log-sum-exp trick

when I implement models with discrete variables (which actually happens more than one can think), I always end up estimating this value:
\[ V = \log \left( \sum_i e^{b_i} \right) \]
Why ? This usually happens at the denominator of a Bayes formula for example. I try to keep \(log\)-probabilities all the time so that not to have to deal with very small numbers and to do additions instead of multiplications. By the way, I was looking at the time and latency of floating-point instructions in the latest processors (like Intel Core i7 for example), and I realized that still in 2010, additions are faster than multiplications (even with SSEx and the like).

Therefore, use \(log\)

In this expression, \(b_i\) are the log-probabilities and therefore \(e^{b_i}\) are very small or very big yielding to overflow or underflow sometimes. A scaling trick can help using numbers in a better range without loss of accuracy and for a little extra cost as follows:
\[ \begin{array}{rcl} \log \left( \sum_i e^{b_i} \right)&=& \log \left( \sum_i e^{b_i}e^{-B}e^{B} \right)\\ ~ &=& \log \left( \left( \sum_i e^{b_i - B }\right)e^{B} \right)\\ ~ &=& \log \left( \sum_{i} e^{b_i - B} \right) + B \end{array} \]

And that's it. For the value of \(B\), take for instance \(B=\max_i b_i\).
So the extra cost is to find the max value and to make a subtraction.

Just for those of you who wants to know how to put formulas in Blogger, I used this link here : http://watchmath.com/vlog/?p=438

Pretty straighforward. It uses a public LaTeX server to render the formulas. Very pretty !

This is my first post on this blog. And to be honest, this is the first time I'm gonna try to blog my thoughts. So, I'll do it on what I like these days: Artificial Intelligence and Machine Learning.

The idea is to post thoughts, tricks, ideas, etc... In the hope people will read it and comment too.

And, oh yes, I just installed in function to include math formulas. I don't know if it works so let's try it now with a simple version of the Bayes formula:
\[ P(A|B) = \frac{P(B|A).P(A)}{P(B)}\]