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In general, the implied surface exists, because the market does not believe in the Black-Scholes equation, i.e. they use different volatilies for different strikes and maturities. What does this tell you? Consider a certain maturity. Then, you only have a volatility smile/skew. If you have a particular steep curve on the left, OTM options are very expensive, ...


A call and a put option with the same strike and same expiry should have the same implied volatility by put-call parity indeed. A call and a put are essentially the same when you hedge the initial delta in terms of greek exposures


In swaptions, there is the expiration of the swaption into an underlying swap. When the dealers provide the vol surface, in the first column, they typically put the expiry of the swaption from earliest to farthest. Along the top row, they put maturity of the underlying swap from shortest to farthest. So when the dealers describe the upper left having high ...


I have found the answer to my own question during the last month where the question have been unanswered The main question: Look at page 9 from "the ODE can be rearranged to $f'(y)=....=F(y,f)$". For given $\alpha, \beta, rho, \epsilon$ and $\gamma $ we have all the relevant information to solve this ODE, hence finding $f(y(t))$ which is denoted as $f(y)$ ...

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