Gray Markov model matlab implementation

Gray prediction GM(1,1)

• Treat the prediction sequence X 0 = { x 0 , 1 , x 0 , 2 , ⋯ , x 0 , n } 0 , n } \}X0​={ x0,1​,x0,2​,⋯,x0,n​} , generate$X_0$An accumulation sequence of $X_1=\{x_{1,1},x_{1,2},\cdots ,x_{1,n}\}$,其中：
  $$x_{1,k}=\sum _{t=1}^k{x}_{0,i}(k=1,2,\cdots ,n)$$
• Conduct a rank comparison test on the original data. First calculate the level ratio $r_k$序列：
  $$r_k=\frac{x_{0,k-1}}{x_{0,k}}(k=2,\cdots ,n)$$
• Then judge $r_k$Whether they are all in the admissibility coverage interval $\partial =(e^{-2/(n+1)},e^{2/(n+1)})$. If so, the corresponding data sequence can establish gray$GM(1,1)$ model; otherwise, the appropriate constant$b$ Perform translational transformation processing on this set of data, so that the processed data sequence$Y_0=y_{0,1},y_{0,2},\cdots ,y_{0,n}$The level ratio of falls within the admissibility coverage interval, and its translation transformation process is:
  $$y_{0,k}=x_{0,k}+b$$
  passes through the accumulation sequence$X_1$, establish Nansihu gray GM (1, 1) GM (1, 1)First-order differential equation of$GM$$($$1$$,$$1$$)$
  $$\frac{d{X}_1}{dt}+\alpha X_1=q$$
  where:$a,q$ are the development coefficient and gray action amount respectively.
• 设$a=(a,q{)}^T$ uses the least squares method to solve:
  $$a=(a,q{)}^T=(B^{T}B{)}^{-1}{B}^{T}D$$
  其中$$B=\left[\begin{array}{ll}-0.5(x_1+x_2)& 1\\ \cdots & \cdots \\ (-0.5x_{n-1}+x_n)& 1\end{array}\right]D=\left[\begin{array}{l}{x}_{0,2}\\ x_{0,0,2}\end{array}\right]$$
• Obtain the gray GM(1,1) model:
  $${\hat{x}}_{1,k+1}=(x_{0,1}-\frac{q}{a})e^{-and}+\frac{q}{a}$$
• The accumulated value ${\hat{x}}_{1,k+1}$After a cumulative reduction, it is restored to the predicted value ${\hat{x}}_{0,k+1}$
  $${\hat{x}}_{0,k+1}={\hat{x}}_{1,k+1}-{\hat{x}}_{1,k}$$

Model testing

In order to test the credibility of the model, the posterior difference test needs to be performed on the predicted values. Establish a first-order residual sequence:
E 0 = { e 0 , 1 , e 0 , 2 , ⋯ , e 0 , n } = { x ^ 0 , 2 − x ^ 0 , 2 , x ^ 0 , 3 − x ^ 0 , 3 , ⋯ , x ^ 0 , k − x ^ 0 , k , x ^ 0 , k } E _ { 0 } = \{ e _ { 0 , 1 } , e _ { 0 , 2 } , \cdots , e _ { 0 , n } \} =\\ \{ \hat{x}_ { 0 , 2 } - \hat{x}_ { 0 , 2 } , \hat{x}_ { 0 , 3 } - \hat{x}_ { 0 , 3 } , \cdots , \hat{x}_ { 0 , k } - \hat{x}_ { 0 , k } , \hat{x}_ { 0 , k } \}E0​={ e0,1​,e0,2​,⋯,e0,n​}={ x^0,2​−x^0,2​,x^0,3​−x^0,3​,⋯,x^0,k​−x^0,k​,x^0,k​}
The variance of the original data sequence is$s_1$, the residual sequence $E_0$The variance of is $s_2$, respectively calculate the posterior ratio $c$ and small error probability$p$:
c = s 2 s 1 p = { 1 4 0 , k − A 0 ∣ < 0.6745 s 1 } c = \frac { s _ { 2 } } { s _ { 1 } }\\ p = \{ 14 _ { 0 , k } - A _ { 0 } | \lt 0.6745 s _ { 1 } \} c=s1​s2​​p={ 140,k​−A0​∣<0.6745 s1​}
wherepp$p$ and$The\; size\; of\; c$ jointly determines the model accuracy level. The table gives the model accuracy levels of 4 levels: good, qualified, basically qualified and unqualified. model$The\; smaller\; c$ is,$The\; larger\; p$ is, the higher the model accuracy is. $The\; smaller\; c$ is, then$s_1$The larger, $s_2$The smaller the value, that is, the greater the discrete degree of the original data sequence, the smaller the discrete degree of the residual sequence, and the smaller the difference between the predicted value obtained by the model and the original data, $A\; larger\; p$ value indicates that the predicted values ​​are more uniform. If the inspection accuracy level meets the requirements, the established gray$The\; GM(1,1)$ model can directly predict the data; if the accuracy level does not meet the requirements, the predicted data will be corrected.

