Conditional probability table

In statistics, the conditional probability table (CPT) is defined for a set of discrete and mutually dependent random variables to demonstrate conditional probability of a single variable with respect to the others. For example, assume there are three random variables $x_{1},x_{2},x_{3}$ where each has $K$ states. Then, the conditional probability table of $x_{1}$ provides the conditional probability values $P(x_{1}=a_{k}\mid x_{2},x_{3})$ – where the vertical bar $|$ means “given the values of” – for each of the K possible values $a_{k}$ of the variable $x_{1}$ and for each possible combination of values of $x_{2},\,x_{3}.$ This table has $K^{3}$ cells. In general, for $M$ variables $x_{1},x_{2},\ldots ,x_{M}$ with $K$ states for each of them, the CPT for any one of them has size $K^{M}.$

With only two variables, a CPT can be put into matrix form. For example, the values of $P(x_{1}=a_{k}\mid x_{2}=b_{j})=T_{kj},$ with k and j ranging over K values, create a K×K matrix. This matrix is a stochastic matrix since the columns sum to 1; i.e. $\sum _{k}T_{kj}=1$ for all j. For example, suppose that two binary variables x and y have the joint probability distribution given in this table:

	x=0	x=1	P(y)
y=0	4/9	1/9	5/9
y=1	2/9	2/9	4/9
P(x)	6/9	3/9	1

Each of the four central cells shows the probability of a particular combination of x and y values. The first column sum is the probability that x =0 and y equals any of the values it can have – that is, the column sum 6/9 is the marginal probability that x=0. If we want to find the probability that y=0 given that x=0, we compute the fraction of the probabilities in the x=0 column that have the value y=0, which is 4/9 ÷ 6/9 = 4/6. Likewise, in the same column we find that the probability that y=1 given that x=0 is 2/9 ÷ 6/9 = 2/6. In the same way, we can also find the conditional probabilities for y equalling 0 or 1 given that x=1. Combining these pieces of information gives us this table of conditional probabilities for y:

	x=0	x=1
P(y=0 given x)	4/6	1/3
P(y=1 given x)	2/6	2/3
Sum	1	1