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4

Just wanted to add to @StackG's great answer using a different approach. Please, double-check my solution as well because I'm not 100% sure. Let $\sigma_t$ be sufficiently regular such that $\dot{\sigma}_t \stackrel{def}{=}\frac{d \sigma}{dt}$ is well defined. Then, Ito's lemma: $$ d(\sigma_t W_t) = \dot{\sigma}_t W_t dt + \sigma_t dW_t $$ which in integral ...


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What a great question! I've had a go at it below, I'd say I'm about 75% sure of the result I've got to but I'd love feedback from others. I'm going to use the definition of the Ito integral, \begin{align} \int^t_0 \sigma_s dW_s = \lim_{n \to \infty} \sum_{i=1}^n \sigma_{t_{i-1}} \bigl( W_{t_i} - W_{t_{i-1}} \bigr) \end{align} where $t_n = t$. Then, using the ...


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