$ – Did Apr 8 '15 at 16:12
  • $\begingroup$ Thanks for the comment. I guess I could define $f(Y_t) = e^{Y_t}$ then $dZ_t = df(Y_t) = f'(Y_t) dY_t + \frac12 f''(Y_t)d\langle Y \rangle_t$ which would result I guess result in $Y_t'f(Y_t) + \frac12 (Y_t')^2 f(Y_t) d\langle Y \rangle_t$, but what is $Y_t'$ in this case? I'm a bit confused. $\endgroup$ – user155670 Apr 8 '15 at 16:27
  • $\begingroup$ Of course $Y_t'=1$, thanks got it. $\endgroup$ – user155670 Apr 8 '15 at 16:36
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    $\begingroup$ No $Y'_t$ is not $1$, not at all. $\endgroup$ – Did Apr 8 '15 at 16:38
  • $\begingroup$ Notation-ally, I guess indeed it is incorrect. But as defined above $f'(Y_t)=f(Y_t)$ as it is an exponential, right? $\endgroup$ – user155670 Apr 8 '15 at 16:42

As @Did points out, by definition of $Y_t$ in terms of stochastic integrals,

$$ \text dY_t = \frac{1}{X_t}\text dX_t-\frac{1}{2}\frac{1}{X_t^2}\text d\langle X\rangle_t\tag{1} $$ And $$ Z_t^{-1}\text dZ_t = Z_t^{-1}\text de^{Y_t}=Z_t^{-1}e^{Y_t}(\text dY_t+\frac{1}{2}\text d\langle Y\rangle_t) = \text dY_t+\frac{1}{2}\text d\langle Y\rangle_t\tag{2} $$

Using (1) in (2), conclude.

  • $\begingroup$ I suppose that the second equality in (2) follows from Ito's formula. Then by rewriting it is easily seen that $dY_t = \frac{1}{Z_t} dZ_t - \frac12 d \langle Y \rangle_t$. To prove that $Z=X$ we should have that $d\langle Y \rangle_t = \frac{1}{Z_t^2}d\langle Z \rangle_t$, is there any easy way to show this? $\endgroup$ – user155670 Apr 8 '15 at 16:55
  • $\begingroup$ @Rodel, Yes, the second equality follows from Ito applied to $f(Y_t)$, where $f(y)=e^{y}$. To prove $Z_t\stackrel{\text{law}}{=}X_t$, read off both $\text dY_t$ and $\text d\langle Y \rangle_t$ (in terms of $X_t$) directly from (1), and plug these into (2): you should get something very nice and obvious when you do ;) $\endgroup$ – ki3i Apr 8 '15 at 18:28
  • $\begingroup$ I'm sorry but how do I from (1) get $d \langle Y \rangle_t$ in terms of $X_t$? $\endgroup$ – user155670 Apr 8 '15 at 19:02
  • $\begingroup$ @Rodel, Formally, $$d\langle Y \rangle_t = dY_t\cdot dY_t = \frac{1}{X_t^2}d\langle X \rangle_t$$ The "$d\langle X \rangle_t$" term, and associated products of this with the "$dX_t$" term, do not contribute. $\endgroup$ – ki3i Apr 8 '15 at 19:13
  • $\begingroup$ @Rodel, You are welcome. $\endgroup$ – ki3i Apr 8 '15 at 19:33

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