Given martingale $M$, $Y:=\mathcal{E}(X)$, does $\langle M,X \rangle = Y^{-1} \langle M,Y\rangle$ hold?

Question

Given semimartingale $X(t)$, its Doléans-Dade exponential $$Y(t):=\mathcal{E}(X(t))(t):=\exp\left(X(t) - \frac12\langle X,X \rangle(t)\right)$$ and a martingale $M$, is it true that $$\langle M,X \rangle = Y^{-1} \langle M,Y\rangle$$? Here $\langle X,Y\rangle$ denotes the quadratic variation.

I read this in a notes but I'm not sure this is correct or I missed some prerequisite.

My thought goes as this:

First, for quadratic variation we have $d\langle X,Y\rangle = dX\cdot dY$ for any semimartingale $X$ and $Y$. Second, for Doléans-Dade exponential we have $d\mathcal{E}(X) = X dX$ for any semimartingale $X$.

Now $\langle M,X \rangle = Y^{-1} \langle M,Y\rangle$ is equivalent to $Y \langle M,X \rangle = \langle M,Y\rangle$, to get this we need ensure

$$d\left(Y\langle M,X \rangle\right) = d(\langle M,Y\rangle)$$

However,

$$RHS = d(\langle M,Y\rangle) = dM\cdot dY =Y \cdot dM \cdot dX$$

$$LHS = d\left(Y\langle M,X \rangle\right) = dY \cdot \langle M,X \rangle + Y \cdot d\left(\langle M,X \rangle\right) + dY \cdot d\left(\langle M,X \rangle\right) $$ $$= \langle M,X \rangle \cdot X dX + Y \cdot dM \cdot dX + dY \cdot dM \cdot dX $$ $$= \langle M,X \rangle X \cdot dX + Y \cdot dM \cdot dX + Y \cdot dM \cdot dX \cdot dX$$

If we write $M = \mu dW$ and $dX = \theta dt + \sigma dW$, I can see that $Y \cdot dM \cdot dX \cdot dX = Y \cdot \mu dW \cdot \sigma^2 dt = 0 $ , so

$$LHS = \langle M,X \rangle X \cdot dX + Y \cdot dM \cdot dX $$

To make $LHS = RHS$, we need

$$\langle M,X \rangle X \cdot dX = 0$$

But this does not sounds hold for any martingale $M$ and semimartingale $X$.

So am I right that $$\langle M,X \rangle = Y^{-1} \langle M,Y\rangle$$ does not hold, or I missed something?

Answer 1

The identity you have shown seems to have a typo. To see why the current version is not correct, notive that $\left\langle M,X\right\rangle$ has bounded variation, whilst $Y^{-1}\left\langle M, Y\right\rangle$ does not. Instead, what is true is the following.

Recall that if $M$ is a martingale and $H$ is in $L^2(M)$, then the stochastic integral of $H$ against $M$, $H \cdot M$, is the unique $L_0^2(M)$ process such that $$\left\langle H \cdot M , N \right\rangle = H \cdot \left\langle M, N \right\rangle = \int H_s d\left\langle M, N \right\rangle_s$$

for all $L^2$-bounded martingales $N$.

Then, a bit of symbol pushing shows that $$\left\langle Y, M \right\rangle = \left\langle Y \cdot X , M \right\rangle = Y \cdot \left\langle X , M\right\rangle$$