Define $E(X)_t=\exp(X_t-\frac{1}{2} \langle X \rangle _t)$ where $(X_t)$ is an adapted continuous semimartingale. Then it is trivial that this is a continuous semimartingale and is the unique solution to the SDE:
$$dZ_t = Z_t dX_t \text{ with } Z_0 = 1$$
To show this I know $2$ solutions:
$1)$ Apply Ito's formula to $\frac{1}{Z_t}$
$2)$ Consider a solution of the form $\exp(X_t+V_t)$ and apply Ito's formula again to get the result.
However, I came across another solution and I'm not really sure why it is true:
Applying Ito to the semimartingale $X_t −\frac{1}{2} \langle X \rangle_t$
and the exponential $x \rightarrow \exp(x)$ shows
$$dE(X)_t=E(X)_td(X_t −\frac{1}{2} \langle X \rangle_t)+\frac{1}{2}E(X)_td \langle X_t −\frac{1}{2} \langle X \rangle_t \rangle=E(X)_tdX_t$$
I'm not really sure about the last inequality. It says that (with a bit of an imagination instead of the dots)
$$d(\dots)+d \langle \dots \rangle =d(\dots)$$
To be honest, I have no idea why this holds. I'd be really grateful if you could be of an assistance.
Best Answer
Let's consider the first and the second term on the right-hand side of your equation separately:
First term: If $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are semimartingales, then
$$\int_0^T f(t) \, d(X_t-Y_t) = \int_0^T f(t) dX_t - \int_0^T f(t) dY_t$$
for any (nice) mapping $f$, and therefore
$$d(X_t-Y_t) = dX_t - dY_t.$$
This implies that
$$E(X_t)_t d(X_t-\tfrac{1}{2} \langle X \rangle_t) = E(X_t) \, dX_t - \frac{1}{2} E(X_t) \, d\langle X \rangle_t. \tag{1}$$
Second term: By assumption, $X_t = X_0 + M_t+A_t$ is a semimartingale, and this implies that $$Y_t := X_t - \frac{1}{2} \langle X \rangle_t = X_0 + \underbrace{M_t}_{\text{martingale part}} + \underbrace{\left( A_t - \frac{1}{2} \langle X \rangle_t \right)}_{\text{finite variation part}}$$ is also a semimartingale. By the very definition of the square bracket, $N_t = \langle Y \rangle_t$ is the unique finite variation process such that $M_t^2 - N_t$ is a continuous local martingale. Hence, $\langle Y \rangle_t = \langle X \rangle_t$, i.e.
$$\langle X- \tfrac{1}{2} \langle X \rangle \rangle_t = \langle X \rangle_t.$$
Consequently, we get for the second term in your equation that
$$\frac{1}{2} E(X_t) \, d\langle X- \tfrac{1}{2} \langle X \rangle \rangle_t = \frac{1}{2} E(X_t) \, d\langle X \rangle_t. \tag{2}$$
Combining $(1)$ and $(2)$ we find that
$$E(X_t)_t d(X_t-\tfrac{1}{2} \langle X \rangle_t) + \frac{1}{2} E(X_t) \, d\langle X- \tfrac{1}{2} \langle X \rangle \rangle_t = E(X_t) \, dX_t.$$