# Tag Info

14

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*} ...

12

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. ...

8

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 ...

5

Note that total implied variance defined as $$V(T,K) = T\Sigma(T,K)^2$$ should be an increasing function of $T$. Otherwise you have a calendar arbitrage (sell the call with shorter expiry and buy the cheap longer one). If you interpolate linearly your implied volatility is $$\Sigma(T,K) = w\Sigma(T_i,K) + (1-w)\Sigma(T_{i+1},K)$$ with weight $w = ... 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 ... 4 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 ... 4 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? 4 please go to {drvd} BVOL Equity Implied Volatilities Calculations paper. Disclamer: I was working for Bloomberg, that is as far we disclosed. 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 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! 3 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 ... 3 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 ... 3 Generally speaking, in the real world, you'd always want to use the correct implied vol. But you should think of your question in terms of: (1) Vega mark-to-market (m2m) PnL vs. theta/gamma profile (2) Change in risk and PnL due to higher order risks (vanna, volga) Vega mark-to-market PnL vs. theta/gamma profile In a simple, pure Black Scholes world ... 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

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

The answer of AFK is very good and accurate in a BS setting. Thinking of jumps I would add the following: If we assume that stocks sometimes move in jumps (usually downwards) then it is clear that ATM or OTM options shortly before expiration with a price that accounts for the possibility of a jump - which is therfore quite high - only fit in the BS ...

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

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 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 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 I'll address your questions in order: 1a) For TSRV constructed using high frequency returns from NYSE market open to market close on a single day, the output should be numbers on the order of magnitude of 1e-4 to 1e-5. In other words, your numbers look about right. I got these number from calculating TSRV for IBM data myself using Kevin Sheppard's MatLab ... 2 It is all a matter of frequency. For instance if you want to get annual realized volatility you multiply your last expression by \sqrt{(N*251)} or the second to last expression by \sqrt{(251)}. In other words, your last expression is the 5-min realized volatility whereas the second to last expression is the daily realized volatility. 2 I don't believe you will necessarily find a cite-able source as, I believe, this comes from a practical rather than theoretical motivation. As you know option prices are a function of: future prices, discount rates and implied volatility, volatility surface skew and other supple/demand factors. So when you are trading these instruments, you need to ... 2 It implies negative forward variance. I have the book, and went through the section following your quote. In math terms, he is making a proof by contradiction. He first assumes that you can interpolate Iinearly, and comes to the conclusion that it is not a good assumption. The argument does involve some calculus. I don't think I have a better explanation, so ... 1 This has been asked many times already. Volatility always refers to a model. And unless stated otherwise this model is the Black-Scholes model. In this model the volatility is the standard deviation of the log-returns divided by the square-root of time:$$ \log(\frac{S_{t}}{S_0}) = (r - \frac{1}{2}\sigma^2)t + \sigma W_t \sim \mathcal{N}\left( (r - ...

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