New answers tagged greeks
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 ...
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!
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 ...
I think I have figured this out. The key to the understanding is to think of the options' vegas as "key-strike vegas" compared to the var swap/replication portfolio's vega, which is analogous to "key rate durations of a bond portfolio" to the total effective duration of the portfolio.
The variance swap's Vega that is equal to the variance notional refers to the realized variance. The Black-Scholes vega refers to the market implied volatility. Now if you want, you can estimate the realized variance at expiry from the volatility of the options (for instance taking the atm variance arbitrarily), and that's often what people do. But that's ...
most models in financial maths are linear so prices and Greeks just add. This is in particular true of Black--Scholes so Yes. However, once one starts taking into account value adjustments non-linearities appear and it is a lot more complicated.
I think you are answering your own question. Hull states: "When $\Theta$ is large and positive, $\Gamma$ tends to be large and negative and vice versa." In practice, you can expect $r(V-S \Delta)$ to be quite small.
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