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Volatility

  • Implied Volatility

Option price in the Black-Scholes-Merton formula can be seen as a function of volatility \sigma if interest rate r and strike price K are known. Market quote of option price gives the volatility which is called implied volatility. Market convention is annualized volatility.

  • Realized Volatility

Observing the prices of a stock every day in a year, we can compute the realized volatility.

Suppose the stock price follows geometric Brownian Motion, i.e.,

    \[S(t)=S_0e^{\alpha-\frac{1}{2}\sigma^2t + \sigma W(t)},\]

where W(t) is the standard Brownian Motion, and time t takes year as unit.

Then, for n=1,2, \ldots, 251, the number of stock trading days of NYSE in a year.

    \[\frac{S_{n}}{S_{n-1}} = e^{\frac{\sigma^2}{251}+\sigma\left(W\left(\frac{n}{251}\right)-W\left(\frac{n-1}{251}\right)\right)} \overset{\sD}{=} e^{\frac{\sigma^2}{251}}e^{\sigma W\left(\frac{1}{251}\right)}.\]

    \[\log(\frac{S_{n}}{S_{n-1}}) = \frac{\sigma^2}{251} + \sigma W\left(\frac{1}{251}\right)\]

Hence,

    \[\sigma = \sqrt{251 \cdot \mbox{imperical variance}{\log(\frac{S_{n}}{S_{n-1}})}}.\]

Sometimes, people use \log(x)\sim x-1 near x=1. In this case,

    \[\sigma = \sqrt{251 \cdot \mbox{imperical variance}{\frac{S_{n}}{S_{n-1}}}}.\]

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Geometric Brownian Motion

Denote the stock price at time t by S(t) for t\geq 0. {S(t),t\geq 0} is a stochastic process adapted to a filtration (\mathcal{F}_t)_{t\geq 0}. W(t) is the one-dimensional standard Brownian motion. We assume S(t) satisfies the following stochastic differential equation(SDE):

(1)   \begin{equation*}\mathrm{d} S(t) = \alpha S(t) \mathrm{d} t + \sigma S(t) \mathrm{d} W(t),\end{equation*}

where \alpha is the return rate of the stock, and \sigma represent the volatility of the stock. The left side of the equation represents the change of stock price, and the right side of the equation is the sum of return and noise that are proportional to the stock price S(t).

The solution to (1) is a geometric Brownian motion. It can be solved by the following way. Solution to ODE y'=\alpha y is y(0)e^{\alpha t}. It is reasonable to guess the solution to (1) is S(t)=S(0)e^{at+bW(t)} with a,b to be determined. We then apply Ito’s formula to f(x,y)=e^{ax+by}.

    \[\begin{aligned}\mathrm{d} S(t) = &\ f_x(t,S(t))\mathrm{d} t + f_y(t,S(t))\mathrm{d} W(t) \\& + \frac{1}{2}f_{xx}(t,S(t))\mathrm{d} \langle t,t \rangle + \frac{1}{2} f_{xy}(t,S(t)) \mathrm{d} \langle t,W(t) \rangle + \frac{1}{2}f_{yy}(t,S(t)) \mathrm{d}\langle W,W\rangle_t\\= &\ ae^{at+bW(t)} \mathrm{d} t + be^{at+bW(t)} \mathrm{d} W(t) \\ & + \frac{1}{2}f{xx}(t,S(t)) \mathrm{d} 0+\frac{1}{2}f_{xy}(t,S(t)) \mathrm{d} 0+\frac{1}{2} b^2 e^{at+bW(t)} \mathrm{d} t \\= & \ (a+\frac{1}{2}b^2)e^{at+bW(t)} \mathrm{d} t + be^{at+bW(t)} \mathrm{d} W(t) \\= &\ (a+\frac{1}{2}b^2)S(t) \mathrm{d} t + bS(t) \mathrm{d} W(t)\end{aligned}\]

By letting a+\frac{1}{2}b^2 = \alpha and b= \sigma, and solving for a,b, we will get

(2)   \begin{equation*}S(t) = S(0)\exp((\alpha-\frac{1}{2}\sigma^2)t + \sigma W(t)).\end{equation*}