Is it possible to decompose a conditional probability with three or more elements (i.e. events) into conditional probability of only two elements or the marginal probability of one element? Knowing this decomposition, it would help to solve higher order Markov Chain mathematically. I also know that this decomposition can be solved if we add assumption of conditional independent.
To make it concrete here is a negative example:
P(c│a,b)=(P(a,b│c)∙P(c))/(P(a│b)∙P(b) ).
Notice that the RHS still contains a conditional probability with three elements P(a,b│c).
Assuming conditional independent on c, we have P(a,b│c)=P(a│c)∙P(b│c). Thus, the conditional probability decomposition becomes
P(c│a,b)≅(P(a│c)∙P(b│c)∙P(c))/(P(a│b)∙P(b) )
My question is whether this type of conditional probability decomposition into one or two element is possible without making assumption. If it is really unsolvable problem, then at least we know that the assumption of conditional independent is a must.
For my opinion we can use any product formulas only under condition of independence. For example the formula that you used P(a,b│c)=P(a│c)∙P(b│c) is valid if a and b are independent each from other. Both may be dependent from c. But it can be the case of so called "false dependence". For example a is the usage of energy, b is sale of warm clothes, c is decreasing of tempetature.
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