# Tag Info

## Hot answers tagged option-pricing

3

Simply put, no. Vega depends on a variety of factors (including the level/price of the underlying asset). However, vomma/volga/vega convexity (whatever you want to call dVega/dIV) is always positive. So as IV increases, the vega of an option increases - I think this might have been what you were getting at. It's important to understand that IV is an input ...

3

Your characterisation is correct but incomplete. 1) The most important part of Black-Scholes is not the model but the more general framework of dynamic hedging: you can replicate your payoff by continuously trading the underlying and the amount (delta) you should hold is the derivative of the current premium with respect to the current spot. This is a much ...

2

Might not be the answer you're looking for, but just some thoughts that immediately come to mind... As you've alluded to, the BS model is much more than just a tool to price options. But there's no need to get into this here. Prior to publication of the BS model, option prices already traded at more-or-less the price implied by B.S. (part of the ...

2

IV is one of the inputs for your option pricing model, vega measures the actual impact (e.g. in Dollars, Euros...) of any change in IV. Intuitively IV is the price of the option while vega is the sensitivity to IV. Bottom line: There is a clear distinction!

2

Because there are several non-linearities involved this depends very much on where you are concerning the level of volatility and time to expiry. But I think what you really want is to get some feel for the sensitivities involved, right? With the following demonstration you can play with all kinds of combinations of all parameters to get some intuition for ...

1

By no arbitrage, $S(0)=(Su q+Sd (1-q))/(1+r)$ and $C(0)=((Su-k)^+ q+(Sd-k)^+(1-q))/(1+r)$. Simplifying and rearranging (and assuming $Su>k$), $$\left[\begin{array}{c} S(0) \\ C(0) \end{array} \right]=\frac{1}{1+r}\left[\begin{array}{cc} Su & Sd \\ (Su-k) & 0 \end{array} \right]\left[\begin{array}{c} q\\ 1-q \end{array} \right]$$ Clearly, ...

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