Robust Volatility Estimation with and Without the Drift Parameter

Muneer Shaik; S. Maheswaran

doi:10.1007/s40953-018-0129-4

Journal of Quantitative Economics

pp 1–35 | Cite as

Robust Volatility Estimation with and Without the Drift Parameter

Authors
Authors and affiliations

Muneer Shaik
S. Maheswaran

Original Article

First Online: 22 March 2018

Abstract

We find the closed form solution for the joint probability of the running maximum and the drawdown of the Brownian motion with a non-zero drift parameter at a random time that is exponentially distributed and independent of the Brownian motion. This characterization leads us to come up with a robust method of estimating volatility using open, high, low and closing prices. We rigorously show the independence of robust volatility estimators based on extreme values of asset prices relative to the standard robust volatility estimator based on closing price alone. We further prove that the proposed robust volatility ratio is unbiased with no drift parameter. Moreover, we find that the robust volatility ratio with a non-zero drift parameter has only a second order effect. We have shown that our proposed extreme value robust volatility estimator is 2–3 times relatively more efficient when compared to the classical robust volatility estimator based on Monte Carlo simulation experiment. On the empirical side, we test the proposed robust volatility ratio based on high and low prices on different asset classes like stock indices, exchange rate and precious metals.

Keywords

Robust volatility modeling Extreme value estimators Radon Nikodym derivative Brownian motion Drawdown Absolute returns

JEL Classification

C13 C51 C58 G12

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Appendix A

Reflection Principle for Standard Brownian Motion

Theorem 1

The closed form solution for the joint probability density of the stochastic process $X_{t}$ and the running maximum $M_{t}$ that follows the standard Brownian motion with the drift $\mu =0$ and the variance $\sigma ^2$ at a fixed time t is given by

$$\begin{aligned} \begin{aligned} f_{X_{t},M_{t}} (x,b;\mu =0,\sigma ^2,t) = \frac{2u}{\sigma ^3 t^{\frac{3}{2}}} \cdot \phi ( \frac{u}{\sigma .\sqrt{t}}){\bigg |_{\scriptstyle {u=2b-x}}} \end{aligned} \end{aligned}$$

(45)

Proof

Let $\{X^*_{t}, M^*_{t}\}$ denote the value of the stochastic process and the running maximum of the standard Brownian motion at a fixed point in time t.

Now let us define the random time of the Brownian motion for specified level b defined as

$$\begin{aligned} T_{b} = \inf \{t\ge 0,X_{t}\ge b\} \;for\; b\>0 \end{aligned}$$

By using the symmetric property of the Brownian Motion over the set $(t\ge T_{b})$ we have,

$$\begin{aligned} P(X^*_{t}\le x \mid \mathcal {F}_{T_{b}}) = P(X^*_{t}\ge u \mid \mathcal {F}_{T_{b}}) \end{aligned}$$

(46)

Based on the above Eq. (46), we further solve to get the joint probability of the stochastic process and the running maximum for $x\le b, b\ge 0$ where $\{x,b\}$ are the specific levels on the Brownian motion path , as shown below.

$$\begin{aligned} P\{X^*_{t}\le x,M^*_{t} \ge b\} = P\{X^*_{t} \ge u \}{\bigg |_{\scriptstyle {u=2b-x}}} \end{aligned}$$

(47)

Let us consider

$$\begin{aligned} R.H.S&= P\{X^*_{t} \ge u \}{\bigg |_{\scriptstyle {X^*_{t} \sim N(0,t) , u=2b-x}}} \\&= P\{\frac{1}{\sqrt{t}}X^*_{t} \ge \frac{u}{\sqrt{t}} \} = P\{ Z \ge \frac{u}{\sqrt{t}} \} = 1 - \varPhi (\frac{u}{\sqrt{t}}) = \overline{\varPhi }(\frac{u}{\sqrt{t}}) \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} P\{X^*_{t}\le x,M^*_{t} \ge b\} = P\{X^*_{t} \ge u \}{\bigg |_{\scriptstyle {u=2b-x}}} = \overline{\varPhi }(\frac{u}{\sqrt{t}}){\bigg |_{\scriptstyle {u=2b-x}}} \end{aligned} \end{aligned}$$

(48)

We further solve the joint probability density function by applying the partial differential equations and using some properties⁴,⁵,⁶ as shown in the footnote.

