I am trying to understand a Lemma in Olav Kallenberg's book "Foundations of Modern Probability" (Lemma 26.19 in the second edition or 23.19 in the first edition). 


The part of the lemma that I do not understand goes as follows. Let $M$ be a (not necessarily continuous) local martingale, $a \in \mathbb{R}$ and define $X = e^{M - a[M]}$. The author then claims that the following is a consequence of Itô's formula:
$$X^{-1}\_- \cdot X = M - (a-1/2)[M]^c + \sum \left\{e^{\triangle M - a(\triangle M)^2 }- 1 - \triangle M\right\}.$$
In trying to prove this, I proceeded as follows. Putting $f(x) = \log(x)$, we have $f(X) = M - a[M]$ and then, by Itô's formula,
$$\begin{aligned}
M - a[M] = f(X) &= f'(X\_-) \cdot X + \frac{1}{2} f''(X\_-) \cdot [X]^c + \sum\left\{\triangle f(X) - f'(X\_-)\triangle X\right\} \\
&= X^{-1}\_- \cdot X -\frac{1}{2} X^{-2}\_- \cdot [X]^c + \sum \left\{\triangle (M - a[M]) - X^{-1}\_- \triangle X \right\}
\end{aligned}$$
or, rearranging,
$$X^{-1}\_-\cdot X = M - a[M] + \frac{1}{2} X^{-2}\_- \cdot [X]^c - \sum \triangle M + a \sum \triangle [M] + \sum X^{-1}\_- \triangle X.$$
Now, I noticed the following:


$\bullet\; [M] - \sum \triangle [M] = [M]^c.$


$\bullet\;$ $\sum X^{-1}\_- \triangle X = \sum \left( e^{\triangle M - a \triangle [M]} - 1\right) = \sum \left( e^{\triangle M - a (\triangle M)^2} - 1\right)$.


Using these, the above becomes
$$X^{-1}\_-\cdot X = M - a[M]^c + \frac{1}{2} X^{-2}\_- \cdot [X]^c + \sum \left\{e^{\triangle M - a(\triangle M)^2 }- 1 - \triangle M\right\}.$$
So what seems to be missing is the equality
$$X^{-2}\_- \cdot [X]^c = [M]^c.$$
Can anyone clarify why this is true or otherwise where I have gone wrong?



An identity for the exponential of a martingale
I think your last equality is true: in your notation,
$[M]^c=[M-a[M]]^c=[f(X)]^c=[f´(X\_{-})\cdot X]^c=[(X\_{-})^{-1} \cdot X]^c=(X\_{-})^{-2} \cdot [X]^c$ using your Ito expansion of $f(X)$. Hope it helps.


