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## Hot answers tagged stochastic-volatility

8

This is an interesting and not so easy question. Here's my 2 cents: First, you should distinguish between mathematical models for the dynamics of an underlying asset (Black-Scholes, Merton, Heston etc.) and numerical methods designed to calculate financial instruments' prices under given modelling assumptions (lattices, Fourier inversion techniques etc.). ...

7

$X_t$ being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function $f$ of $t$ and $X_t$. Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming $X_t$ is here a continuous, square integrable stochastic process) $$df(t,X_t) = \frac{... 6 [Short answer] No closed-form formula in general. You need to resort to numerical methods. Monte Carlo is preferred by most practitioners but you could also use Finite Difference schemes (and sometimes even Fourier inversion techniques depending on the model used and the instruments to be priced). [Long answer] One usually distinguishes between 2 classes ... 5 This effect is coming from the supply and demand in the options markets. Many portfolio managers want (or need) to buy out of the money put options, and many are willing to sell out of the money call options (thereby funding the purchase of put options). Now, when the market goes down, dealers find themselves short vol and they need to buy options to cover ... 5 Let$$q (S) := \frac{d\mathbb {Q}(S_T \leq S)}{dS} denote the probability density function of the stock price at time T>0 under the risk-neutral measure. By definition, the price of a European call then writes \begin{align} C (K,T) &= P (0,T) E_0^{\mathbb {Q}}[(S_T-K)^+] \\ &= P (0,T) \int_K^\infty (S - K) q (S) dS \end{align} with P (0,... 4 Quick summary: Your model should still be well specified, as long as: 1) You do the analysis on a heavily traded asset, e.g. IBM on NYSE, and 2) You use heteroskedasticity-consistent standard errors in your estimation framework, e.g. White's standard errors. I'm going to start the long answer by re-stating the question to make sure I've got it right. Let ... 3 Your problem probably comes from the notations used. Let the Moment Generating Function (MGF) of a random variable X be defined as M_X(u) := E[e^{uX}] $$From this definition, it entails that$$ E(X^n) = M_X^{(n)}(u=0) = \frac{d^{n} M_X}{ d u^{n}}(u=0) $$Knowing this, the function$$ f_{\lambda}(t,r)=E[e^{-\lambda {r_{T}}}|r_t=r] $$can be ... 3 Here's my 2 cents: a) Conditional expectations can always be seen as martingales (this is a direct consequence of the tower property). Thus, we here have that$$ M_t := E^*[e^{-\lambda {r_{T}}}|r_t] $$is a martingale. Applying Itô's lemma to M_t = f_{\lambda}(t,r_t) as you did is a good starting point. But doing this, leaves you with an SDE, not a PDE.... 3 Let$$ f_{\lambda}(t,r)=E^{(t,r)}\left[e^{-\lambda r_{T}}\right] $$where E^{(t,r)} denotes the expectation conditional on r_{t}=r. We assume f is smooth for the remainder. Let \theta=T\wedge\inf\left\{ s>t\colon\left|r_{s}-r\right|>1\right\} . By the Markov property of \{r_{t}\},$$ f_{\lambda}(t,r)=E^{(t,r)}\left[f_{\lambda}(\left(t+h\right)...

3

Even though it's a straightforward extension, it took me a while (a year? yikes!); but now you can easily incorporate Bayesian ar(1) (or more generally, Bayesian regression) in joint estimation by using designmatrix = "ar(1)" as an argument to svsample. It's not well documented yet (except in the help files), but I nevertheless hope easy to use. From the ...

2

Diffusion brings about a standard deviation which increases with the square root of time (just like in Brownian motion), while jumps add variability proportional to time (since the jump times are a Poisson process). So they are quite different. Experience shows that sharp stock market moves do occur (in connection with big news events for example), so ...

2

The problem is that what some mean when they say "volatility" is BS implied vol from an option price. What some others mean when they say "volatility" is some diffusion parameter from a drift diffusion model (with or without jumps). These are the same value in the log normal model of stock prices but different for many other models including those with jumps....

2

It is difficult to gain intuition by just looking at the price surface, and it is also easier to calibrate models on the volatility surface rather than on the price surface because with the later you are dealing with numbers of very different sizes (depending on the moneyness and maturity) which is not good for minimization algorithms. However low and high ...

2

The answer is yes. In fact, there always exist a 'Black Scholes like' formula. Easy to show too. If the risk neutral distribution of the price has cumulative density $P$ and probability density $p$, then $$E(S-K)^+=E((S-K)\ 1_{S>K})=E(S\ 1_{S>K})-K\ E(1_{S>K})$$ The second expectation is just $P(K)$, ie the probability that the option ends up in ...

