## 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.

A B C D P
0 0 0 0 0.0096
0 0 0 1 0.0144
0 0 1 0 0.0224
0 0 1 1 0.0336
0 1 0 0 0.004
0 1 0 1 0.004
0 1 1 0 0.006
0 1 1 1 0.006
1 0 0 0 0.216
1 0 0 1 0.144
1 0 1 0 0.216
1 0 1 1 0.144
1 1 0 0 0.0756
1 1 0 1 0.0324
1 1 1 0 0.0504
1 1 1 1 0.0216

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

A B C D Unnormalized P
0 0 0 0 360000
0 0 0 1 600
0 0 1 0 36000
0 0 1 1 3600
0 1 0 0 600
0 1 0 1 100000
0 1 1 0 60
0 1 1 1 600000
1 0 0 0 6000
1 0 0 1 10
1 0 1 0 900
1 0 1 1 90
1 1 0 0 600
1 1 0 1 100000
1 1 1 0 90
1 1 1 1 900000

### 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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