# Bayes theorem and Radon-Nikodym derivative

I am dealing with introduction to information theory and there is a problem I am not able to deal with. I will use the author's notation.

As reported in [Information theory and statistics, Kullback], he rewrites, on $$(\Omega, \mathcal{f})$$ in which two absolute continuous measure functions $$\mu_1$$ and $$\mu_2$$ are defined, the Bayes theorem as $$\mathbb{P}(H_i|x)=\frac{\mathbb{P}(H_i)f_i(x)}{\mathbb{P}(H_1)f_1(x)+\mathbb{P}(H_2)f_2(x)},\qquad i=1,2$$ where $$f_i(x)$$ is the Radon-Nikodym derivative with respect to the measure $$\mu_i$$: $$\mu_i(E)=\int_E f_i(x)d\lambda(x)$$, where $$\lambda$$ is a probability measure which is equivalent to $$\mu_1$$ and $$\mu_2$$.

I can't understand why the author rewrites the conditional probability $$\mathbb{P}(x|H_i)$$ as the Radon-Nikodym derivative $$f_i(x)$$. I am pretty sure that the "abstract" Bayes formula (with conditional expectations) is involved, but actually I cannot understand why $$f_i(x)=\mathbb{P}(x|H_i)$$.

Many thanks.

• For context - how familiar are you with measure theoretic probability? In particular, have you studied the R-N derivative before this? If not, are you comfortable with the notion of a probability density? Commented Jun 27, 2019 at 14:39
• Consider that I am not a mathematician and I studied Economics. I have knowledge in measure theory, I know what is a Lebesgue measure and how the Lebesgue integral is constructed. I know the formal definition of probability density function and I am confortable with probability measures. However, you should also consider that I studied (and I am studying) these arguments by myself for my personal interest, so that my knowledge about it is likely to be pretty weak and possibly biased. Commented Jun 28, 2019 at 15:06
• I posted an intuitive development of the expression, which follows a standard heuristic about continuous random variables. Please ask me to elaborate on anything you'd like. As an aside, if you're reading Kullback to learn Information Theory, may I suggest a more modern treatment? For instance, Cover and Thomas gets to the heart of the subject with no more machinery than undergrad discrete probability. If you're specifically interested in the interplay b/w Information Theory and Statistics, again, more modern treatments are available that may be better to read. Commented Jun 28, 2019 at 16:31
• (contd.) These texts tend to be pitched at beginning grad students, so an important advantage, beyond the decades of streamlining of the subject, is that they do not use probabilitic machinery that a complete treatment would require, but which detracts from the subject at hand. Commented Jun 28, 2019 at 16:32

The RN derivative is basically the same as a pdf (this is exactly true when $$\lambda$$ is the Lebesgue measure and $$X$$ is a continuous RV). The standard heuristic about continuous random variables is that since $$P(X = x)$$ is $$0$$ for every $$x,$$ an observation of $$x$$ should really be interpreted as $$X \in x \pm \delta x$$ for $$\delta x \to 0.$$ By definition of the pdf, $$P(X \in x \pm \delta x) = f(x) \cdot (2\delta x) + o(\delta x),$$ where $$f$$ is the pdf.
With this in hand, we can see that the above is just an instantiation of the usual Bayes' law - for any events $$A$$ and $$B_1,B_2$$ (all with non-zero probability), $$P(B_i|A) = \frac{P(A|B_i) P(B_i)}{ P(A|B_1) P(B_1) + P(A|B_2) P(B_2)}.$$
Indeed, the events $$B_1$$ and $$B_2$$ are just $$H_1$$ and $$H_2$$ above, and $$X$$ has the pdf $$f_1$$ under $$H_1$$ and $$f_2$$ under $$H_2$$. From the above heuristic, we should have $$P(H_i|X=x) = \lim_{\delta x \to 0}\frac{(f_i(x) (2\delta x) + o(\delta x))P(H_i)}{(f_1(x) (2\delta x) + o(\delta x))P(H_1) + (f_2(x) (2\delta x) + o(\delta x))P(H_2)} \\ = \lim_{\delta x \to 0} \frac{f_i(x) P(H_i) + o(1)}{f_1(x) P(H_1) + f_2(x) P(H_2) + o(1) } = \frac{f_i(x) P(H_i)}{f_1(x) P(H_1) + f_2(x) P(H_2)}$$
Pertaining specifically to the problem in your Q, $$P(x|H_i),$$ or more precisely $$P(X = x|H_i)$$ is not interpreted as $$f_i$$, but the ratio $$P(X = x|H_1)/P(X = x|H_2)$$ can formally be expressed as $$f_1(x)/f_2(x)$$. (Of course, this is a $$0/0$$ form, so must be interpreted in some limiting sense. The sense appropriate is well expressed by the above heuristic).