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Since $W_{2t}-W_{t}$ is independent of $W_t$ and has the same law as $W_{2t-t}=W_t$ we only have to compute $$P(X(X+Y)<0)$$ where $(X,Y)$ follows a bivariate normal distribution (with zero correlation). From there you can split the probability in two cases : either $X<0$ and $X+Y>0$ or the opposite. The two events have the same probability since ...


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Your posting has an error, that is, the identity should be \begin{align*} -P(0, T) \mathbb{P}(S_T > K) = \frac{\partial C}{\partial K}. \end{align*} The derivation below is based on this assumption. We denote by $f(x)$ the density function for $S_T$. Then \begin{align*} \mathbb{P}(S_T > K) = \int_K^{\infty} f(x) dx, \end{align*} and \begin{align*} C(K, ...


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Your decomposition is correct. I will show here the computation for one term: \begin{align*} P(W_t < 0, W_{2t} >0) &= E(W_t < 0, W_{2t}-W_t > -W_t)\\ &= E\Big(E\big(\mathbb{1}_{\{W_t < 0\}}\mathbb{1}_{\{W_{2t}-W_t > -W_t\}}\mid W_t\big)\Big)\\ &= E\Big(\mathbb{1}_{\{W_t < 0\}} \Phi\big(W_t/\sqrt{t}\big)\Big) \\ ...



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