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1

It is because theta is not premium / days to expiration. Theta is a "local" decay, measure of current rate of option decay, which is not assumed to stay constant. In the example you provided, theta will be closer to zero (decay rate will slow down) as you approach expiration.


2

The price of the April option will be more than $5.00, correct. How much more depends on the implied volatility ($\sigma$) of the option and the interest rates ($r$). The higher $\sigma$ and $r$ are, the higher the time value of money and the value of the April option. I highly recommend playing around with this calculator to gain an intuitive ...


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 ...


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, ...


0

Gatheral (Amazon) has a quite extensive discussion on that, and dives into calibration issues. In summary, what you describe appears to be less of a modeling issue, and more of a calibration problem. This is primarily because the model functions (such as the Heston model) are not by nature convex in their input parameters. This is simply result of the fact ...


0

Your question concerns the term structure of volatilty. In this case, USO's vol term structure is inverted (downward sloping) since far-dated IV is less than current. Please keep in mind that the market determines the shape of the term structure and it can and will change over time. Currently, USO has seen a steep correction and there is some geopolitical ...


0

Yes, there is a unique time homogeneous local vol model. This is proven in http://www.sciencedirect.com/science/article/pii/S0304414912002487. There is a slight generalization required that if the option-implied density is zero somewhere, the corresponding local vol is infinite in that region, giving a "gap diffusion". No, there is no nice formula for the ...


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 ...


0

I tried to use the BPV/delta relashionship $\Delta = \frac{ \frac{\partial Z}{\partial r}-\frac{\partial P(t,T)}{\partial r}\frac{Z}{P(t,T)} } {\frac{\partial P(t,S)}{\partial r}}$ but it doesn't work as well.


0

No. In practice the local volatility model has a finite number of slices, so a single slice works as well. Now the problem is : how to compute the time derivative ? Well without adding any information you know that $$ C(0,K) = (S_0-K)_+ $$ so you could try $$ C_\tau = \frac{C(\tau,K)-C(0,K)}{\tau} $$ but it is a very crude approximation. What you may want ...


1

If someone wants simple intuition, here is what happened to the drift. It did go into the formula, believe it or not, but it came into the B-S math sort of in two ways so it cancelled out in the end. It disappears because Black-Scholes assumes people may have different preferences for risk, but at least everyone is consistent on their own preference. ...


0

To me this aims at computing a daily implied volatility surface. For some stocks/indices you may have either vanillas options or american options quoted in the market. If your implied volatility is computed from hybrid vanilla/american call/put options then your implied volatility computation methodology should be as close as possible. You should not ...


0

If you are using a model such as BS where the distribution of $S_{j}(T)/S_{j}(0)$ does not depend on $S_{j}(0)$ it really makes very little difference. Just take the initial stock prices to be $1$ and away you go.



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