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## New answers tagged delta-hedging

4

First we write dynamic of ${{x}_{t}}=\ln ({{S}_{t}})$ \begin{align} & d{{x}_{t}}=({{r}_{t}}-\delta -\frac{1}{2}\sigma _{t}^{2})t+{{\sigma }_{t}}d{{W}_{1}}(t) \\ & d{{\sigma }_{t}}=a({{\sigma }_{t}},t)dt+b({{\sigma }_{t}},t)d{{W}_{2}}(t) \\ & d{{r}_{t}}=\alpha ({{r}_{t}},t)dt+\beta ({{r}_{t}},t)d{{W}_{3}}(t) \\ \end{align} Let \begin{align} ...

0

The PDE only holds in t he continuation region, in the excerise region, P is just the pay off of the function. Let $\tau$ be the first time you enter the stopping region, then by the martingale property of the option price up to the first stopping time ...

2

Assuming zero interest, the put option has the price \begin{align*} KN(-d_2)-S_0N(-d_1), \end{align*} and delta $-N(-d_1)$. When $N(-d_1)$ units of stocks are shorted and invested in bonds, the total value in bonds is $KN(-d_2)$, which is indeed greater than the option price. However, as you have shorted $N(-d_1)$ units of stocks, your portfolio value is ...

1

A perfectly hedged portfolio should not make any profits different from the risk free interest rate. However, you won't be able to hedge perfectly in the real world. Delta hedging for example requires continous trading and adjusting (this is one way to derive the black -scholes formula: thex hedge the stock perfectly and therefore obtain a risk -free rate ...

3

An Investment Bank earns a profit by selling you an option at a slightly higher price than the theoretical price, or buying it back from you at a slightly lower price. They call this "earning a spread". Then they hedge the option, so as not to make any [further] gains or losses on it (other than the risk free rate). Another way they could earn a profit is ...

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