$$\begin{aligned} f^*_{X^*,M^*}(x,b)&= P\{X^*_{t} \in dx, M^*_{t} \in db \} = -\frac{\partial ^2}{\partial x \partial b} P\{X^*_{t} \le x, M^*_{t} \ge b \} \\&= -\frac{\partial ^2}{\partial x \partial b} \overline{\varPhi }(\frac{u}{\sqrt{t}}){\bigg |_{\scriptstyle {u=2b-x}}} = 2\cdot \frac{d^2}{du^2}\overline{\varPhi }(\frac{u}{\sqrt{t}}) = 2\cdot \frac{d}{du} \Big [ \frac{d}{du} \overline{\varPhi }(\frac{u}{\sqrt{t}}) \Big ] \\&= 2\cdot \frac{d}{du} \Big [ \frac{d}{du} \int _{\frac{u}{\sqrt{t}}}^{\infty } \phi (w).dw \Big ] = 2\cdot \frac{d}{du} \Big [ - \phi (\frac{u}{\sqrt{t}}) \cdot \frac{1}{\sqrt{t}}\Big ] \\&= 2\cdot (-1)\cdot \frac{1}{\sqrt{t}} \cdot (-\frac{u}{\sqrt{t}}).\phi (\frac{u}{\sqrt{t}}).\frac{1}{\sqrt{t}} = 2\cdot \frac{u}{t^{\frac{3}{2}}} \cdot \phi (\frac{u}{\sqrt{t}}) \end{aligned}$$

Thus the joint probability density of the stochastic process and the running maximum of the standard Brownian motion at a fixed point in time t is defined as

$$\begin{aligned} \begin{aligned} f^*_{X^*,M^*}(x,b)&= 2\cdot \frac{u}{t^{\frac{3}{2}}} \cdot \phi (\frac{u}{\sqrt{t}}){\bigg |_{\scriptstyle {u=2b-x}}}\\&= P\{X^*_{t} \in dx, M^*_{t} \in db \} \equiv \;P_{\mu =0,\sigma ^2=1}\{X_{t} \in dx, M_{t} \in db \} \end{aligned} \end{aligned}$$

(49)

Now let us multiply $X^*_{t} $ and $M^*_{t} $ with a scalar variable $\sigma $.

Let us define $\{X_{t}, M_{t} \}$ as $ X_{t} = \sigma X^*_{t} $ and $ M_{t} = \sigma M^*_{t} $.

Now we define the generic random variables $\{X_{1},X_{2}\}$ for $\{X^*_{t}, M^*_{t} \}$ as,

$$\begin{aligned} X_{1} = X^*_{t} \;\; {\text {and}} \;\; X_{2} = M^*_{t} \end{aligned}$$

Now, the transformed variables can be defined as,

$$\begin{aligned} Y_{1} = \sigma X_{1} \;\; {\text {and}} \;\; Y_{2} = \sigma X_{2} \end{aligned}$$

Now let us apply the inverse transformation and we get,

$$\begin{aligned} X_{1} = \frac{1}{\sigma } Y_{1} \;\;{\text {and}}\;\; X_{2} = \frac{1}{\sigma } Y_{2} \end{aligned}$$

(50)

The matrix notation for the Eq. (50) can be written as,

$$\begin{aligned} \left[ \begin{array}{c} X_1 \\ X_2 \end{array} \right] = \begin{bmatrix} \frac{1}{\sigma }&0 \\ 0&\frac{1}{\sigma }\end{bmatrix} \times \left[ \begin{array}{c} Y_1 \\ Y_2 \end{array} \right] \\ \end{aligned}$$

The Jacobian of this transformation is given by

$$\begin{aligned} J= \begin{bmatrix} \frac{1}{\sigma }&0 \\ 0&\frac{1}{\sigma } \end{bmatrix} \\ \end{aligned}$$

Thus we have, $|det(J) |= \frac{1}{\sigma ^2}$.

Also we have defined u as $u=2b-x = \frac{2y_{2}-y_{1}}{\sigma }. $ Since we have $b=X_{2} = \frac{1}{\sigma } Y_{2}$ and $ x= X_{1} = \frac{1}{\sigma } Y_{1} $. By using Eq. (49) and the determinant of the jacobian of the transformed variables, we further solve the joint probability density as shown below

