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Here's the question:

Let $X(t)$ be a stochastic process such that

$$ dX = a(t)Xdt + b(t)XdW(t) $$

where $a(t)$ and $b(t)$ are $\mathcal{F}_t$-adapted processes with "good" properties. Furthermore $b(t)$ is non-zero almost surely. $W(t)$ is a wiener process with respect to a measure $P$. For which $a(t)$ is $X(t)$ a $P$-martingale?

My attempt:

A martingale cannot have a drift term so it must be of the form $dX = v(t,X)dW(t)$ for some $v$. Therefore $a(t) \equiv 0$.

This was incorrect, but no feedback was given as to why this is not right. Could anyone point out the mistake and/or give me a hint in the right direction?

R. Rayl
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    When the drift is null you can only say that it is a local "martingale", so null dirft is necessary but not sufficient for it to be a true martingale you can check this post: https://math.stackexchange.com/questions/38908/criteria-for-being-a-true-martingale/38947#38947 – TheBridge Feb 28 '20 at 10:05

1 Answers1

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You only find one solution but not a general one.

Let $Y$ be a stochastic process such that $$dY = b(t)dW(t) + \left(a(t) - \frac{1}{2}b^2(t)\right) dt$$ Then $$X(t) = e^{Y(t)}$$ is a solution to your SDE. Additionally we know that $X(t)$ is a so called exponential martingale iff $$Y(t) = M_t - \frac{1}{2}<M>_t$$ for a martingale $M$.

But we have: $$Y(t) = \int_0^t b(u) dW(u) + \int_0^t \left(a(u) - \frac{1}{2}b^2(u)\right) dt$$ so our candidate for $M$ is: $$M_t = \int_0^t b(u) dW(u)$$

What's now $\frac{1}{2}<M>_t$ and which condition is necessary for $a$ that it holds $$\frac{1}{2}<M>_t = \int_0^t \left(\frac{1}{2}b^2(u) - a(u)\right) dt$$ because for all those $a$ your process $X$ is a martingale.

Gono
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