# Tag Info

12

Let $t_0, t_1, \ldots, t_n$ be observation dates, where $0=t_0 < \cdots < t_n = T$, and $\{S_t \mid t \geq 0\}$ be the equity price process without dividend payments. Then the realized variance is defined by \begin{align*} \frac{252}{n}\sum_{i=1}^n \ln^2 \frac{S_{t_i}}{S_{t_{i-1}}}. \end{align*} Note that, for sufficiently small $x$, \begin{align*} ...

11

It seems that you are thinking of the volatility as some sort of standard deviation of your stock price. It is not. In the BS model, $\sigma\sqrt{T}$ is the standard deviation of the log-return $\log(\frac{S_T}{S_0})$. There is no mathematical upper bound to its standard deviation. There is also no mathematical problem with returns being negative either. ...

7

Upon close reading, this appears to be 3 (interesting) questions, not one. I'm not sure if the mods have the tools needed to split it up, so I'm just going to write down the three questions as I see them and then deal with them one by one. Note, it is simpler for me to talk about variance instead of volatility. This has no material impact on the answer. ...

5

No. Implied volatility isn't a historical measure of standard deviation. Implied volatility is used to relate a market price to some model, be that Black-Scholes or something more sophisticated. Another way to phrase it, implied vol is that single vol input into a model, such that the model reproduces the market prices. Different models will have ...

5

For How VIX works you can read this wonderful blog : http://onlyvix.blogspot.com/2011/09/intuitive-understanding-of-vix-formula.html It provide wonderful non mathematical explanation of the how vix is actually computed. Now comes to your last answer why vix is inversely related to market movement ? In simple words, if market is more volatile then ...

4

The method described in Hallerbach (2004) always worked well for me. We derive an estimator for Black-Scholes-Merton implied volatility that, when compared to the familiar Corrado & Miller [JBaF, 1996] estimator, has substantially higher approximation accuracy and extends over a wider region of moneyness.

4

A very popular choice for mean reversion is the Ornstein–Uhlenbeck process (here in discretized form): $$L_{t+1}-L_t=\alpha(L^*-L_t)+\sigma\epsilon_t$$ Here you see that the level change is governed by some parameter $\alpha$, the mean reversion rate (or speed), and the distance between the long run mean $L^*$ and the actual level $L_t$ plus some noise. A ...

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 ...

3

There are lots of papers online and here are a few I would suggest math.umn riskworx G. Dimitroff, J. de Kock Nowak, Sibetz I you have matlab there is an step step example to calibrate SABR model. Since it uses the financial toolbox of matlab for a few functions I dont think you can replicate it in any other language. There must be C++ code available ...

3

First, as far as I can tell, you are not taking into account dividends. Second, If you simply take the forward price of the SPX @ $5.5\%$ which is what you are using, you get $1411 \cdot \text{exp}(0.055 \cdot 2.99) = 1663$. Given a strike of $1300$, the call should have an intrinsic value of $1663-1300= 363$. You have a price of $272$. The price is less ...

3

Since American style options allow early exercise, put-call parity will not hold for American options (unless they are held to expiration). In practice, there is also a difference between calls and puts for European options as well. The full description is here: What causes the call and put volatility surface to differ?

2

In my mind volatility (SD) of a stock and implied volatility (IV) are two quite different things: volatility is usually measured backwards looking. The common methods (empirical, GARCH, ..) look into the past. Measuring the risk of owning the stock in the future is often based on these backwards looking observations. We try to measure risk in the real ...

2

All option pricing formulas except this one and this one use some sort of historical volatility . I can't see how you can use the Black Sholes framework and not use some sort of historical volatility uses an order book uses geometric shapes and volume

2

If you want to estimate volatility from historical data, the only best linear unbiased estimator (BLUE) is $$\sigma=\sqrt{\frac{1}{T-1}\sum_{i=1}^T (r_i-E(r_i))^2}$$ Any other estimator will hence either be biased or not consistent. Another approach could be to estimate volatility via a GARCH model, which has shown good empirical results in the past. It is ...