Gray Markov prediction model

对$E_0$Create gray $GM(1,1)$ results:
$${\hat{e}}_{1,k+1}=(e_{0,2}-\frac{q^{\prime }}{a^{\prime }})e^{-a^{\prime t}}+\frac{q^{\prime }}{a^{\prime }}(k=2,3,\cdots ,n)$$
Perform cumulative reduction and reduction of the model to obtain the residual correction value${\hat{e}}_{0,k+1}$:
$${\hat{e}}_{0,k+1}={\hat{e}}_{1,k+1}-{\hat{e}}_{1,k}$$
Use the residual correction value ${\hat{e}}_{0,k+1}$The prediction values ​​of the traditional GM(1,1) model are corrected and the corrected values ​​are obtained:
$${\hat{x}}_{0,k}^{\prime }={\hat{x}}_{0,k}+m_{0,k}{\hat{e}}_{0,k+1}$$
For example:
m 0 , k = { 1 ( x 0 , k − x 0 , k > 0 ) − 1 ( x 0 , k − x 0 , k < 0 ) m _ { 0 , k } = \{ \begin{ array}{ll}{1}&{(x_{0,k}-x_{0,k}\gt0)}\\{-1}&{(x_{0,k}-x_ { 0 , k } \lt 0 ) } \ end { array }m0,k​={ 1−1​(x0,k​−x0,k​>0)(x0,k​−x0,k​<0)​
Introduce the gray Markov model to determine $m_{0,k}$positive or negative. It is suitable for predicting random and irregular data, making up for the traditional $The\; GM(1,1)$ model has the disadvantage of low prediction accuracy for volatility and trend data. The calculation process is as follows:

• According to $E_0$Division status. This article divides two states, state 1 means the residual is positive, and state 2 means the residual is negative.
• ．Find from state $i$ moves to statejjThe probability pij p_{ij}$of\; the\; number\; of\; times\; j$ passes$p_{ij}$:
  $$p_{ij}=\frac{M_{ij}}{M_i}(i=1,2;j=1,2)$$
  In formula:$M_{ij}$for state $i$ moves to state$The\; number\; of\; times\; j$ passes;$M_i$for state $The\; total\; number\; of\; times\; i$ appears. Obtain the state transition matrix$P$
  $$P=\left[\begin{array}{ll}{P}_{11}& P_{12}\\ P_{21}& p_{22}\end{array}\right]$$
• Select the state of the last value of the residual sequence as the initial state vector $m_0$。 Let $m_0=(m_{0,1},m_0,2)$ , among which$m_{0,1},m_{0,2}$represent the probabilities of being in state 1 and state 2 respectively. That is, if the last residual value is positive, $m_0=(1,0)$ ; if negative,$m_0=(0，1)$ 。
• According to $m_t=m_0{P}^t$ is found after$After\; t$ state transitions, the$The\; state\; probability\; of\; t$ times. Select the state with the highest probability as the final result. If the probabilities of the two states are equal, take the result of the previous calculation:

Simulation results

matlab source code

https://mbd.pub/o/bread/YpWWkphs source code