# Proving conditional independence from mutual independence

Suppose I have a Bayesian inference problem where $E$ is the result of a chemistry experiment. The chemistry experiment consists of a controlled sequential digestion of one end of the protein using a specific enzyme that has known statistical properties. The sequence of the protein is unknown at the beginning of the experiment. The data collected is the number of reactions and the times in which they occur. The goal is to infer sequence information using the enzyme.

$P\left(x_{1},\, x_{2},\,\ldots,\, x_{n}|E\right)$ = $\dfrac{P\left(x_{1},\, x_{2},\,\ldots,\, x_{n}\right)P\left(E|x_{1},\, x_{2},\,\ldots,\, x_{n}\right)}{P\left(E\right)}$

Suppose also that the discrete random variables $X_{1},\, X_{2},\,\ldots,\, X_{n}$ of the posterior are known to be mutually independent a priori. These random variables of the posterior give sequence information (the 20 amino acids) for the first n positions in the protein. This Bayesian inference calculation is performed for all amino acid sequences of length n. The marginal distributions are then taken to infer the unknown sequence.

How do I prove the conditional independence of the posterior random variables given the evidence:

$P\left(x_{1},\, x_{2},\,\ldots,\, x_{n}|E\right)$ = $P\left(x_{1}|E\right)P\left(x_{2}|E\right)\cdots P\left(x_{n}|E\right)$ .

• I revised the question by including how I am using the variables and the evidence (i.e. the event). – user44821 May 2 '14 at 7:13
• What do you mean by 'of the posterior' in 'Suppose also that the random variables $X_1,X_2,\ldots,X_n$ of the posterior are mutually independent'. Do you mean that you are assuming the conclusion? Or do you mean to assume that the random variables $X_1,\ldots,X_n$ are independent? In the latter case, the conclusion is not in general true. – Juho Kokkala May 2 '14 at 9:56
• I answered by a counterexample based on the latter interpretation. If that was not intended, you need to clarify the question. – Juho Kokkala May 2 '14 at 10:05
• You need to define what 'these random variables cannot be a part of $E$ in a counter example' means. What does your result $E$ look like? Is it a single number, or do you perhaps have one measurement per each component of $X$? – Juho Kokkala May 2 '14 at 11:06

If you mean that the assumption is independence of $X_1,\ldots,X_n$ (a priori), the conclusion is not true. For a counter-example, let $X_1,X_2$ be independent Bernoulli random variables: $$P(X_i=1) = 0.5,~P(X_i=0) = 0.5,~i\in\{1,2\}.$$ Let $E$ be the event that $X_1=X_2$. Clearly $X_1$ and $X_2$ are not conditionally independent conditional on $E$.

• The assumption is that X_1,...,X_n are known to be mutually independent a priori. In the situation I am considering the random variables X_1,...,X_n are not part of E. E is the result of an experiment that counts the number and times of chemical reactions. – user44821 May 2 '14 at 10:38
• I don't know if to be 'part of $E$' means anything. Surely $E$ somehow depends on the $X$s, otherwise the experiment is not informative at all. – Juho Kokkala May 2 '14 at 10:41
• What I mean is that your counter example does not apply to my situation. I do not condition on the X_1,...,X_n random variables. – user44821 May 2 '14 at 10:58
• What, then, is $P(E\mid x_1,\ldots,x_n)$ in your question? If your $E$ is independent of the $X$s, then the prior is equal to the posterior. – Juho Kokkala May 2 '14 at 11:01
• The E is the result of a experiment on a single protein molecule. The measured quantities are the number of reactions and when the reactions occur. X_1 takes on one of the twenty amino acids at position one in the protein. X_2 does the same thing at position two, and so on until position n is reached. – user44821 May 2 '14 at 11:13

Recall the multiplication rule of probability $$P(A_{1}\cap ...\cap A_{n})=P(A_{1})P(A_{2}|A_{1})...P(A_{n}|A_{1}\cap...\cap A_{n-1})$$ Well you could extend this for conditional probability so $$P(A_{1}\cap ...\cap A_{n}| E)=P(A_{1}|E)P(A_{2}|A_{1}\cap E)...P(A_{n}|A_{1}\cap...\cap A_{n-1}\cap E)$$ where since all $A_{i}$'s are mutually independent then the only "valuable" information (i.e. changes the conditional probability) is $E$ so we have $$P(A_{1}\cap ...\cap A_{n}| E)=P(A_{1}|E)P(A_{2}|E)...P(A_{n}| E)$$ Though this is not really a proof maybe it can give some insight to why this is your proposition is true