Simultaneous estimation of the parameters of the Hurst–Kolmogorov stochastic process

Abstract

Various methods for estimating the self-similarity parameter (Hurst parameter, H) of a Hurst–Kolmogorov stochastic process (HKp) from a time series are available. Most of them rely on some asymptotic properties of processes with Hurst–Kolmogorov behaviour and only estimate the self-similarity parameter. Here we show that the estimation of the Hurst parameter affects the estimation of the standard deviation, a fact that was not given appropriate attention in the literature. We propose the least squares based on variance estimator, and we investigate numerically its performance, which we compare to the least squares based on standard deviation estimator, as well as the maximum likelihood estimator after appropriate streamlining of the latter. These three estimators rely on the structure of the HKp and estimate simultaneously its Hurst parameter and standard deviation. In addition, we test the performance of the three methods for a range of sample sizes and H values, through a simulation study and we compare it with other estimators of the literature.

Appendices

Appendix A: Proof of Eqs. 8 and 9

From Eq. 7 we obtain:

$$ p\left( {{\varvec{\uptheta}}|{\mathbf{x}}_{n} } \right) \, = \, {\frac{1}{{(2\pi )^{n/2} }}}{\frac{1}{{\sigma^{n} }}}\left[ {\det \left( {\mathbf{R}} \right)} \right]^{ - 1/2} \exp \left[ { - {\frac{1}{{2\sigma^{2} }}}\left( {{\mathbf{e}}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{e}}\left( {\mu - {\frac{{x_{n}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{e}}}}{{e^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{e}}}}}} \right)^{2} + {\frac{{{\mathbf{e}}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{ex}}_{n}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{x}}_{n} - \left( {{\mathbf{x}}_{n}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{e}}} \right)^{2} }}{{{\mathbf{e}}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{e}}}}}} \right)} \right] $$

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Since e ^T R ⁻¹ e > 0 (R is positive definite matrix) the maximum of p(θ|x _n ) is achieved when

$$ \hat{\mu } = {\frac{{{\mathbf{x}}_{n}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{e}}}}{{{\mathbf{e}}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{e}}}}} $$

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For that value of μ, taking the logarithm of the posterior density we obtain:

$$ { \ln }[p\left( {{\varvec{\theta}}|{\mathbf{x}}_{n} } \right)] = - \left( {n/ 2} \right){ \ln }\left( { 2\pi } \right) - n\,{\ln }\sigma - \left( { 1/ 2} \right){ \ln }\left[ {{ \det }\left( {\mathbf{R}} \right)} \right] - {\frac{1}{{2\sigma^{2} }}}\left( {{\mathbf{x}}_{n} - \hat{\mu }{\mathbf{e}}} \right)^{\text{T}} \, {\mathbf{R}}^{ - 1} \left( {{\mathbf{x}}_{n} - \hat{\mu }{\mathbf{e}}} \right) $$

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$$ {\frac{{\partial { \ln }[p\left({{\varvec{\theta}}|{\mathbf{x}}_{n} } \right)]}}{\partial \sigma}} = - {\frac{n}{\sigma }} + {\frac{1}{{\sigma^{3} }}}\left({{\mathbf{x}}_{n} - \hat{\mu }{\mathbf{e}}} \right)^{\text{T}} {{\mathbf{R}}^{ - 1} ({\mathbf{x}}_{n} - \hat{\mu}{\mathbf{e}}})$$

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Thus, the logarithm of the maximum posterior density is maximized when ${\frac{{\partial { \ln }[p\left( {{\varvec{\uptheta}}|{\mathbf{x}}_{n} } \right)]}}{\partial \sigma }} = 0$. The solution of this equation proves Eq. 8 and gives the ML estimator of σ.

