# On the stochastic exponential in stochastic calculus

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.

• What is your definition of the square bracket $\langle \cdots \rangle$?
– saz
Commented Oct 28, 2018 at 6:04
• If $X_t=X_0+M_t+A_t$, then <$X$> $_t=$ < $M$ > $_t$ being the unique finite variation process, such that $M_t^2-$ < $M$ > $_t$ is a continuous local martingale
– asdf
Commented Oct 28, 2018 at 7:33

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

$$E(X_t)_t d(X_t-\tfrac{1}{2} \langle X \rangle_t) + \frac{1}{2} E(X_t) \, d\langle X- \tfrac{1}{2} \langle X \rangle \rangle_t = E(X_t) \, dX_t.$$

• Nice! Completely forgot about the usage of the definition of quadratic variation for the semimartingale until you asked for it. Thanks!
– asdf
Commented Oct 28, 2018 at 8:50
• @asdf You are welcome.
– saz
Commented Oct 28, 2018 at 9:25