Tag Info

Let's suppose $P$ is total annual deposits made continuously, then the change in value of total deposits $dV_t$ is (assuming no condition on additional deposits) $$dV_t= V_t r dt + P dt$$ where we assumed $r$ is constant. Solving above differential equation, we have: $$V_T = V_0 e^{rT} + \frac{P}{r} (e^{rT} -1)$$ Assuming $t_1$ is the time period at ...
Since $Y=e^{(r-\frac{\sigma^2}{2})\tau + \sigma \sqrt{\tau}Z}$, then \begin{align*} xY > K \Leftrightarrow Z > -d_2, \end{align*} where \begin{align*} d_2 = \frac{\ln \frac{x}{K} + (r-\frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}. \end{align*} Consequently, \begin{align*} e^{-r\tau}\mathbb{E}\big(Y \mathbb{1}_{\{xY >K\}} \big) &= ...