Substituting the values of μ and σ from Eq. 8, we obtain:

$$ { \ln }\left[ {p\left( {{\varvec{\theta}}|{\mathbf{x}}_{n} }\right)} \right] = \frac{n}{2}\ln \left( {{\frac{n}{2\pi }}}\right) - \frac{n}{2} - \frac{n}{2}\ln \left[ {\left({{\mathbf{x}}_{n}- {\frac{{{\mathbf{x}}_{n}^{\text{T}}{\mathbf{R}}^{ - 1} {\mathbf{e}}}}{{{\mathbf{e}}^{\text{T}}{\mathbf{R}}^{ - 1} {\mathbf{e}}}}}{\mathbf{e}}}\right)^{\text{T}} {\mathbf{R}}^{ - 1} \left( {{\mathbf{x}}_{n}-{\frac{{{\mathbf{x}}_{n}^{\text{T}} {\mathbf{R}}^{ - 1}{\mathbf{e}}}}{{{\mathbf{e}}^{\text{T}} {\mathbf{R}}^{ - 1}{\mathbf{e}}}}}{\mathbf{e}}} \right)} \right] - \frac{1}{2}\ln\left[ {\det \left( {\mathbf{R}} \right)} \right] = \frac{n}{2}\ln\left( {{\frac{n}{2\pi }}} \right) - \frac{n}{2} + g_{1} (H) $$

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which is a function of H through the matrix R. So we maximize the above single-variable function, or equivalently the function g ₁(H), and find $\hat{H}$.

We may observe that it is not necessary to form the entire matrix R and invert it to compute g ₁(H) (It suffices to form a column (ρ₀ … ρ_n−1)^T). Since R is a positive definite Toeplitz matrix we can use the Levinson–Trench–Zohar algorithm as described in Musicus (1988). This algorithm can solve the problem of calculating R ⁻¹ e and ln[det(R)] using only O(n ²) operations and O(n) storage. In contrast, standard methods such as Gaussian elimination or Choleski decomposition generally require O(n ³) operations and O(n ²) storage. This is of critical importance when the time series size is large and computer memory capacity restricts its ability to solve the problem.

Appendix B: Proof of the bracketing of the H in (0, 1] in the LSV solution

In order to examine the behaviour of $\hat{\sigma }$ and g ₂(H) from Eqs. 21 and 22 we calculate the following limits:

$$ \mathop {\lim }\limits_{H \to 1} {\frac{{\alpha_{12}^{2} (H)}}{{\alpha_{11} (H)}}} = \left[ {\sum\limits_{k = 1}^{k'} {{\frac{{\ln (n/k)k^{2} s^{2(k)} }}{{k^{p} }}}} } \right]^{2} \mathord{\left/ {\vphantom {{} {}}} \right. \kern-\nulldelimiterspace}\sum\limits_{k = 1}^{k'} {{\frac{{\ln (n/k)k^{2} }}{{k^{p} }}}} > 0\;{\text{and}}\;\mathop {\lim }\limits_{H \to 1} {\frac{{\alpha_{12} (H)}}{{\alpha_{11} (H)}}} = \infty $$

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Therefore, there is a possibility that g ₂(H) could have a minimum for H = 1 and σ = ∞, when σ tends to infinity from this path: $\sigma = \sqrt {\alpha_{12}({H})/\alpha_{11} ({H}) }$

Then $\begin{gathered}\mathop {\lim }\limits_{H \to 1} {{g}}_{ 2}({{H) = }}\sum\limits_{k = 1}^{k'} {{\frac{{s^{4(k)} }}{{k^{p} }}}- \left( {\sum\limits_{k = 1}^{k'} {{\frac{{\ln (n/k)k^{2}s^{2(k)} }}{{k^{p} }}} } } \right)^{2}\mathord{\left/ {\vphantom{{} {}}} \right. \kern-\nulldelimiterspace}\left( {\sum\limits_{k= 1}^{k'} {{\frac{{\ln (n/k)k^{2} }}{{k^{p} }}}} } \right) }\hfill \\ \hfill \\ \end{gathered}$

Now we prove e ²(σ, H) attains its minimum for H ≤ 1. The proof is given bellow:

