# Tag Info

## Hot answers tagged skew

11

Great question. Let me try provide some insights and thoughts regarding your points and questions raised. It may not be a full answer but hopefully it helps to connect the contents in the paper/book with some trading intuition: From a theoretical perspective, I don't see any mistake in your thinking regarding skew decay but two questions arise on my end: ...

8

VG belongs in the family of variance-mean mixture models. Given a horizon $T$ the distribution of log-returns $f$ is a mixture of Gaussians $f_G$ with randomised mean and variance. The randomisation density is $g$ and its mean and variance increase with $T$. For the VG process this randomised factor is Gamma-distributed. More concretely, denote with $f_G(x;\... 8 If you take Quantuple's stuff a little further, you can really see whether you're long skew. You can pretty easily see the dependence on convexity too (though it should be obvious that you're long convexity). So first off, we need some smile parametrisation that lets us easily control convexity and skew. I just went with a made up one; $$\mathrm{convexity} ... 7 As I've mentioned in a comment, it would be wrong to think that entering a variance swap specifically amounts to being "long skew". What you can say however is that, in the absence of jumps (i.e. in a pure diffusion framework, see here and here for further info), the fair variance strike K_{var} at which a variance swap with notional N and payoff$$ N ... 6 Basically, the author is saying that the delta of an option,$dC/dS = \frac{\partial C}{\partial S} + \frac{\partial C}{\partial v}\frac{\partial v}{\partial S}$, where the$\frac{\partial C}{\partial S}$is the delta assuming constant volatility, the$\frac{\partial C}{\partial v}$is the vega of the option, and the$\frac{\partial v}{\partial S}$... 6 Let $$\ln\left(S_T/S_t\right)$$ have mean$\mu_\tau$and standard deviation$\sigma_\tau$, where$\tau=T-t$, and density of its standardized form $$X= \frac{\ln(S_T/S_t)-\mu_\tau}{\sigma_\tau}$$ approximated by Gram-Charlier expansion $$f_X(x) = \phi(x) - \gamma_{1\tau} \frac{1}{3!} D^3 \phi(x) + \gamma_{2\tau} \frac{1}{4!} D^4 \phi(x),$$ with$\phi$... 5 Well the terminal FX rate is lognormally distributed and lognormals are skewed. So this is not surprising. 4 You do not state whether your evaluations will result in potentially implementing multiple strategies or just one of them. This matters because if you are going to be combining multiple ones then you need some reasonable capital allocation assumptions, which increases complexity immensely. Let's take the simpler case where you just want to choose one. ... 4 It is not the fact that volatility is time varying that creates the skew per se, but the fact that volatility is negatively correlated with the spot. That is to say, as the stock/index price declines volatility will tend on average to increase, and vice versa. Time varying volatility itself would create a more symmetric 'smile'. Edit: Suppose that you ... 3 Let's focus on the volatility contract price. Generalisation to cubic and quartic contracts is straightforward. Following the paper's notations, the evaluation date is$t$and the (European) contracts all expire at$T = t+\tau$. A volatility contract is specifically associated to the payoff function $$H[S] = R(t,\tau;S)^2 = \left(\ln S(t+\tau) - \ln S(t) ... 3 First of all, quanto options (options denominated in FOR currency but whose value we wish to determine is in DOM currency) are mainly traded over-the-counter, hence their prices are not likely available, as opposed to non-quanto options (options denominated in DOM currency with payoff also in DOM currency) for wich we have some quotes. The idea of the ... 3 The value of a call option at expiry is V=\mathrm{max}(0, S_t-K). If you set K=0, then you have V=\mathrm{max}(0, S_t), and since S\geqslant0, \mathrm{max}(0, S_t) = S_t - i.e. ie's equivalent to holding the stock, which at expiry you'll expect to be worth the whatever the forward is. 3 The answer is that by definition, if the underlying stock obeys a lognormal distribution with std deviation parameter sigma, then the implied vol of options priced using this model will be sigma. Of course in the market we observe that options of different strikes have different vols - this just means that the underlying distribution is not perfectly ... 3 You have to remember that implied volatility comes from a "wrong" model to give the right answer. Option prices are determined by supply and demand (subject to a few arbitrage bounds). A higher implied volatility for OTM/ITM options relative to ATM options simply means that the prices of these options are higher than the Black-Scholes model would imply (... 3 Why don't you just use SSVI (https://arxiv.org/abs/1204.0646) or maybe even eSSVI (https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2971502)? With this parametric approaches an arbitrage free volatility surface is guaranteed and you only need a handfull of parameters. Gatheral and Jacquier even give you the calibration procedure which should be simple to ... 3 My preferred measure for skew would be the difference between the implied volatilities corresponding to the strikes where the Black-Scholes d_2 = 0 and d_1 = 0. The reason I prefer this to others you often come across is that when the smile is symmetric the d_2=0 implied volatility equals the d_1 = 0 implied volatility. Furthermore, the implied ... 3 The usual answer is that most risk assets tend to exhibit left-skew, with correlations ->1 into the left tail (ie diversification breaks down). And so positively skewed assets have attractive portfolio features, over and above their own intrinsic fundamentals. The more formal answer is that for two distributions with the same mean and standard deviation, ... 2 In the public domain, there is SVI (stochastic volatility inspired) curve invented by Jim Gatheral. If you need curves which can fit very liquid names or handle W-shape curves (e.g. on earning), you should look at the curves by Vola Dynamics. 2 In optimiazation system, you have to weight the price for the different maturities in a way that reflect your confidence in each data point (influenced by liquidity). One way to do so is to weight, each price by its Black-Scholoes Vega (see Tankov (2003)). So when minimazing the squared differences of the sum your weighted option prices, you can use the ... 2 Old and golden question, and maybe a new perspective: As the previous answers have pointed out, distinction needs to made between "skewness" and "skew". The former is the third moment of returns, and the latter is what volatility traders/portfolio managers usually associate with the difference between two implied volatilities straddling ... 2 put call parity guarantees that the implied volatility of a call and put with the same strike is the same. So the smile graph is the same as well and so are all quantities derived for it. In more detail,$$ C(K) = P(K) + F(K)$$The value of$F(K)$is model independent and does not depend on volatility. So knowing the implied of$C(K)\$ gives you the price ...

