0
$\begingroup$

Zipf's law states that given some corpus of natural language utterances, the frequency of any word is inversely proportional to its rank in the frequency table.

So I understand "inversely proportional" to mean the frequency of a word of a given rank $r$ is $f(w) = K / r(w)$ (where $K$ is a constant, the frequency of the top-ranked word, and $r(w)$ is the rank of word $w$).

However, in my corpus I found $f(w) = K * 0.9^{r(w)}$. I'm pretty certain this can't be said to be "inversely proportional".

Here is my question: How would you describes in words the correspondence represented by $f(w) = K * 0.9^{r(w)}$? Is there a good phrasing for this? Is it "frequency is exponentially proportional to rank"? Wikipedia seems to suggest so, but because $a < 1$ values shrink instead of exploding like "exponentially proportional" suggests to my ears...

$\endgroup$

1 Answer 1

2
$\begingroup$

$$ K \cdot 0.9^{r(\omega)} = K \cdot \left( \frac{10}{9} \right)^{- r(\omega)}$$

It gets smaller because the base is less than one. As $r(\omega)$ increases, $f(\omega)$ decays exponentially, and as $r(\omega)$ decreases, $f(\omega)$ grows exponentially.

$\endgroup$
2
  • $\begingroup$ Ah, so I'm understanding that I should describe it as "frequency decays exponentially proportional to rank", yes? $\endgroup$
    – Sled
    Mar 16, 2014 at 21:17
  • 1
    $\begingroup$ @ArtB It feels a bit unnatural to force the word "proportional" into the expression. It'd be more common to say "the frequency decays exponentially with the rank". Perhaps this is because the parameter of interest tends to be the base of the exponent rather than the constant of proportionality? $\endgroup$
    – Erick Wong
    Mar 16, 2014 at 21:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .