Stock Price Forecasting Based on Multi-Input Hamacher T-Norm and ANFIS

Abstract

This study proposed the expression and some feathers of multi-input Hamacher T-norm first, then constructed the multi-input Hamacher T-norm based ANFIS, finally forecasted the stock price of Pingan Bank (000001) from Jan. 4, 2014 to AUG. 28, 2014, by constructing a model based on Multi-input Hamacher T-norm and ANFIS (Adaptive Neuro-Fuzzy Inference System). 5-cross fold validation has been used in this study to recognize the its best performance in MSE (Mean Square Error), MAE (Mean Absolute Error) and MAPE (Mean Absolute Percentage Error) belongs to high prediction accuracy, while it’s also superior to regular ANIFS. Therefore, Multi-input Hamacher T-norm could improve the performance of ANFIS in stock price forecasting.

Appendices

Appendix A

Proof

When $0 \le j < n\mathrm{{ }}1 \le i \le n$,

${{\chi }^{j}}({A_n}) = \sum \limits _{{c_1}, \ldots ,{c_j} \in \{ 1, \ldots ,i - 1,i + 1, \ldots ,n\} ,{c_1} \ne \cdots \ne {c_j}} ({a_{{c_1}}}{a_{{c_2}}}{a_{{c_3}}} \cdots {a_{{c_j}}})$

$= \sum \limits _{{c_1}, \ldots ,{c_j} \in \{ 1, \ldots ,i - 1,i + 1, \ldots ,n\} ,{c_1} \ne \cdots \ne {c_j}} ({a_{{c_1}}}{a_{{c_2}}}{a_{{c_3}}} \cdots {a_{{c_j}}})$

$+ \,{a_i}\sum \limits _{{c_1}, \ldots ,{c_j} - 1 \in \{ 1, \ldots ,i - 1,i + 1, \ldots ,n\} ,{c_1} \ne \cdots \ne {c_{j - 1}}} ({a_{{c_1}}}{a_{{c_2}}}{a_{{c_3}}} \cdots {a_{{c_{j - 1}}}})$

$= {\chi ^j}({A_n}\backslash {a_i}) + {a_i}{\chi ^{j - 1}}({A_n}\backslash {a_i})$.

Which completes the proof.

Proof

When $j \ne n$,

${\chi ^n}({A_n}) = {a_1}{a_2}{a_3} \cdots {a_n}$ $= {a_n}({a_1}{a_2}{a_3} \cdots {a_n})$ $= {a_n}{\chi ^n}({A_n}).$

Which completes the proof.

Proof

When $0 \le j < n,1 \le i \le n$,

$$\frac{{\partial {\chi ^j}({A_n})}}{{\partial {a_i}}} = \frac{{\partial {\chi ^j}({A_n}\backslash {a_i})}}{{\partial {a_i}}} + \frac{{\partial ({a_i}{\chi ^{j - 1}}({A_n}\backslash {a_i}))}}{{\partial {a_i}}}= {\chi ^{j - 1}}({A_n}\backslash {a_i}).$$

Which completes the proof.

Appendix B

Proof

${\lambda _1},{\lambda _2} \in [0, + \infty ]$ and ${\lambda _1} < {\lambda _2}$, when $n = 1, {T_{{\lambda _1}}}({A_2}) = {T_{{\lambda _1}}}({a_1},{a_2}) \ge {T_{{\lambda _2}}}({a_1},{a_2})$ $\ge {T_{{\lambda _2}}}({A_2})$. Especially, when ${a_1},{a_2} \ne 1$ and ${a_1},{a_2} \ne 0, {T_\lambda }({A_n})$ is strictly decreasing with respect to $\lambda$. The proposition is confirmed.

Assume when $n=t-1$,the proposition is right too, then when $n=t, {T_{{\lambda _1}}}({A_{t + 1}}) = {T_{{\lambda _1}}}({T_{{\lambda _1}}}({A_t}),{a_{t + 1}}) \ge {T_{{\lambda _1}}}({T_{{\lambda _2}}}({A_t}),{a_{t + 1}}) \ge {T_{{\lambda _2}}}({T_{{\lambda _2}}}({A_t}),{a_{t + 1}}) = {T_{{\lambda _2}}}({A_{t + 1}})$. Especially, when $\forall i \in [0,n + 1],\mathrm{{ }}{a_i} \ne 1$ and ${a_i} \ne 0,\mathrm{{ }}{T_\lambda }({A_n})$ is strictly decreasing with respect to $\lambda$.

Proof

The proof is given below: Let

$${\mathrm{{Q}}_n} = {\lambda ^{n - 1}} + \sum \limits _{j = 1}^{n - 1}, {\lambda ^{n - j - 1}}{(1 - \lambda )^j}{\chi ^j}({A_n}) - \sum \limits _{i = 1}^{n - 1} {(1 - \lambda )^i}{\chi ^n}({A_n}),$$

so ${T_\lambda }({A_n}) = \frac{{{\chi ^n}({A_n})}}{{{Q_n}}}.$ When $n=2$,

$$\begin{aligned} {T_\lambda }({A_2})&~= \frac{{{a_1}{a_2}}}{{\lambda + (1 - \lambda )({a_1} + {a_2} - {a_1}{a_2})}} \\&~= \frac{{{a_1}{a_2}}}{{\lambda + (1 - \lambda )({a_1} + {a_2}) + (\lambda - 1){a_1}{a_2}}} \\&~= \frac{{{\chi ^2}({A_2})}}{{{\lambda ^{2 - 1}} + \sum \limits _{j = 1}^{2 - 1} {\lambda ^{1 - j}}{{(1 - \lambda )}^j}{\chi ^j}({A_2}) - \sum \limits _{i = 1}^{2 - 1} {{(\lambda - 1)}^i}{\chi ^2}({A_2})}} \\&= \frac{{{\chi ^2}({A_2})}}{{{Q_2}}} \end{aligned}$$

The proposition is right.

Assume when $n=t$, proposition is right too. So, ${T_\lambda }({A_t}) = \frac{{{\chi ^t}({A_t})}}{{{Q_t}}},$

$$\begin{aligned} {T_\lambda }({A_{t + 1}})&~ = \frac{{\frac{{{\chi ^t}({A_t})}}{{{Q_t}}}{a_{t + 1}}}}{{\lambda + (1 - \lambda )(\frac{{{\chi ^t}({A_t})}}{{{Q_t}}} + {a_{t + 1}}) - (1 - \lambda ){a_{t + 1}}\frac{{{\chi ^t}({A_t})}}{{{Q_t}}}}}\\&~ = \frac{{{\chi ^{t + 1}}({A_{t + 1}})}}{{\lambda {Q_t} + (1 - \lambda ){\chi ^t}({A_t}) - (1 - \lambda ){\chi ^{t + 1}}({A_{t + 1}}) + (1 - \lambda ){a_{t + 1}}{Q_t}}}, \end{aligned}$$

$\lambda {Q_t} = \lambda \sum \limits _{j = 0}^{t - 1} {\lambda ^{t - j - 1}}{(1 - \lambda )^j}{\chi ^j}({A_t}) + \lambda \frac{{{{(1 - \lambda )}^t} - (1 - \lambda )}}{\lambda }{\chi ^t}({A_t})$

$= \sum \limits _{j = 0}^{t - 1} {\lambda ^{t - j}}{(1 - \lambda )^j}{\chi ^j}({A_t}) + [(1 - \lambda {)^t} - (1 - \lambda )]{\chi ^t}({A_t})$,

$(1 - \lambda ){a_{t + 1}}{Q_t} = \sum \limits _{j = 0}^{t - 1} {\lambda ^{t - j - 1}}{(1 - \lambda )^{j + 1}}{a_{t + 1}}{\chi ^{j + 1 - 1}}({A_t}) + \frac{{[(1 - \lambda {)^{t + 1}} - {{(1 - \lambda )}^2}]}}{\lambda }{a_{t + 1}}{\chi ^t}({A_t})$

$= \sum \limits _{j = 1}^t {\lambda ^{t - j}}{(1 - \lambda )^j}{a_{t + 1}}{\chi ^{j - 1}}({A_t}) + \frac{{[(1 - \lambda {)^{t + 1}} - {{(1 - \lambda )}^2}]}}{\lambda }{\chi ^{t + 1}}({A_{t + 1}})$

$\lambda {Q_t} + (1 - \lambda ){\chi ^t}({A_t}) - (1 - \lambda ){\chi ^{t + 1}}({A_{t + 1}}) + (1 - \lambda ){a_{t + 1}}{Q_t}$

$= {\lambda ^t} + \sum \limits _{j = 1}^{t - 1} {\lambda ^{t - j}}{(1 - \lambda )^j}{\chi ^j}({A_t}) + \sum \limits _{j = 1}^t {\lambda ^{t - j}}{(1 - \lambda )^j}{a_{t + 1}}{\chi ^{j - 1}}({A_t})$

$+\,[(1 - \lambda {)^t} - (1 - \lambda )]{\chi ^t}({A_t}) + \frac{{[(1 - \lambda {)^{t + 1}} - {{(1 - \lambda )}^2}]}}{\lambda }{a_{t + 1}}{\chi ^{t + 1}}({A_{t + 1}})$

$+\,(1 - \lambda ){\chi ^t}({A_t}) - (1 - \lambda ){\chi ^{t + 1}}({A_{t + 1}}))$

$= {\lambda ^t} + \sum \limits _{j = 1}^{t - 1} {\lambda ^{t - j}}{(1 - \lambda )^j}({\chi ^j}({A_t}) + {a_{n + 1}}{\chi ^{j - 1}}({A_t})\,+ {(1 - \lambda )^t}f({a_{t + 1}})\chi _{t - 1,f( \cdot )}^{t - 1}({A_t}) + {(1 - \lambda )^t}{\chi ^t}({A_t}) - (1 - \lambda ){\chi ^{t + 1}}({A_{t + 1}})$

