`library(rje)`

First let’s generate a joint probability distribution for a \(2 \times 2 \times 2 \times 2\)-table.

```
set.seed(123)
= rprobdist(2, 4) p
```

We can easily calculate the marginal distribution for the first two variables:

`marginTable(p, 1:2)`

```
## [,1] [,2]
## [1,] 0.1095329 0.4286592
## [2,] 0.2429444 0.2188636
```

Note that the function `base::margin.table()`

performs the
same function as `marginTable()`

, but is not as fast. Output
is ordered according to how the variables are entered into the
function:

`marginTable(p, 2:1)`

```
## [,1] [,2]
## [1,] 0.1095329 0.2429444
## [2,] 0.4286592 0.2188636
```

but this can be over-ridden by setting the argument
`order=FALSE`

.

We can also obtain conditional distributions:

`conditionTable(p, 3, 1)`

```
## [,1] [,2]
## [1,] 0.3399897 0.5162332
## [2,] 0.6600103 0.4837668
```

`conditionTable()`

orders output with the ‘free’ variables
first (as ordered in the argument `variables`

) followed by
the conditioning variables. Sometimes it’s useful to keep a conditional
(or marginal) distribution in the same form as the original table, even
for variables which are removed

`conditionTable2(p, 3, 1)`

```
## , , 1, 1
##
## [,1] [,2]
## [1,] 0.3399897 0.3399897
## [2,] 0.5162332 0.5162332
##
## , , 2, 1
##
## [,1] [,2]
## [1,] 0.6600103 0.6600103
## [2,] 0.4837668 0.4837668
##
## , , 1, 2
##
## [,1] [,2]
## [1,] 0.3399897 0.3399897
## [2,] 0.5162332 0.5162332
##
## , , 2, 2
##
## [,1] [,2]
## [1,] 0.6600103 0.6600103
## [2,] 0.4837668 0.4837668
```

In causal inference it is common to want to know what happens if we
*intervene* on a variable under a certain causal ordering. This
is effectively just knowing about a joint distribution after dividing by
a particular conditional distribution.

```
= interventionTable(p, 3, 1:2)
p_int ## check this is p(1,2) * p(4 | 1, 2, 3)
= conditionTable2(p, 1:2, c())*conditionTable2(p, 4, 1:3)
p_int2 all.equal(p_int, p_int2)
```

`## [1] TRUE`

`#rprobMat(100, dim=2, d=4)`

When dealing with margins of multivariate distributions, it can be useful to be able to repeat probabilities to match the pattern of a joint distribution. In particular if we are given various conditional distributions (say from a Bayesian network model), we may wish to multiply them together to obtain the joint distribution.

For example, the model in which \(X_2\) is independent of \(X_3\) given \(X_1\) might be stored as the conditional probability tables \(P(X_1)\), \(P(X_2 | X_1)\) and \(P(X_3 | X_1)\). In order to reconstruct the joint distribution \(P(X_1,X_2,X_3)\), one needs to multiply \[ P(X_1=x_1,X_2=x_2,X_3=x_3) = P(X_1=x_1) \cdot P(X_2 = x_2 | X_1=x_1) \cdot P(X_3 = x_3 | X_1=x_1), \] so that the values of \(x_1,x_2,x_3\) match.

To use R’s vectorization for this we must turn the probability tables into vectors indexed by \((x_1,x_2,x_3)\), regardless of which variables are actually represented in the table; if a variable is not represented then values will be repeated. The indexing should be in reverse lexicographical order (i.e. first index changes fastest: 000, 100, 010, 110, …, 111), which is the way arrays are stored in R.

For example, if \(X_1,X_2,X_3\) are all binary (i.e. take values in \(\{0,1\}\)) then we’d transform the table of \(X_3 | X_1\) into \[ P(X_3 = 0 | X_1 = 0), \, P(X_3 = 0 | X_1 = 1), P(X_3 = 0 | X_1 = 0), \, P(X_3 = 0 | X_1 = 1)\\ P(X_3 = 1 | X_1 = 0), \, P(X_3 = 1 | X_1 = 1), P(X_3 = 1 | X_1 = 0), \, P(X_3 = 1 | X_1 = 1). \] Now, suppose we already have a vector for \(P(X_3 = x_3 | X_1 = x_1)\) indexed by \((x_1, x_3)\) in reverse lexicographical order: \[ P(X_3 = 0 | X_1 = 0), \, P(X_3 = 0 | X_1 = 1), P(X_3 = 1 | X_1 = 0), \, P(X_3 = 1 | X_1 = 1), \] we need the first and second entries repeated, followed by the third and fourth entries:

`patternRepeat0(c(1,3), c(2,2,2))`

`## [1] 1 2 1 2 3 4 3 4`

`patternRepeat0()`

requires us only to specify the
elements present and the dimension of the full distribution. The
existing order of the distribution is assumed to be reverse
lexicographic, regardless of the order given in the first argument, but
this can be over-ridden.

`patternRepeat0(c(3,1), c(2,2,2))`

`## [1] 1 2 1 2 3 4 3 4`

`patternRepeat0(c(3,1), c(2,2,2), keep.order=TRUE)`

`## [1] 1 3 1 3 2 4 2 4`

Another way to think about this is that if we take the possible
indices for a 3 dimensional array and match them to the indices of just
the first and third dimensions, `patternRepeat0()`

tells us
which point should be matched.

Let’s generate some conditional probability tables.

```
set.seed(134)
= c(rdirichlet(1,c(1,1)))
p1 .1 = c(rdirichlet(2,c(1,1)))
p2.1 = c(rdirichlet(2,c(1,1)))
p3
= p1*p2.1
p12 ## get joint distribution
= p12*p3.1[patternRepeat0(c(1,3), c(2,2,2))]
p123
## put into array to verify this has correct
## conditional distribution
dim(p123) = c(2,2,2)
conditionTable(p123, 3, 1)
```

```
## [,1] [,2]
## [1,] 0.5956466 0.6709426
## [2,] 0.4043534 0.3290574
```

```
## can also get conditional distribution indexed by all variables
.1[patternRepeat0(c(1,3), c(2,2,2))] p3
```

```
## [1] 0.5956466 0.6709426 0.5956466 0.6709426 0.4043534 0.3290574 0.4043534
## [8] 0.3290574
```

`c(conditionTable2(p123, 3, 1))`

```
## [1] 0.5956466 0.6709426 0.5956466 0.6709426 0.4043534 0.3290574 0.4043534
## [8] 0.3290574
```