$$\begin{aligned} f_{Y_{1},Y_{2}} (y_{1},y_{2})&= |det(J) |. f^*_{X_{1},X_{2}} (x_{1},x_{2}){\bigg |_{\scriptstyle {x_{1}=\frac{1}{\sigma } y_{1} , x_{2} = \frac{1}{\sigma } y_{2}}}} = \frac{1}{\sigma ^2} \cdot 2\cdot \frac{u}{t^{\frac{3}{2}}} \cdot \phi (\frac{u}{\sqrt{t}}){\bigg |_{\scriptstyle {u=\frac{2y_{2}-y_{1}}{\sigma }}}}\\&= \frac{1}{\sigma ^2} \cdot 2 \cdot \frac{2y_{2}-y_{1}}{\sigma .t^{\frac{3}{2}}} \cdot \phi \Big [\frac{2y_{2}-y_{1}}{\sigma .\sqrt{t}}\Big ] = \frac{2}{\sigma ^3 t^{\frac{3}{2}}} \cdot \widetilde{u} \phi ( \frac{\widetilde{u}}{\sigma .\sqrt{t}}){\bigg |_{\scriptstyle {\widetilde{u} = 2y_{2}-y_{1}}}} \end{aligned}$$

Therefore the closed form solution for the joint probability density of the stochastic process $X_{t}$ and the running maximum $M_{t}$ that follows the standard Brownian motion with the drift $\mu =0$ and the variance $\sigma ^2$ at a fixed time t is given by

$$\begin{aligned} \begin{aligned} f_{X_{t},M_{t}} (x,b;\mu =0,\sigma ^2,t) = \frac{2u}{\sigma ^3 t^{\frac{3}{2}}} \cdot \phi ( \frac{u}{\sigma .\sqrt{t}}){\bigg |_{\scriptstyle {u=2b-x}}} \end{aligned} \end{aligned}$$

(51)

Hence proved the theorem. $\square $

Definition 1

For all $x\in \mathbb {R}$, we have

$$\begin{aligned} \int _{0}^{\infty } e^{-\lambda t} \Big \{\frac{|x |}{t^{\frac{3}{2}}}\Big \} \phi \Big ( \frac{x}{\sqrt{t}}\Big ) dt = exp\Big \{-\sqrt{(}2 \lambda ) \cdot |x |\Big \} \end{aligned}$$

Proof

Let us suppose that the daily return $x_{t}$ conditional on the unobserved stochastic volatility for day t is distributed normally with mean 0 and variance $\sigma ^2=Y_{t}$.

$$\begin{aligned} (x_{t}\mid Y_{t})\sim N(0,\sigma ^2=Y_{t}) \end{aligned}$$

(52)

Now let us assume that the unobserved stochastic volatility $Y_{t}$ is distributed exponentially with parameter⁷ $\lambda $. That is

$$\begin{aligned} Y_{t} \sim Exp(\lambda ) \end{aligned}$$

(53)

We call this setup as the Exponential Mixture of Normal distribution. We know that the exponential mixture of the normal distribution will be equal to the double exponential probability density function. That is to say we have,

$$\begin{aligned} \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big \{\frac{1}{\sqrt{t}}\varPhi \Big (\frac{x}{\sqrt{t}}\Big )\Big \}dt = \frac{1}{2}\beta {e^{- \beta (|x|)}} \end{aligned}$$

(54)

In order to solve the claim, we make use of the above Eq. (54). We have 2 special cases

Case (i): Let us consider the case of $x\ge 0$.

The L.H.S of the Eq. (49) is given as,

$$\begin{aligned} L.H.S = \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big \{\frac{1}{\sqrt{t}}\varPhi \Big (\frac{x}{\sqrt{t}}\Big )\Big \}dt \end{aligned}$$

Let us differentiate the above equation, we get,

$$\begin{aligned} \frac{d}{dx} (L.H.S)&= \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big \{\frac{\partial }{\partial x}\frac{1}{\sqrt{t}}\varPhi \Big (\frac{x}{\sqrt{t}}\Big )\Big \}dt\\&= \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big (\frac{1}{\sqrt{t}}\Big ) \Big (\frac{-x}{\sqrt{t}}\Big ) \varPhi \Big (\frac{x}{\sqrt{t}}\Big ) \Big (\frac{1}{\sqrt{t}}\Big ) dt \\&= - \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big \{ \frac{x}{t^{\frac{3}{2}}} \varPhi \Big (\frac{x}{\sqrt{t}}\Big ) \Big \} dt \end{aligned}$$

That is to say for $x\>0$, we have

$$\begin{aligned} \begin{aligned} -\frac{d}{dx} (L.H.S) = \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big \{ \frac{x}{t^{\frac{3}{2}}} \varPhi \Big (\frac{x}{\sqrt{t}}\Big ) \Big \} dt \end{aligned} \end{aligned}$$

(55)

