Suggest edit — inference

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title: "inference"
source: https://www.jemoka.com/posts/kbhinference/
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inference is the act of updating the distribution of a random variable based on distribution of actually observed variables:
\begin{equation} P(X|Y) \end{equation}
where \(Y\) is observed, and we want to know how likely \(X\) would therefore be.
We call the set \(X\) the &ldquo;query variables&rdquo;, \(Y\) as &ldquo;evidence varibales&rdquo;, and anything that we didn&rsquo;t use which connects the two variables as &ldquo;hidden variables&rdquo;.
If things are not in the right order of \(X\) and \(Y\), consider the Bayes rule.
Inference is Hard mix of continuous and discrete distribution results could be either a PMF or a PDF Example Suppose we&rsquo;d like to know \(P(b^{1} | d^{1}, c^{1})\), where \(b^{1}\) is considered a query variable, and \(c^{1}\) is considered evidence varibales. The definition of the conditional probability gives us:
\begin{equation} p(b^{1} | d^{1}, c^{1}) = \frac{p(b^{1}, d^{1}, c^{1})}{p(d^{1}, c^{1})} \end{equation}
To compute \(p(b^{1}d^{1}c^{1})\), we first compute:
\begin{equation} p(b^{1}, d^{1}, c^{1}, E, S) \end{equation}
and then, use the law of total probability to get:
\begin{equation} p(b^{1}, d^{1}, c^{1}) = \sum_{e=E} \sum_{s=S} p(b^{1}, d^{1}, c^{1}, E, S) \end{equation}
you will note this is very expensive computationally O(es) &mdash; and if you have like a 1000 hidden variables you will die.
We therefore introduce sum-product elimination.
sum-product elimination You will note the summation in the example above has a lot of interlocking for loops. You can &ldquo;factor them out&rdquo; via the sum-product elimination algorithm.
Suppose you are interested in:
\begin{equation} P(b | d&rsquo;, c&rsquo;) \end{equation}
Step 1: write down factors Write down all factors associated with this computation:
\begin{equation} \phi_{1}(B), \phi_{2}(S), \phi_{3}(E,B,S), \phi_{4}(D,E), \phi_{5}(C,E) \end{equation}
we have evidence at two variables: \(D, C\).
Step 2: performing factor conditioning for all evidence variables Therefore, \(\phi_{4}\) and \(\phi_{5}\) can be replaced by the factor conditioning as we observed \(d, c\), so we no longer need \(d, c\) as input because we know them:
now we have, to replace \(\phi_{4}, \phi_{5}\):
\begin{equation} \phi_{6}(E), \phi_{7}(E) \end{equation}
Step 3: using the law of total probability and factor product, get rid of hidden variables We then choose an ordering of the hidden variables and apply a factor product using the law of total probability to get rid of them:
First get rid of any hidden variables Then use factor product to combine results \begin{equation} \phi_{8}(B,S) = \sum_{E=e} \phi_{3}(E,B,S) \phi_{6}(e) \phi_{7}(e) \end{equation}
\begin{equation} \phi_{9}(B) = \sum_{S=s} \phi_{2}(s) \cdot \phi_{8}(B,S) \end{equation}
We now only have two factors left: \(\phi_{1}(B)\phi_{9}(B)\). We finally apply factor product again:
\begin{equation} \phi_{10} (B) = \phi_{9}(B) \cdot \phi_{1}(B) \end{equation}
Approximate Inference See Approximate Inference
Gaussian Inference See Inference for Gaussian Models