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Hot answers tagged stochastic-volatility

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Well, what you find is that the introduction of stochastic vol changes the delta of your options. So what does this mean? If the new delta reduces the variance of your hedged portfolio versus the pure local vol model , then it means that the introduction of stochastic vol has resulted in a better description of market dynamics versus the pure local vol ...

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Because vanilla derivatives with European exercise depend only on total variance , not on it's dynamics in time. If you have a simpler model (like interpolation of these total variances from your volatility surface) you don't have as much of unobservable parameters stochastic volatility models have. Having more parameters (which many times would need to be ...

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This depends largely on the model as well as the market, so there is no one-size-fits-all approach. Let us take the Stochastic Alpha Beta Rho (SABR) model, which has four parameters, as an example: $\alpha$, the initial instantaneous volatility of an ATM option. This is calibrated frequently and often intra-day due to the importance of ATM options. $\beta$, ...

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Maybe it would help you to think of it the following way. The strike $\sigma^2(T)$ of a fresh-start variance swap of maturity $T$ in the Heston model only depends on parameters $(v_0,\theta,\kappa)$, see related question here. More specifically \begin{align} \sigma^2(T) &= \Bbb{E}_0^\Bbb{Q}\left[ \frac{1}{T} \langle \ln S\rangle_T \right] \\ &= \...

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The following source contains detailed answers to your questions in a research paper from ETH Zürich. van der Weijst, Roel (2017). "Numerical Solutions for the Stochastic Local Volatility Model" http://resolver.tudelft.nl/uuid:029cbbc3-d4d4-4582-8be2-e0979e9f6bc3

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I agree with the comment made by will: for a given model, you can potentially compute a Delta according to any "stickiness rule" depending on which data you decide to bump vs. keep constant. That being said, if you look at the following quantity $$\Delta = \left. \frac{\partial V}{\partial S_0} \right\vert_{\Theta}$$ that we could call the in-model Delta ...

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In short , this claim does not hold under all circumstances. There are a few ways to break down such approximation. The options under consideration have very long expiry, i.e. $T$ is very large As expiration date approaches, the volatility smile becomes more pronounced, i.e. $v$ becomes relatively large. Under extreme market condition, the magnitude of $\... 2 If you want to use the normal SABR ($\beta=0$), my paper, Hyperbolic normal stochastic volatility model (arXiv | SSRN | DOI) might give you a solution. It reports an exact closed-form MC simulation scheme for the normal SABR model. Better than that, it shows that Johnson's$S_U$distribution ($\sinh$transformation of normal variate) is a close cousin of the ... 2 Both @alexprice and @FunnyBuzer have some good points, and I have upvoted them. I think I have enough to add here that I'll make another answer entry. First off, @AFK was fairly correct that you do not need stochastic volatility for vanilla (European exercise) option pricing, since (as he says and alexprice elaborates) you just interpolate the surface of ... 2 I think that the main advantage of using a stochastic volatility model is to produce a consistent volatility smile. Let's consider the pricing formulas for the normal and lognormal volatilities:$$dS_t=\sigma dW_t\Rightarrow \mathbb{E}[(S_T-K)^+]=(S_t-K)\Phi\left(\frac{s-S_t}{\sigma\sqrt{\Delta t}}\right)+\sigma\sqrt{\Delta t}\phi\left(\frac{s-S_t}{\sigma\... 2 First and foremost it is important to clarify that the underlying is not necessarily normal/lognormal but for the special cases of$\beta$the underlying is normal/lognormal Conditioned on a realization of the volatility. As mentioned in the answer by @ilovevolatility. Simple stochastic calculus will show the properties you mentioned. For realized ... 2 You don't want to use the SABR (or an extension) to price equity options or FX options. The lag of mean-reversion in the model's volatility dynamics leads to explosive behavior and to a implied distribution that is absolutely not in line with empirics -- especially on longer time horizons. To my knowledge people use it mostly for interest rate derivatives. ... 2 These papers are interested in modelling the stochastic volatility, so implicitely they model the dynamics of the forward price which has zero drift under the corresponding terminal measure. This makes the exposition much simpler. It is then easy to switch back to the stock price: european options: an option with expiry$T$on the stock price$S_t$is ... 2 It is a big topic but here is a simplistic recipe! The starting point would be to check the distribution of the historical returns. Histogram would give an idea of how the shifts are distributed. Have a look at the tails, if the tails are fat or don’t ‘tail-off’ then that would be indicative of jumps or non constant volatility. If you decide that a simple ... 1 No, the simulation is not exact in general, precisely for the reason you mentioned. By "exact", it is meant that there is no discretization error in time. Of course, there will always be a Monte-Carlo sampling error. For the Black-Scholes model, the simulation is exact if you simulate the log asset, as it is a standard arithmetic Brownian motion, and then ... 1 Please see, if the below serves as a counter-example - Consider,$\Sigma= \rho S\sigma - \rho K S\sigma^2 +S\sigma$So,$\Sigma_\rho = S\sigma - K S \sigma^2$There exists$K$, such that$K= \frac 1 \sigma$where$\Sigma_\rho = 0$. Evaluating$\Sigma_{S \sigma}$below -$\Sigma_S= \rho \sigma - \rho K \sigma^2 +\sigma \Sigma_{S \sigma} = \rho - 2 ...

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volatility of the volatility controls convexity of the skew/smile => more vol of vol generates more convex function ( = more smile) mean revertion and correlation between brownian motions both control ATM skew. long term variance controls overall level of skew (moves whole skew graph higher) In practice these parameters are calibrated to market quotes of ...

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We need to categorise the types of models before we consider the term $\dfrac{\partial^2 \Sigma}{\partial S \partial \sigma}$. I will only consider local volatility models and stochastic volatility models. Local volatility models The local volatility function is, of course, $\sigma^2(K,T)=2 \dfrac{\partial_T C_{KT}}{\partial_{K}^2 C_{KT}}$ This can be ...

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Given the main uses of the VaR relate to risk management such as limit management, and measurement of P&L volatility, it is usually calculated under the physical/real world measure. Reason being that the risk measure are normally used to predict or explain the P&L movements from one day to another, which one can relate to their historical movements. ...

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my two cent's worth. It depends ultimately what you are trying to achieve, and calibrate to. The key tests for implied vols on option prices, relate to their pdf (probability density functions). The Hagan expansion allows us to have a analytical form, by which one can compute the implied vols, and subsequently feed that into a Black76 equation. If the ...

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Given (conditional on) a realisation of the volatility, it is normal for $\beta = 0$ and lognormal for $\beta = 1$

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By investing the amount of $\alpha_t$ at time $t$ in the risky asset, the wealth is given by \begin{align*} X_t = \frac{\alpha_t}{S_t} S_t, \end{align*} where $\frac{\alpha_t}{S_t}$ is the units of the risky asset. For $\Delta$ sufficiently small, the wealth at time $t+\Delta$ becomes \begin{align*} X_{t+\Delta} = \frac{\alpha_t}{S_t} S_{t+\Delta}. \end{...

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Here is a practical hedge for forward volatility swaps using only straddles with a certain strike, and with a notional that is determined by the skew at that magic strike. The same method for spot/seasoned volatility swaps will be posted online in due course as well. https://papers.ssrn.com/sol3/papers.cfm?abstract_id=3354408

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The answer given so far, by Mats Lind, to the first question is not in the spirit of the paper. I am referring to Question: In the paper they say one should really integrate from $y$ to $\infty$... What they mean is that you should not use any information on what happens from $-\infty$ to $y$, which would correspond to the standard cumulative distribution ...

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