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Under the condition $r=\frac{\sigma^2}{2}$, it is true that $S_t = S_0e^{\sigma W_t}$. Since \begin{align*} E\Big( S_T - \min_{0 \le t \le T} S_t\Big) = E\big( S_T\big) - E\Big(\min_{0 \le t \le T} S_t\Big), \end{align*} what you need is the expectation $E\big(\min_{0 \le t \le T} S_t\big)$. Note that \begin{align*} \min_{0 \le t \le T} S_t = S_0e^{\sigma \...


To add a bit to Will Gu's answer: Compute $\mathbb{E} \left[ \left. S_T \right| S_T > K \right]$ using the fact that $S_T$ is lognormally distributed with mean $ln(S_0) + (r - \sigma^2/2)T$ and variance $\sigma^2 T$. Then find the pdf of the lognormal distribution on, e.g., Wikipedia, and compute the expectation integral. You may find the following ...

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