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I am learning some basic stochastic calculus and came across the following exercise:

Consider a local martingale $M$ with continuous trajectories. Let $Z_t = \exp(M_t −0.5[M]_t)$. Show that Z satisfies the equation $$Z_t = Z_0 +\int_0^tZ_sdMs$$ Is $Z$ a local martingale? Compute $Z$ for the case where $M_t = σB_t$ for a standard Brownian motion $B$.

What confuses me, is the presence of quadratic variation in the function of a martingale. I've tried to apply standard Ito's formula:

$f(X) = f(X_0)+\int f^\prime(X)\,dX + \frac{1}{2}\int f^{\prime\prime}(X)\,d[X]$

But I am not sure how to treat the $[M]_t$ element in the function. Could someone help? Thanks!

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1 Answer 1

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I assume that the Ito formula that you wrote refers to a semi-martingale $X_t$. A semi-martingale is the sum of a local martingale and a process with bounded variation: $X_t = M_t + A_t$. But the process with bounded variation is irrelevant in the computation of the quadratic variation, namely $[X]_t = [M]_t$. So in your case, if $$ X_t = \sigma B_t - \frac12 \sigma^2 t, $$ then $$ [X]_t = \sigma^2 [B]_t = \sigma^2 t. $$

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