$+ \frac{{[(1 - \lambda {)^{t + 1}} - {{(1 - \lambda )}^2}]}}{\lambda }{\chi ^{t + 1}}({A_{t + 1}})$

$= {\lambda ^t} + \sum \limits _{j = 1}^{t - 1} {\lambda ^{t - j}}{(1 - \lambda )^j}{\chi ^j}({A_{t + 1}}) + {(1 - \lambda )^t}({\chi ^t}({A_{t + 1}}) - {\chi ^t}({A_t}) + {(1 - \lambda )^t}{\chi ^t}({A_t}) + \frac{{{{(1 - \lambda )}^{t + 1}} - {{(1 - \lambda )}^2} - \lambda + {\lambda ^2}}}{\lambda }{\chi ^{t + 1}}({A_{t + 1}})$

$= {\lambda ^t} + \sum \limits _{j = 1}^{t - 1} {\lambda ^{t - j}}{(1 - \lambda )^j}{\chi ^j}({A_{t + 1}}) + {(1 - \lambda )^t}({\chi ^t}({A_{t + 1}}) - {\chi ^t}({A_t}) + {(1 - \lambda )^t}{\chi ^t}({A_t}) + \frac{{{{(1 - \lambda )}^{t + 1}} - {{(1 - \lambda )}^2} - \lambda + {\lambda ^2}}}{\lambda }{\chi ^{t + 1}}({A_{t + 1}})$

$= {\lambda ^t} + \sum \limits _{j = 1}^t {\lambda ^{t - j}}{(1 - \lambda )^j}{\chi ^j}({A_{t + 1}}) + \frac{{{{(1 - \lambda )}^{t + 1}} - (1 - \lambda )}}{\lambda }{\chi ^{t + 1}}({A_{t + 1}})$

$= {\lambda ^t} + \sum \limits _{j = 1}^t {\lambda ^{t - j}}{(1 - \lambda )^j}{\chi ^j}({A_{t + 1}}) - \sum \limits _{i = 1}^t {(1 - \lambda )^i}{\chi ^{t + 1}}({A_{t + 1}})$

$= {Q_{t + 1}}$.

So, ${T_\lambda }({A_{t + 1}}) = \frac{{{\chi ^{t + 1}}}}{{{Q_{t + 1}}}}$.

In conclusion, when $n \in {N^ + },{T_\lambda }({A_n}) = \frac{{{\chi ^n}({A_n})}}{{{\lambda ^{n - 1}} + \sum \limits _{j = 1}^{n - 1} {\lambda ^{n - j - 1}}{{(1 - \lambda )}^j}{\chi ^j}({A_n}) - \sum \limits _{i = 1}^{n - 1} {{(1 - \lambda )}^i}{\chi ^n}({A_n})}}$.

Which completes the proof.

Proof

$\frac{{\partial {Q_n}}}{{\partial \lambda }} = (n - 1){\lambda ^{n - 2}} + \sum \limits _{i = 1}^{n - 1} (n - i - 1){\lambda ^{n - i - 2}}{(1 - \lambda )^i}{\chi ^i}({A_n}) - {\lambda ^{n - i - 1}}{(1 - \lambda )^{i - 1}}$ $\,{\chi ^i}{A_n} + \sum \limits _{i = 1}^{n - 1} i{(1 - \lambda )^{i - 1}}{\chi ^n}({A_n})$

$= (n - 1){\lambda ^{n - 2}} + \sum \limits _{i = 1}^{n - 1} (n - i - 1){\lambda ^{n - i - 2}}{(1 - \lambda )^i}{\chi ^i}({A_n}) - {\lambda ^{n - i - 1}}{(1 - \lambda )^{i - 1}}{\chi ^i}{A_n} + \sum \limits _{i = 1}^{n - 1} i{(1 - \lambda )^{i - 1}}{\chi ^n}({A_n}) + + \sum \limits _{i = 1}^{n - 1} i{(1 - \lambda )^{i - 1}}{\chi ^n}({A_n})$

$(n - 1){\lambda ^{n - 2}} + \sum \limits _{i = 1}^{n - 1} {\lambda ^{n - i - 2}}{(1 - \lambda )^{i - 1}}((n - 1) - (n - 1)\lambda - i){\chi ^i}({A_n}) + \sum \limits _{i = 1}^{n - 1} i{(1 - \lambda )^{i - 1}}$

$\,{\chi ^n}({A_n}) = {R_n}$.

$$\frac{{\partial {T_n}({A_n})}}{{\partial \lambda }} = \frac{{\partial \frac{{{\chi ^n}({A_n})}}{{{Q_n}}}}}{{\partial \lambda }} = \frac{{ - {\chi ^n}({A_n}){R_n}}}{{Q_n^2}}.$$

Which completes the proof.