## PGM Assignment 2

### Question 3

Now we have four binary variables A, B, C, D, and two cases of their joint distribution.

Notice that in TABLE 2, we only give you an unnormalized probability distribution. But you can transfrom these to a normalized distribution by dividing the sum of all unnormalized Ps.

Check all independent and conditionally independent relations among A, B, C, D. (Hint: do you need a normalized distribution for this problem?)

Table 1: Joint distribution of A, B, C, D.

ABCDP
00000.0096
00010.0144
00100.0224
00110.0336
01000.004
01010.004
01100.006
01110.006
10000.216
10010.144
10100.216
10110.144
11000.0756
11010.0324
11100.0504
11110.0216

Table 2: Joint distribution of A, B, C, D.

ABCDUnnormalized P
0000360000
0001600
001036000
00113600
0100600
0101100000
011060
0111600000
10006000
100110
1010900
101190
1100600
1101100000
111090
1111900000

### Solution

$x0$ 代表 $x=0$

$x0y0$ 代表事件 $(x=0, y=0)$

## Table 1

### 2个元素独立性

$\therefore \ A \ \bot \ B$

### 4个元素独立性

#### 结果

$\therefore \ (C\ \bot\ D \ | \ A,\ B)$

## Table 2

### 3个元素独立性

#### 结果

$\therefore (B \ \bot \ C \ | \ D ), \quad (\ A \ \bot \ D \ | \ B)$

### 4个元素独立性

#### 结果

$\therefore \ (A\ \bot\ D \ | \ B,\ C), \quad (B\ \bot\ C \ | \ A,\ D)$

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