Suppose that H ₂ > 1 and σ ₂ > 0 (It’s easy to prove that an estimated $\hat{\sigma } > 0$ always). Now for any H ₁ ∈ (0, 1) we can always find a σ ₁ > 0, such that c _k (H ₁) $k^{{ 2H_{1} }}$ $\sigma_{1}^{2}$ − s ^2(k) < 0 for every k. For these values of H ₁ and σ ₁: | c _k (H ₁) $k^{{ 2H_{1} }}$ $\sigma_{1}^{2}$ − s ^2(k) | < | c _k (H ₂) $k^{{ 2H_{ 2} }}$ $\sigma_{2}^{2}$ − s ^2(k) | for every k. This proves that e ²(σ₁, H ₁) < e ²(σ₂, H ₂). Thus, e ²(σ, H) attains its minimum for H ≤ 1.

Appendix C: Calculation of Fisher Information Matrix’s elements

We can easily calculate the I ₁₂(θ), I ₁₃(θ) and I ₂₃(θ) elements of the Fisher Information Matrix (Robert 2007, p. 129):

$$ {\frac{{\partial { \ln }\left[ {p\left( {{\varvec{\uptheta}}|{\mathbf{x}}_{n} } \right)} \right]}}{\partial \mu }} = -{\frac{1}{{\sigma^{2} }}}\left( {{\mathbf{e}}^{\text{T}} {\mathbf{R}}^{ - 1} e\mu - {\mathbf{x}}_{n}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{e}}} \right) $$

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$$ {\frac{{\partial { \ln }\left[ {p\left( {{\varvec{\uptheta}}|{\mathbf{x}}_{n} } \right)} \right]}}{\partial \sigma }} = - {\frac{n}{\sigma }} + {\frac{1}{{\sigma^{3} }}}\left( {{\mathbf{x}}_{n} - \mu {\mathbf{e}}} \right)^{\text{T}} {\mathbf{R}}^{ - 1} \left( {{\mathbf{x}}_{n} - \mu {\mathbf{e}}} \right) $$

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$$ {\frac{{\partial^2 { \ln }\left[ {p\left( {{\varvec{\uptheta}}|{\mathbf{x}}_{n} } \right)} \right]}}{\partial \mu\, \partial \sigma }} = {\frac{2}{{\sigma^{3} }}}\left( {{\mathbf{e}}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{e}}\mu - {\mathbf{x}}_{n}^{\text{T}} {\mathbf{R}}^{ - 1} {\mathbf{e}}} \right) $$

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$$ {\frac{{\partial^2 { \ln }\left[ {p\left( {{\varvec{\uptheta}}|{\mathbf{x}}_{n} } \right)} \right]}}{\partial \mu\, \partial H}} = {\frac{1}{{\sigma^{2} }}}\left( {\mu {\mathbf{e}}^{\text{T}} {\mathbf{R}}^{ - 1} {\frac{{\partial {\mathbf{R}}}}{\partial H}}{\mathbf{R}}^{ - 1} {\mathbf{e}} - {\mathbf{x}}^{\text{T}}_{n} {\mathbf{R}}^{ - 1} {\frac{{\partial {\mathbf{R}}}}{\partial H}}{\mathbf{R}}^{ - 1} {\mathbf{e}}} \right) $$

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$$ {\frac{{\partial^2 { \ln }\left[ {p\left( {{\varvec{\uptheta}}|{\mathbf{x}}_{n} } \right)} \right]}}{\partial \mu\, \partial H}} = - {\frac{1}{{\sigma^{3} }}}({\mathbf{x}}_{n} - \mu {\mathbf{e}})^{\text{T}} {\mathbf{R}}^{ - 1} {\frac{{\partial {\mathbf{R}}}}{\partial H}}{\mathbf{R}}^{ - 1} ({\mathbf{x}}_{n} - \mu {\mathbf{e}}) $$

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The expectations of the above expressions are easily calculated and give the corresponding elements of the Fisher Information Matrix I(θ).