2

There's no best method. The question is : what is the behavior of the volatility structure (atm and skew) when the underlying moves? Each method assumes something different. In the real market, one method might work well for a period of time (in the sense that it minimizes residual p/l), but then another method might take over as best. Practitioners ...

2

When I saw these curves they seemed very strange to me. I believe it is a data-quality issue.I went to Bloomberg and I retrieved the implied vols for 70 near ATM strikes of the weekly SPX options expiring November 27 2015 (I believe that is the yellow curve in your diagrams i.e. November 4th week). This was today 2015-oct-27 at about 15:00 New York time. As ...

2

It also depends on at what levels of the spot the higher vol gets realized. In your example: if you buy an option on a 40 vol expiring in a month and over the next month stock the average vol of the stock is 60 and you dynamically hedge, are you guaranteed to make money? If not could you please give me a simple example perhaps where you'd wind up ...

2

The market does not follow Black-Scholes assumptions, as you clearly know : there is a skew and vol levels change. Neither does it follow any other particular known model. So when you say "dynamically hedge" you have to understand this as an approximate hedge that still leaves some significant risk. Vols will move, and not always together and in the way ...

2

Basically there are three steps to accomplish this. 1 - collect time series of options for several expirations and strikes. 2 - calculate implied volatility surface for every time period, and use model-based or model-free interpolation to create continuum of strikes / expirations. 3 - from the continuous surfaces you can calculate series of any specific ...

2

In black-scholes world, correlation between volatility and spot is zero. From the above details you can estimate how the implied volatility for a given option (note options have FIXED strikes) might change for a given move in spot. If when stock goes up, the option's implied vol goes down, this would be a violation of the black-scholes model (which scenario ...

2

Suppose you were to price 2 instruments: a strongly OTM put and a strongly OTM Call. In the standard BS settings, instantaneous volatility is assumed to be constant. Consequently, the implied volatility of these 2 instruments will be the same, resulting in an absence of implied volatility skew. Now, assume a negative spot/instantaneous volatility ...

2

Well, the probabilities implied by the market are not equal. If you believe they should be equal, then go ahead and express yourself in the market. The point is , it is not an objective fact that it must be 50/50- that's your subjective opinion.

2

The use of forwards is just another method to look at the underlying. The Black-Scholes options model utilizes Spot and handles the carry as an interest rate in the model. On the other hand the Black Model uses forwards instead. Since the forward price would take into account the carry, both models should yield the same result if one is accounting for all ...

Only top voted, non community-wiki answers of a minimum length are eligible