2

You almost get there. However, you ca not conclude that $\rho^2$ is a constant based on $(10)$. Note that, from your $(7)$ and $(8)$, \begin{align*} \frac{\rho(z_t)^2}{\beta} e^{\beta \tau} (e^{\beta \tau} - 1) = -h'(\tau)+e^{\beta \tau}h'(0). \end{align*} Taking derivative with respect to $\tau$ on both sides, we obtain that \begin{align*} \frac{\rho(...

2

From Equation (6), $B(t,T)=-t+c(T)$ for some function $c(T)$. $1=P(t,t)=e^{-A(t,t)-(c(t)-t)r_t}$ or $A(t,t)+(c(t)-t)r_t=0,\,\forall (r_t,t)$. So $c(t)=t, A(t,t)=0,\forall t$. For Equation (8) you have missed the square on $\sigma$ and a factor of $\frac13$. Then you just need to substitute in the function for $b(s)$ and integrate the following to get the ...

2

For starters, the short rate model you mention in equation (1) is Cox-Ingersoll-Ross while the bond price in equations (2)-(4) correspond to the Vacisek model. So there is a problem somewhere, I would go for a typo in (1). Second, what you wrote seems fine to me, so there must definitely be yet another typo in your solution manual. Note that if there is no $... 2 Loosely speaking, it can be seen as inserting an additional degree of freedom in the underlying's dynamics. This can be useful from a static perspective: with an additional lever to play on, one can hope to better capture the short term implied volatility smile, which "naive" stochastic volatility models (single volatility factor, no jumps) are known to be ... 2 I think,the additional volatility factor,$v_2(t)$, provides more flexibility in modeling the volatility surface.We know$\rho$controls the slope of the implied volatility.In the single-factor Heston model,$\rho$is constant over maturities,In deed $$Corr[{dS}/{S\,,\,dv]}\;=\rho \,$$ which means that model has trouble providing an adequate fit to market ... 2 Two volatility processes yield a higher flexibility of the model. This is of greater importance if one tries to price derivatives with different maturities in one single model. A additional volatility component helps to capture the term structure of volatility, which can depend greatly on time to maturity. See for example the VIX term structure from CBOE: ... 2 It comes from Heat Kernel expansion and differential geometry. See Theorem 6 and Section 8 of http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1717676&download=yes 2 Great reads to further explore and better understand stochastic volatility models are the series of articles "Smile Dynamics" by Lorenzo Bergomi. As the name indicates the idea is to study stochastic volatility models not only as "smile models" (in the sense that SV models can be used to capture the state of the vanilla market by correctly accounting for ... 1 Local vol model gives a "too shallow" forward skew. Derivatives of which the price are depending on the forward skew will be mispriced. If i remember correctly, Hagan's paper 1 Chan, Karolyi, Longstaff, and Sanders (1992) compares empirically the performance of the main interest rates model. The first page also provides the main references on those models. 1 Generally, the Wishart stochastic volatility model identifies the volatility of the asset as the trace of a Wishart process. Contrary to a classic multifactor Heston model, this model allows to add degrees of freedom with regard to the stochastic correlation. Thanks to its flexibility, this model enables a better fit of market data than the Heston model. ... 1 Your adjusted scheme is correct. Basically, taking a maturity$T$, you can consider the forward price process$F_t^T = S_t e^{r(T-t)}$. You apply the Andersen scheme to$F_t^T\$ and then note that \begin{align*} S_{t+\Delta} &= F_{t+\Delta}^T e^{-r(T-(t+\Delta))}\\ &=F_t^T \exp(\ \Box \ ) e^{-r(T-(t+\Delta))}\\ &=S_t e^{r(T-t)}\exp(\ \Box \ ) e^{-...

1

I suggest taking a look at optimal hedge Monte Carlo and by extension the garam model from http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1428555. The basic issue is that a risk premium exists in option markets which is like Vol also unobserved. The better your guess of the future realized vol distribution, the better your guess of the risk premium ...

1

Different methods exists to compute implied vol from the same option prices, eventually it's prices that matters to calibration. But if you can reproduce same option prices accurate to the cent by fitting implied vol, I think it doesn't matter.

1

Hans Buehler investigated this in some detail, including in his doctoral thesis. When I tried it out some years ago, back when volatility exotics were more liquid, I found the models nearly impossible to calibrate to my satisfaction, even for the SP500 complex. I think the mathematical analogy is fair, and enjoyed Buehler's work, but in practice it won't ...

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