Volume of elements mapped to the same codeword is $2^{H(X|\hat{X})}$ In this paper by Tishby, Pereira and Bialek they mention on page 4 in the Relevant quantization chapter the setting is the following; Given some signal space $X \sim p(x)$ and a quantized codebook $\hat{X}$. They seek a possibly stochastic mapping characterized by the pdf $p(\hat{x}|x)$ from every value $x \in X$ to a codeword $\hat{x} \in \hat{X}$. 

This has made me wonder about the following:

1) They mention that the average volume of the elements of $X$ that are mapped to the same codeword is $2^{H(X|\hat{X})}$ - why is this? 

2) They mention that for ease of exposition both $X$ and $\hat{X}$ are finite, so howcome $p(\hat{x}|x)$ is a p.d.f and not a p.m.f?
If I understand right what "average volume" means here, I don't think this is correct. For example, let's say you map $n$-bit strings (under uniform distribution) to $n$-bit strings as follows: Given an input $x$, if the first bit is 0, output $x$, and otherwise output "1". 

The average volume would be $\frac{1}{2}(2^{n-1}+1)$, whereas the $2^{H(X|\hat{X})} = \sqrt{2^{n-1}}$.

The correct expression for the average volume should perhaps be the expectation $\sum p(\hat{x}) 2^{H(X|\hat{X}=\hat{x})}$.

You are right about p.d.f. vs p.m.f, this kind of mix-up of terminology is quite common (but shouldn't be).