Abstract
The purpose is to establish an automated investing strategy which can imitate an advisor’s behaviour in financial market. In view of the above needs, we review previous studies of Markov regime-switching model whose duration is geometrically distributed, propose a semi-Markov regime-switching model whose duration has a general distribution. By extending the state space of the semi-Markov chain, the model is transformed to a Markov regime-switching model. As the full information of the semi-Markov regime-switching model is available in the issue, we propose a divide-and-conquer and computationally tractable algorithm to estimate parameters. Experiments with empirical datasets show that the automated investing strategy based on estimated parameters behaves like the investment advisor. For an investment advisor, the automated investing strategy can help the advisor to avoid boring routines, and evaluate the advisor’s advice thoroughly.
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Funding
This work was supported by the National Natural Science Foundation of China (Grant No. 11991023 & 12271433), Author Liu J.R. has received the funding of teaching and researching program of Northwest University (Grant No. XM05190195), and the fund of Shaanxi province (21BZ023).
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All authors contributed to the study conception and design. Material preparation, data collection and analysis were performed by QD and JL. The first draft of the manuscript was written by JL and all authors commented on previous versions of the manuscript. All authors read and approved the final manuscript.
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Appendices
Appendix A: Proof of Theorem 3.1
The proof needs the following Lemma A.1.
Lemma A.1
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1)
Assume that \(z_k\ne z_{k+1}\). For \(r= (z_{k+1},1)\), we have that \(P\left( Y_{k+1}=r|A\right) =1\). The conditional probability is 0 for any value of \(r\ne (z_{k+1},1)\).
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2)
Assume that \(m>k\), \(z_k\ne z_{k+1}\) and \(z_m\ne z_{m+1}\). We have that
$$\begin{aligned} P\left( Y_l,Y_{l+1}|A\right) =P\left( Y_l,Y_{l+1}|Y_{k+1}=r,Z_{k+1}=z_{k+1},\ldots ,Z_{m+1}=z_{m+1},Y_{m+1}=s\right) , \end{aligned}$$where \(r=(z_{k+1},1)\) and \(s= (z_{m+1},1)\).
Proof
As \(z_k\ne z_{k+1}\) and \(\mathcal{G}(Y_k)=Z_k\) as well as \(\mathcal{G}(Y_{k+1})=Z_{k+1}\), it follows from the event \(\{ Z_k\ne Z_{k+1}, Z_{k+1}=z_{k+1}\}\) that \(\{\mathcal{G}(Y_k)\ne \mathcal{G}(Y_{k+1})\}\). Note that \(P_{r,s}=0\) when \(\mathcal{G}(r)\ne \mathcal{G}(s)\) and \(\mathcal{H}(s)\ne 1\). Hence if \(z_k\ne z_{k+1}\), \(P(\mathcal{H}(Y_{k+1})= 1|A)=1\). Then 1) follows from \(\mathcal{G}(Y_{k+1})=Z_{k+1}=z_{k+1}\). Now we turn to prove 2). Under the assumptions, similar to the preceding proof, we have that for an event \(B=\{Y_{k+1}=r,,Y_{m+1}=s\}\),
And hence for any \(r',s'\in R_e\),
Therefore,
And the result follows from the Markovian property of \(\{Y_k,Z_k\}\). \(\square \)
Proof of Theorem 3.1
Let \(\log \mathcal{L}\) be
where the last two terms are Lagrange multipliers.
It follows from(6) that for \(i,j \in R_g\) such that \(i\ne j\),, \(\frac{\partial P_{r,s}}{\partial q_{i,j}}=1\) when \(\mathcal{G}(r)=i, \mathcal{G}(s)=j \) and \(\mathcal{H}(s)= 1\). For other cases, we have that \(\frac{\partial P_{r,s}}{\partial q_{i,j}}=0\). Then it follows from
(7) and 1) of Lemma A.1 that for \(i\ne j\)
and the preceding partial derivative is zero when \(i=j\). For a given \(i\in R_g\), we can solve \(\frac{\partial \log \mathcal{L}}{\partial q_{i,j}}=0\) for all \(j\in R_g\) and obtain \(\hat{q}_{i,j}\) as given in 1).
Similarly, we can prove 2) by using 2) of Lemma A.1. \(\square \)
Appendix B: Estimator \(\hat{p}_r^c,\hat{p}_r^e,\hat{p}_r^n\)
Theorem 3.1 ensures that \(\hat{p}_r^c,\hat{p}_r^e,\hat{p}_r^n\) can be calculated group by group, where in one group, every substate \(r=(i,k)\) has the same i. In the following, we manage to establish the estimator for a group of \(r=(i,k)\), \(k=1,2,\ldots ,K\). Without any loss of generality, write the estimator as \(\hat{p}_k^c,\hat{p}_k^e,\hat{p}_k^n\) for \(r=(i,k)\). From 2) of Theorem 3.1, let \(t_l\) denote \(f_l-s_l\) for the given i and suppose that \(l=1,2,\ldots ,L\).
For the random vector \((T_1, \ldots , T_L)\) corresponding to \((t_1, \ldots , t_L)\), it follow from the Markovian property of \(\{Y_n\}\) that \(T_1, \ldots , T_L\) are independent. In fact, \(T_\cdot \) draws from a discrete coxian distribution (Duan & Liu, 2020) with parameter \(p_m^c,p_m^e,p_m^n\), \(m=1,2,\ldots ,K\). Further, we can study an extended coxian random vector (T, I) determined by the sequence \(\{Y_n\}\) such that \(I=1\) when \(Y_{T}=(i,K)\) and \(I=0\) otherwise. That is, I indicates whether the the last state i of time series \(Y_n\), \(n=1,\ldots ,T\) has the substate (i, K).
Then using a similar discussion of Appendix A, we can establish the following Theorem.
Theorem B.1
Suppose (S, J), (T, I) and (U, H) are three extended coxian random vectors. (T, I) draws from the distribution with parameter \(p_m^c,p_m^e,p_m^n\), \(m=1,2,\ldots ,K\), (S, J) is with parameter \(p_m^c,p_m^e,p_m^n\), \(m=1,2,\ldots ,k-1\), and (U, H) is with parameter \(p_m^c,p_m^e,p_m^n\), \(m=1,2,\ldots ,k\), respectively. And V draws from a geometric distribution with parameter \(p_k^c\). For given \(t_1, \ldots , t_L\), we have that
and
where \(W_1=P(S+U=j,W=t_l-j,J=0|T=t_l)\) and \(W_2=P(S=j,W=t_l-j,J=0|T=t_l).\) Moreover, \(\hat{p}_K^e=1\) and if \(k<K,\)
We refer to Appendix of Duan and Liu (2019) the formulae of the probability mass function of a discrete coxian distribution.
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Liu, J., Chen, Z. & Duan, Q. Automation of the Individualized Investing Strategy for an Investment Advisor Established by a Semi-Markov Regime-Switching Model. Comput Econ (2023). https://doi.org/10.1007/s10614-023-10409-z
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DOI: https://doi.org/10.1007/s10614-023-10409-z