Bayesian Estimation Under the t-Distribution for Financial Time Series

Abstract

This chapter studies Student’s t-distribution for fitting serially correlated observations where serial dependence is described by the copula-based Markov chain. Due to the computational difficulty of obtaining maximum likelihood estimates, alternatively, we develop Bayesian inference using the empirical Bayes method through the resampling procedure. We provide a Metropolis–Hastings algorithm to simulate the posterior distribution. We also analyze the stock price data in empirical studies for illustration.

Appendix: Moment Estimates

Using $m_{1}=E(Y)=\mu ,$ we first obtain the moment estimates of $\mu$ given by

$$ \tilde{\mu }_{(c)}=m^{(c)}_1=\frac{1}{n}\sum _{t=1}^nY^{(c)}_t. $$

Based on $m_{2}=E(Y^2)=\mu ^{2}+\sigma ^{2}(\frac{\nu }{\nu -2})$, and $m_{4}=E(Y^4)=\mu ^{4}+6\mu ^{2}\sigma ^{2}(\frac{\nu }{\nu -2})+\sigma ^{4}\frac{3\nu ^{2}}{(\nu -2)(\nu -4)}$, we have

$$ m^{(c)}_2={m^{(c)}_1}^2+\tilde{\sigma }_{(c)}^{2}\left( \frac{\tilde{\nu }_{(c)}}{\tilde{\nu }_{(c)}-2}\right) , $$

and

$$ m^{(c)}_4={m^{(c)}_1}^4+6{m^{(c)}_1}^2\tilde{\sigma }_{(c)}^2\left( \frac{\tilde{\nu }_{(c)}}{\tilde{\nu }_{(c)}-2}\right) +\tilde{\sigma }_{(c)}^{4}\frac{3\tilde{\nu }_{(c)}^{2}}{(\tilde{\nu }_{(c)}-2)(\tilde{\nu }_{(c)}-4)}, $$

where $m^{(c)}_2=\frac{1}{n}\sum _{t=1}^n\left( Y^{(c)}_t\right) ^2$ and $m^{(c)}_4=\frac{1}{n}\sum _{t=1}^n\left( Y^{(c)}_t\right) ^4$. Then, using $\tilde{\sigma }_{(c)}^{2}(\frac{\tilde{\nu }_{(c)}}{\tilde{\nu }_{(c)}-2})=m^{(c)}_2-{m^{(c)}_1}^2$, we obtain

$$ m^{(c)}_4={m^{(c)}_1}^4+6{m^{(c)}_1}^2(m^{(c)}_2-{m^{(c)}_1}^2)+3\bigg (m^{(c)}_2-{m^{(c)}_1}^2\bigg )^2\frac{\tilde{\nu }_{(c)}-2}{\tilde{\nu }_{(c)}-4}. $$

Let

$$ \xi ^{(c)}=\frac{\tilde{\nu }_{(c)}-2}{\tilde{\nu }_{(c)}-4}= \frac{{m^{(c)}_{4}}-{m^{(c)}_{1}}^{4}-6{m^{(c)}_{1}}^{2}({m^{(c)}_{2}}{(c)}-{m^{(c)}_{1}}^{2})}{3({m^{(c)}_{2}}-{m^{(c)}_{1}}^{2})^2} . $$

Therefore

$$\tilde{\nu }_{(c)}=\frac{4\xi ^{(c)}-2}{\xi ^{(c)}-1}.$$

Note that we impose $\xi ^{(c)}=\max \bigg \{\frac{{m^{(c)}_{4}}-{m^{(c)}_{1}}^{4}-6{m^{(c)}_{1}}^{2}({m^{(c)}_{2}}{(c)}-{m^{(c)}_{1}}^{2})}{3({m^{(c)}_{2}}-{m^{(c)}_{1}}^{2})^2},1\bigg \}$ such that $\tilde{\nu }>4$ can be guaranteed. In addition, plugging $\tilde{\nu }_{(c)}=\frac{4\xi ^{(c)}-2}{\xi ^{(c)}-1}$ into $\tilde{\sigma }_{(c)}^{2}(\frac{\tilde{\nu }_{(c)}}{\tilde{\nu }_{(c)}-2})=m^{(c)}_2-{m^{(c)}_1}^2$ implies

$$ \tilde{\sigma }^{2}_{(c)}=({m_2^{(c)}}-{{m^{(c)}_{1}}}^2)\frac{ \xi ^{(c)}}{2\xi ^{(c)}-1}. $$