$n$-th power of stochastic exponential

Question

Let $X$ be a continuous semimartingale with $X_0 = 0$ and $n \geq 1$. Prove that $$ \mathcal{E}(X)_t^n = \mathcal{E}(Y)_t $$ where $Y_t := nX_t + \frac{1}{2}n(n-1)\langle X \rangle_t$

My attempt

If we take $f(\mathcal{E}(X)) = \mathcal{E}(X)^n$, we can apply Ito's formula to $f$ to yield

$$ f(\mathcal{E}(X)_t) - f(\mathcal{E}(X)_0) = \mathcal{E}(X)_t^n - 1 = \int_o^t \frac{\partial f(\mathcal{E}(X)_s)}{\partial(\mathcal{E}(X)_s)} d \mathcal{E}(X)_s + \frac{1}{2} \int_0^t \frac{\partial^2 f(\mathcal{E}(X)_s)}{\partial^2(\mathcal{E}(X)_s)} d \langle \mathcal{E}(X) \rangle_s = \int_0^t n \mathcal{E}_s^{n-1} d \mathcal{E}(X)_s + \frac{1}{2} \int_0^t n(n-1) \mathcal{E}(X)_s^{n-2} d\langle \mathcal{E}(X) \rangle_s$$

At this stage things look promising since the integer coefficients look like they are of the same form as what has to be proved. I also make use of one final trick : $d\mathcal{E}(X) = \mathcal{E}(X) d X$ as $\mathcal{E}(X)$ is a solution to the simple "exponential SDE", so I finally get

$$ \cdots = \int_0^t n \mathcal{E}_s^{n} d X_s + \frac{1}{2} \int_0^t n(n-1) \mathcal{E}(X)_s^{n-2} d \langle \mathcal{E}(X) \rangle_s$$

But this is where I am stuck. Is this the right approach, or how do I proceed from here? Any help would be greatly appreciated.

Update 1

I made some more progress, using the identity $$ \left\langle \int_0^t \xi d X \right\rangle = \int_0^t \xi^2 d \langle X \rangle $$ I can simplify the above to $$ \frac{1}{2} \int_0^t n(n-1) \mathcal{E}(X)_s^{n-2} d \langle \mathcal{E}(X) \rangle_s = \left\langle \frac{1}{2} \int_0^t n(n-1) \mathcal{E}^{\frac{n}{2} - 1}_s d \mathcal{E}(X)_s\right\rangle = \left\langle \frac{1}{2} \int_0^t n(n-1) \mathcal{E}^{\frac{n}{2}}_s d X_s \right\rangle = \frac{1}{2} \int_0^t n(n-1) \mathcal{E}^{n}_s d \langle X_s\rangle$$

At this stage I know I am very close, but I still don't see how to do it...

Answer 1

Recall that for a continuous semimartingale $Z_t$ that starts at zero, its Doléans-Dade exponential is given by $$\mathcal{E}(Z)_t = \exp \left( Z_t - \frac{1}{2}\left\langle Z \right\rangle_t \right)$$

Thus, for $X$, we have $$\mathcal{E}(X)_t^n = \exp \left( nX_t - \frac{1}{2}n\left\langle X\right\rangle_t \right)$$

Now, note that $\left\langle Y \right\rangle_t = \left\langle nX \right\rangle_t = n^2\left\langle X \right\rangle_t$, since the second term in the definition of $Y$ is of locally bounded variation. Thus, $$ \begin{align*} \mathcal{E}(Y)_t &= \exp \left( nX_t + \frac{1}{2}n(n-1)\left\langle X \right\rangle_t - \frac{1}{2}n^2\left\langle X \right\rangle_t\right) \\ &= \exp \left( nX_t - \frac{1}{2}n\left\langle X \right\rangle_t\right) \\ &= \mathcal{E}(X)_t^n \end{align*}$$ As desired.

Answer 2

The following is true: if $X$ does not jump by -1 and $\alpha\neq 0$ is an arbitrary complex number, then $$\mathcal{E}(X)^\alpha = \mathcal{E}(Y), $$ where $$Y = ((1+\mathrm{id})^\alpha-1)\circ X.$$ Here $f\circ X$ denotes the $f$-variation of $X$.

The function $f = (1+\mathrm{id})^\alpha-1$ is analytic at zero, hence for $X$ continuous one has by Émery's formula $$ Y = f'(0)\cdot X + \frac{1}{2}f''(0)\cdot [X,X].$$ For details see https://arxiv.org/abs/2006.11914.

You can also do this by hand, write down the Ito formula for $Z=\mathcal{E}(X)^\alpha$ and observe that it gives $dZ_t = Z_t dY_t$ with $Z_0=1$. Hence $Z=\mathcal{E}(Y)$.