Now let us consider the R.H.S of the Eq. (54), we get

$$\begin{aligned} \begin{aligned} -\frac{d}{dx} (R.H.S)&= -\frac{d}{dx} \Big \{\frac{1}{2}\beta {e^{- \beta (|x|)}}\Big \} \\&= -\frac{d}{dx} \Big \{\frac{1}{2}\beta {e^{- \beta x}}\Big \} = -\frac{1}{2}\beta (-\beta ) {e^{- \beta x}} \\&= \frac{\beta ^2}{2} {e^{- \beta x}}{\bigg |_{\scriptstyle {\beta =\sqrt{2\cdot \lambda }}}} \\&= \lambda e^{-\sqrt{2\lambda }|x |} \end{aligned} \end{aligned}$$

(56)

Thus from Eqs. (55) and (56) for all $x\ge 0$ , we have

$$\begin{aligned} \begin{aligned} \int _{0}^{\infty } e^{-\lambda t} \Big \{\frac{|x |}{t^{\frac{3}{2}}}\Big \} \phi \Big ( \frac{x}{\sqrt{t}}\Big ) dt = exp\Big \{-\sqrt{(}2 \lambda ) \cdot |x |\Big \} \end{aligned} \end{aligned}$$

(57)

Case (ii): Let us consider the case of $x<0$.

We have the L.H.S of the Eq. (54) as

$$\begin{aligned} L.H.S = \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big \{\frac{1}{\sqrt{t}}\varPhi \Big (\frac{x}{\sqrt{t}}\Big )\Big \}dt \end{aligned}$$

Let us differentiate the above equation, we get,

$$\begin{aligned} \frac{d}{dx} (L.H.S)&= \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big \{\frac{\partial }{\partial x}\frac{1}{\sqrt{t}}\varPhi \Big (\frac{x}{\sqrt{t}}\Big )\Big \}dt = \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big (\frac{1}{\sqrt{t}}\Big ) \Big (\frac{-x}{\sqrt{t}}\Big ) \varPhi \Big (\frac{x}{\sqrt{t}}\Big ) \Big (\frac{1}{\sqrt{t}}\Big ) dt \\&= - \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big \{ \frac{|x |}{t^{\frac{3}{2}}} \varPhi \Big (\frac{x}{\sqrt{t}}\Big ) \Big \} dt \end{aligned}$$

That is to say for $x<0$, we have

$$\begin{aligned} \frac{d}{dx} (L.H.S) = \int _{t=0}^{\infty } \lambda {e^{- \lambda t}} \Big \{ \frac{x}{t^{\frac{3}{2}}} \varPhi \Big (\frac{|x |}{\sqrt{t}}\Big ) \Big \} dt \end{aligned}$$

(58)

Now let us consider the R.H.S of the Eq. (54), we get

$$\begin{aligned} \begin{aligned} \frac{d}{dx} (R.H.S)&= \frac{d}{dx} \Big \{\frac{1}{2}\beta {e^{- \beta (|x|)}}\Big \} = \frac{d}{dx} \Big \{\frac{1}{2}\beta {e^{ \beta x}}\Big \} \\ = \frac{1}{2}\beta (\beta ) {e^{ \beta x}} \\&= \frac{\beta ^2}{2} {e^{\beta x}}{\bigg |_{\scriptstyle {x<0}}} = \frac{\beta ^2}{2} {e^{-\beta |x |}}{\bigg |_{\scriptstyle {\beta =\sqrt{2\cdot \lambda }}}} = \lambda e^{-\sqrt{2\lambda }|x |} \end{aligned} \end{aligned}$$

(59)

Thus from Eqs. (58) and (59) for all $x<0$ , we have

$$\begin{aligned} \begin{aligned} \int _{0}^{\infty } e^{-\lambda t} \Big \{\frac{|x |}{t^{\frac{3}{2}}}\Big \} \phi \Big ( \frac{x}{\sqrt{t}}\Big ) dt = exp\Big \{-\sqrt{(}2 \lambda ) \cdot |x |\Big \} \end{aligned} \end{aligned}$$

(60)

Thus based on the results of both the special cases i.e. from Eqs. (57) and (60), we can prove the claim that for all $x\in \mathbb {R}$ we have,

$$\begin{aligned} \begin{aligned} \int _{0}^{\infty } e^{-\lambda t} \Big \{\frac{|x |}{t^{\frac{3}{2}}}\Big \} \phi \Big ( \frac{x}{\sqrt{t}}\Big ) dt = exp\Big \{-\sqrt{(}2 \lambda ) \cdot |x |\Big \} \end{aligned} \end{aligned}$$

(61)

Hence the definition is proved. $\square $

Appendix B

Claim 1:

$$\begin{aligned} P\{Y_{2} \>0\} = \frac{\beta _{+}}{\beta _{+} + \beta _{-}} \end{aligned}$$

(62)

Proof

We know that we have defined $ Y_{2} = x $. This can also be written as

$$\begin{aligned} Y_{2} = x = {M_{\tau } - Y_{\tau }}{\bigg |_{\scriptstyle {M_{\tau }=b, Y_{\tau }=b-x}}} \end{aligned}$$

Therefore $ Y_{2} \>0 $ can also be written as $ M_{\tau } \>Y_{\tau } $.