2

CRR is just a numerical approximation to Black--Scholes. Its main use is in getting American option price. There is no real difference other than slight inaccuracy when using it for Europeans. So no it wouldn't do what you ask. Your questions are philosophical. What is the purpose of the model? if you estimate the volatility from a time series then you can ...

2

You should always think: I buy the one which is to cheap and sell the one that is too expensive and figure it out. The figuring out in this case is noting that: $C\geq 0$ since it will never cost you money The option is strictly better than $S-K$ so has a higher price. Now to your strategy: You buy $C(T_2)$ (the cheap) and sell $C(T_1)$ (the ...

2

Peter Jaeckel wrote a paper just on how to solve this problem: By Implication (July 2006; Wilmott, pages 60-66, November 2006). Probably the most complicated trivial issue in financial mathematics: how to compute Black's implied volatility robustly, simply, efficiently, and fast downloadable from jaeckel.org In my experience the most important thing is to ...

2

If you believe the process $Y_t$ to be stationary, you can try to profit from it via a mean-reversion strategy or any other way that exploits the stationarity. It doesn't matter whether $Y_t$ is obtained as a cointegrational combination of a few non-stationary processes, or as a linear combination of some processes that are stationary themselves. In the ...

2

Based on the example you gave, it seems that indeed your inputs are inconsistent. The intrinsic call value is $S-e^{-rT}K = 286.52355\dots$, which is higher than the market value, implying that there exists an arbitrage. Instead, one of your inputs is probably wrong. Even if the interest rate is set to $0$, the intrinsic call value is still above your bid, ...

2

The most used equity volatility models in the industry are the Black-Scholes model (including its time dependent version) and the local volatility model. It always come along with stochastic rates, discrete dividends and quanto effects (a must-have when pricing even simple payoffs) so the calibration/pricing process is much more involved than what you might ...

2

The central limit theorem guarantees, under fairly general assumptions, that the sum of returns becomes more normally distributed as the number of returns grows (technically, defining a return as $\mathrm{log}(S_{t+\Delta t}/S_t)$, $\sum_i ^n \mathrm{log}(S_{t+\Delta t i}/S_{t+\Delta t (i-1)} \to \mathcal{N}(\cdot,\cdot)$ as $n \to \infty$). Thus, as $T$ ...

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 ...

2

The main thing to keep in mind with all these different option combination strategies is that you are really trading option greeks! I think the answer to why the calender spread is so popular lies in the special combination of gamma and vega risk: Calendar spreads are the one type of trade where gamma can be negative while vega is positive (and vice versa ...

2

If the security has negative correlation with other assets that enjoy attractive risk-free rates, then it can be attractive at a return rate under the risk-free level. It would, of course, never be attractive in a single-security portfolio.

2

I think this extremely hard to do (to the point where I think that every hedge fund that trades vol should be avoided like the plague). The fundamental value of volatility would be a quantity that's related to the speed at which new news comes available to the market, the significance of news, the extent to which this news can be traded, general market ...

2

First note that the price of binary call is related to the price of an ordinary call in any model by $$BinC(T,K) = e^{-rT}\mathbb{E}^{\mathbb{Q}}[1_{S_T>K}] = - \frac{\partial}{\partial K}e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T-K)_+] = - \frac{\partial}{\partial K}C(T,K)$$ Now the volatility smile is implicitly defined by  C(T,K) = ...

2

The main difference is that one approach assumes that a certain dynamical structure properly describes the underlying instrument, while the other approach is really only a re-writing of the price in terms of an implied volatility. Implied volatility Implied volatility really only needs two things: the underlying stock price and the call option price (apart ...

2

The most common use for implied volatility in valuation is for asseing options or option like postions. A volatile instrument is likley to activate or put an option postion in the money just on the basis of its volatility rather than any fundamental change in the intrinsic or fair market value of the underlying. This needs to be taken into account when ...

2

In general, $v = \frac{\partial C}{\partial \sigma} > 0$ and $\theta = \frac{\partial C}{\partial t} < 0$. If maturity $T$ increases than $C$ increases. Suppose volatility is non-constant. Then if $T$ increases, the option value is more volatile, since the stock price is more volatile. Since $v > 0$ the option price must increase. He claims that ...

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