# Tag Info

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If you assume a zero drift for our asset return, this formula is indeed simply a measure for (daily?) volatility.

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From a theoretical point of view, you are supposed to use the correlation calculated under the same measure (i.e. multiplying daily excess returns and weighting that product by your weights). In practice, I have seen both approaches: Weighted correlations and weighted variances, i.e. a weighted covariance matrix, and a mixture: unweighted correlations with ...

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The simplest long vol strategy is to be long an ATM straddle and delta hedge it, the problem is that when it is no longer ATM the exposure to vol weakens. You could then sell that straddle and enter another ATM one. Another solution is the vol swap or variance swap mentioned by Stephane below. It gives constant exposure no matter what the level of S&P. ...

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I do not mean to discourage you, but it sounds like you're a wee bit late for this round of volatility games, for two reasons: You are still trying to figure out how to implement a long vol strategy. The market has already priced the risk in, i.e. buying volatility is already expensive. However, never too late to learn and prepare for a next time. My ...

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The solution was to use the observation per day across the moneyness levels and tenors. In my provided table would line two be the the y-vector in the regression and the values moneyness (see values before "." in header) levels the entries for the explanatory variable and the tenors (values after "." in header) the entries for the tenor variables. Then this ...

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What not to do What you are asking us, without knowing, is related to how to price a variance swap. Well, under a general diffusion process, variance swaps can be priced by forming a suitably weighted portfolio of options over a continuum of strike prices with the entire portfolio maturing on a given date. The intuition is that your exposure to volatility ...

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Suppose that you are riskless asset with return $r_{ft}$ and a risky asset with return $r_t$ and conditional volatility $\sigma_t(r_t) := \sqrt{V_t(r_t)}$. We build a portfolio using weights $(w_1, w_2) \in \mathbb{R}$, or as you wrote it $w_t := w_{1t}$, $w_{2t} := 1 - w_t$. This portfolio will have a time $t$ return of $r_{pt}$. Its volatility is given by $... 0 The basic difference is that for calculating the option's price within the classic BS-framework, you mostly use the historical vol (which is extracted from time series with a model). But this is only a theoretical (arbitrage free) price. At an option's exchange, you will see supply and demand meeting each other. Assuming perfect and efficient capital markets,... 2 Your question makes perfect sense; one has to define volatility. Volatility can be used interchangeably for a number of different metrics. Realized volatility - the observed volatility of the underlying asset (and btw, there are many quite different ways of measuring it). Implied volatility - the number you get when you run your option pricer in reverse. ... 2 Effectively, you take your series of prices and transform those to returns as you would do in your standard approach, as well. step return date weight 1 2020-03-26 0.97 2 2020-03-25 0.97 * 0.97 3 2020-03-24 0.97 * 0.97 * 0.97 ... K T-K+1 097^K For$K$sufficiently large, the sum of the weights$\sum_i \lambda^i$will ... 1 Rebonato called the whole process "the wrong number in the wrong formula to get the right price". We use Black-Scholes much the same way that we look at price-earnings ratio in equities. This is beneficial for traders. First, it translates a fast-moving price into a slow moving valuation metric. Second, it gives us an idea of value. Is this option rich or ... 5 1. Let me first reconcile the Black-Scholes pricing formula with the idea of prices being determined by supply-and-demand. Even if it is not explicitly said this way, from an equilibrium perspective, the Black-Scholes formula defines the unique price of risk that is consistent with the absence of arbitrage. In fact, you explicitly use this price when you ... 1 Each path is evolved based on the vol and a random number. The higher the vol the more the paths will diverge. Paths will diverge if you increase time as well. The solution is to increase the number of paths as vol or time increases to get a standard deviation of terminal values that you are comfortable with. 0 We're going to assume that we have$N$assets in our portfolio with weights$w$and prices$x$. The variance of your portfolio is given by: $$V\left( w'x \right) = w' E\left((x-E(X))(x-E(x))'\right) w = \sum_{i,j=1}^N w_i w_j Cov(x_i, x_j).$$ So, the contribution of one asset is given by: s_i := \frac{w_i^2 ... 2 Because Brownian motion has non zero quadratic variation (as opposed to continuous differentiable function which have 0). Quadratic variation is a defining characteristic of Brownian motion and Brownian motion is central to financial models of the evolution of prices/returns over time. 1 The simple explanation is that in the absence of calendar spread arbitrage, we should observe monotonic option prices with respect to maturity. And option prices are monotonic with respect to increase in volatility. Let$(X_t)_{t \geq 0}$be a martingale,$L>0$and$0\leq t_1, t_2$, then we have $$E[(X_{t_{2}} - L)^{+}] \geq E[(X_{t_{1}}-L)^{+}]$$ for ... 1 A few thoughts on the steps please: 1):You would normally calibrate Dupire based on current option prices; so you won't calculate implied volatility from historical prices of the underlying (assuming this is meant), but the current option prices. The prices are usually quoted in terms of implied volatilities, so in most cases you wont need to calculate the ... 0 Equity market volatility reflects the confidence/certainty about economic future. When it's a low volatility environment, it reflects the underlying psychology of participants (e.g. not a lot of tug of war between the bull and the bear, not much uncertainty of the direction of the economy.). It also reflects relatively stable trading conditions amongst the ... 1 One issue I see: $$2\cdot0.01+0.01\cdot11\cdot0.15 = 0.0365$$ must be replaced by $$2\cdot \left(0.01+0.01\cdot11\cdot0.15\right) = 0.053$$ Edit: (Detailing my comments a bit) Dupire's equation, as you wrote it, is correct (assumes dividends are null):$\$ \frac{\partial C}{\partial T} = \frac{1}{2}\sigma^2 K^2\frac{\partial^2 C}{\partial K^2} -r K \frac{\...

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