4

Using Itô's Lemma, notice that: $$d(tS_t)=tdS_t+S_tdt=dX_t+S_tdt$$ Hence: $$X_t=tS_t-\int S_udu$$ Using independence of Brownian increments, $E(S_udW_u)=E(S_u)E(dW_u)=0$, and the chain rule for the 4th step: $$\begin{align} E(X_t)&=E\left(\int dX_u\right) \\ &=\int uE(dS_u) \\ &=\int u\mu E(S_u)du \\ &=S_0\int u\mu e^{\mu u}du \\ &=S_0\...


1

$$u(x,0)=e^{\frac{k-1}{2}x}v(x,0)$$ and $$v(x,0)=max(e^{x}-1,0)$$ Hence $$u(x,0)=max(e^{\frac{k+1}{2}x}-e^{\frac{k-1}{2}x},0)$$


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