Conditional probability table: Difference between revisions

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In [[statistics]], the '''conditional probability table (CPT)''' is defined for a set of discrete and mutually [[independence (probability)|dependent]] [[random variable]]s to demonstrate [[conditional probability]] of a single variable with respect to the others. For example, assume there are three random variables <math>x_1,x_2, x_3</math> where each has <math>K</math> states. Then, the conditional probability table of <math>x_1</math> provides the conditional probability values for <math>P(x_1=a_k\mid x_2,x_3)</math>for each of the ''K'' possible values <math>a_k</math> of the variable <math>x_1</math> and for each possible combination of values of <math>x_2,\, x_3.</math> This table has <math>K^3</math> cells. In general, for <math>M</math> variables <math>x_1,x_2,\ldots,x_M</math> with <math>K</math> states for each of them, the CPT for any one of them has size&nbsp;<math>K^M.</math><ref name=murphybook>{{cite book|last=Murphy|first=KP|title=Machine learning: a probabilistic perspective|year=2012|publisher=The MIT Press}}</ref>

With only two variables, a CPT can be put into [[matrix (mathematics)|matrix]] form. For example, the values of <math>P(x_1=a_k\mid x_2=b_j)=T_{kj},</math> 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. <math>\sum_k T_{kj} = 1</math> for all ''j''. For example, suppose that two binary variables ''x'' and ''y'' have the [[joint probability distribution]] given in this table:

{|class="wikitable"

|-

! !! 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'':

{|class="wikitable"

|-

! !! x=0 !! x=1

|-

! P(y=0 given x)

| 4/6 || 1/3

|-

! P(y=1 given x)

| 2/6 || 2/3

|-

!

| 1 || 1

|}

==References==