Now let us consider the L.H.S of the claim, we get

$$\begin{aligned} L.H.S&= P\{Y_{2} \>0\} = P\{ M_{\tau } \>Y_{\tau } \} = \int _{y=0}^{\infty } P\{M_{\tau } \>y | Y_{\tau } = y\}\cdot P\{ Y_{\tau } \in dy\} \\&= \int _{0}^{\infty } \Big \{e^{-\beta _{-}\cdot y }\Big \} \cdot \Big \{ \beta _{+}e^{-\beta _{+}\cdot y}\Big \} = \int _{0}^{\infty } \Big \{ e^{-[\beta _{+}+\beta _{-}].y}\Big \}\cdot \beta _{+}\cdot dy \\&= \beta _{+}\cdot \int _{0}^{\infty } \Big \{ e^{-[\beta _{+}+\beta _{-}].y}\Big \} dy = \frac{\beta _{+}}{\beta _{+} + \beta _{-}} = R.H.S \end{aligned}$$

Please check the below footnote⁸ to get details about the above derivation.

Since we have shown $L.H.S = R.H.S$, the claim is proved. $\square $

Note: Similarly we can also show that $P\{Y_{2} <0\} = \frac{\beta _{-}}{\beta _{+} + \beta _{-}}$

Claim 2: $\beta _{+}, \beta _{-}$ and $ \overline{\beta }$ are all strictly positive no matter whether $\theta $ is positive or negative.

Proof

case (i) : Let us consider $\beta _{+} = \sqrt{2 \lambda + {\theta }^2} + \theta $

From the value of $\beta _{+}$ we can clearly say that if $\theta \>0 $ , then $\beta _{+} \>0 $.

Let us look at the value of $\beta _{+}$ when $\theta <0 $.

We have ,

$$\begin{aligned} \beta _{+} = \sqrt{2 \lambda + {\theta }^2} + \theta \end{aligned}$$

Therefore we can write,

$$\begin{aligned}&\sqrt{2 \lambda + {\theta }^2} \>- \theta \\&\Rightarrow \sqrt{2 \lambda + {\theta }^2} \>|\theta |\\&\Rightarrow 2 \lambda + {\theta }^2 \>{|\theta |}^{2} \\&\Rightarrow 2 \lambda + {\theta }^2 \>{ \theta }^{2}\\ \end{aligned}$$

Since $\lambda \>0$, we can say that $ \beta _{+} \>0 $, even when $\theta <0$ as well.

Case (ii): Let us consider $\beta _{-} = \sqrt{2 \lambda + {\theta }^2} - \theta $

From the value of $\beta _{-}$ we can clearly say that if $\theta <0 $ , then $\beta _{-} \>0 $.

Let us look at the value of $\beta _{-}$ when $\theta \>0 $.

We have,

$$\begin{aligned} \beta _{-} = \sqrt{2 \lambda + {\theta }^2} - \theta \end{aligned}$$

Therefore we can write,

$$\begin{aligned}&\sqrt{2 \lambda + {\theta }^2} \>\theta \\&\Rightarrow 2 \lambda + {\theta }^2 \>{ \theta }^{2} \end{aligned}$$

Since $\lambda \>0$, we can say that $ \beta _{-} \>0 $, even when $\theta \>0$ as well.

From both the cases (i) and (ii) , since both $ \beta _{+} {\text {and}} \beta _{-}$ are strictly positive i.e. $ \>0$, no matter whether $\theta \ge 0$ or $\theta <0$. Therefore it follows that their average will also be greater than 0. That is

$$\begin{aligned} \overline{\beta } = \frac{\beta _{+}+\beta {-}}{2} = \sqrt{2 \lambda + {\theta }^2} \>0 \end{aligned}$$

Therefore $\beta _{+}, \beta _{-}$ and $ \overline{\beta }$ are all strictly positive no matter whether $\theta $ is positive or negative. Hence the claim is proved. $\square $

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Muneer Shaik
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S. Maheswaran
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1.Institute for Financial Management and ResearchChennaiIndia

Robust Volatility Estimation with and Without the Drift Parameter

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