I am trying to implement the following:

enter image description here

where the right part returns a probability between 0 and 1. Regarding the product, the authors of the respective paper note:

Due to numerical precision issues with products of probabilities, in our implementation we follow common practice and use summation of log probabilities.

Form what I understand, using the sum of log probabilities helps to prevent underflow. But then I do not get a value between 0 and 1 and the 1- in the formula above does not make sense. What am I missing here? And can I transform the sum of log probabilities back to a value between 0 and 1? When using a large number of probabilities, I still get a very small number, e.g.:

log_probability = math.log(0.9) + math.log(0.3) + math.log(0.9) + math.log(1) + math.log(0.9) + math.log(0.3) + math.log(0.3) + math.log(0.3) + math.log(0.3) + math.log(0.3) + math.log(0.3) + math.log(0.3) + math.log(0.3) + math.log(0.3) + math.log(0.3) + math.log(0.3) + math.log(0.3) 
prob = math.exp(log_probability)

Where log_probality=-15.967728003210647 and prob =1.1622614669999998e-07.

Thank you, I am really hitting the edge of my understanding of stats here...!

  • $\begingroup$ Implementation questions might be more on topic on Stack Overflow. $\endgroup$ May 2 '20 at 16:03
  • $\begingroup$ It looks like you're using Python... Option (1) List comprehensions. Option (2) Built-in map function. Option (3) NumPy-vectorized functions. Option 3 is what I would recommend looking at first.. $\endgroup$ May 2 '20 at 16:04
  • $\begingroup$ For NumPy arrays you can set the datatype, including the float size that is related to the precision. $\endgroup$ May 2 '20 at 16:06
  • $\begingroup$ Yes, I was thinking that this is rather about my (lack of) understanding of the stats which is why I decided to put it here. $\endgroup$
    – Steinroe
    May 2 '20 at 16:06
  • $\begingroup$ If you're really in need of more precision than what NumPy can offer then you might consider mpmath for arbitrary float precision (AFP). AFP is expensive, but sometimes needed. $\endgroup$ May 2 '20 at 16:08

Working in log-probabilties

Yeah, the probability is still small if you apply the inverse transform (exponentiate the sum of log-probability). One can work in log probabilities to avoid really large/small probabilities that can result numerical issues, including underflow. You may or may not transform back to probabilities, as the result of a sequence of operations may not yield a log-probability whose corresponding probability is within your float precision.

A lot of statistics makes use of mathematical optimization, and in many cases if you optimize an expression in terms of log-probability, you are also finding the same (or corresponding) optima for an original problem posed in terms of probability. For example, if we wanted to find the parameter $\theta$ that maximizes $P(Y | X, \theta)$ where $Y$ and $X$ are random variables, we might consider working with $\log \left[ P(Y | X, \theta) \right]$ to find the same optimal value of $\theta$, which we might denote as $\theta^*$.

$$\theta^* = \arg \max_\theta P(Y | X, \theta) = \arg \max_\theta \log \left[ P(Y | X, \theta) \right]$$

  • $\begingroup$ Thanks for your replies! Maybe to provide more context: Its about anomaly detection and the formula above describes the anomaly likelihood. So from what I understand, the above formula with 1-(a product of probabilities) can be "replaced" with the sum of the log probabilities as it behaves similar and is easier to work with in terms of computation? How does the 1- come into play here? With the above formula I threshold the resulting value and if exceeds a certain threshold, an anomaly is given. $\endgroup$
    – Steinroe
    May 3 '20 at 11:18
  • $\begingroup$ If we're looking for optimal parameters, we can drop constant terms when we wish as they preserve argmax/argmin. For example, $f(x) = x^2 + 1$ and $g(x) = x^2$ are both minimized where $x=0$ even though $f(0) = 1$ and $g(0) = 0$ are not equal to each other. $\endgroup$ May 3 '20 at 15:05
  • $\begingroup$ I skimmed the paper, and it looks like $L_t$ is a likelihood function, so it would make sense to apply this type of thinking about optimization. $\endgroup$ May 3 '20 at 15:09
  • $\begingroup$ If your just finding these likelihoods through some estimators of the mean, stdev (which I saw in the paper), then you can still drop the "1" provided that you adjust how you're thresholding accordingly. $\endgroup$ May 3 '20 at 15:15

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