生成关于X,Y,Z的联合分布,使得(X⊥Y),(X⊥Z),(Y⊥Z),而(X,Y⊥Z)不成立.(probabilistic graphic models Exercise 3.1)

probabilistic graphic models Exercise 3.1

起因

学完了一章,看题,怎么都不会?

经过

想了好久

结果

从最简单的开始,令X,Y,Z的取值为{0,1},则只需构造8个项的概率,记作$v_0,v_1,v_2,\cdots ,v_7$,按照X,Y,Z的取值进行编号,X的值对应v下标的最高二进制位,X取0的概率记作$P_x$.
那么
$$\begin{gathered} v_0+v_1+v_2+v_3=P_x \\ {v}_0+v_1+v_4+v_5=P_y \\ {v}_0+v_3+v_5+v_7=P_z \end{gathered}$$
再由已知条件$(X\perp Y),(X\perp Z),(Y\perp Z)$有,
$$\begin{gathered} v_0+v_1=P_a{P}_b \\ {v}_0+v_2=P_a{P}_c \\ {v}_0+v_4=P_b{P}_c \end{gathered}$$
由概率的定义有
$$\sum _{i=0}^7{v}_i=1$$
8个变量,7个方程,于是再加上一个条件,给定$v_0$的值,则可以相应的解出其它的值.

#include <iostream>
#include <matrix.hpp>
#include <mat_algos.hpp>
#include <string>

using namespace std;

int main() {
    
    
    double pa,pb,pc,v0;
    cout<<"find three-variable binary joint distribution."<<endl;
    cout<<"please input P(a0) P(b0) P(c0) P(a0,b0,c0):";
    cin>>pa>>pb>>pc>>v0;
    mine::Matrix m(8,9,{
    
    
        1,  0,  0,  0,  0,  0,  0,  0,  v0,
        1,  1,  1,  1,  1,  1,  1,  1,  1,  
        1,  1,  1,  1,  0,  0,  0,  0,  pa,
        1,  1,  0,  0,  1,  1,  0,  0,  pb,
        1,  0,  1,  0,  1,  0,  1,  0,  pc,
        1,  1,  0,  0,  0,  0,  0,  0,  pa*pb,
        1,  0,  1,  0,  0,  0,  0,  0,  pa*pc,
        1,  0,  0,  0,  1,  0,  0,  0,  pb*pc
    });
    mine::gaussCR(m);//自己写的矩阵库,
    mine::solveUpper(m);//用来求解方程,最后一列即为最终的结果.
    for(int i=0;i<8;++i){
    
    
        printf("%f",m[i][8]);
        if((i+1)%4==0)cout<<endl<<"---------------------"<<endl;
        else if((i+1)%2==0)cout<<endl;
        else cout<<' ';
    }
    return 0;
}

求解结果可能会出现负概率,对输入进行调整,试几次,就会有结果了.

find three-variable binary joint distribution.
please input P(a0) P(b0) P(c0) P(a0,b0,c0):0.1 0.2 0.3 0
0.000000 0.020000
0.030000 0.050000
---------------------
0.060000 0.120000
0.210000 0.510000
---------------------