Suppose $X$ is a random variable taking $k$ values.

$$Val(X) = \{x_1, x_2, x_3, \cdots, x_k\} $$

Then what is the following expression of $N(X)$ called in literature if exists? What does it signify?

$$ N(X) = \prod \limits_{i = 1}^{k} p(x_i)^{p(x_i)}$$

I am using the notation $N(X)$ for the sake of my convenience only.

Background: I am asking this question because of the definition of entropy I encountered. Entropy is calculated as follows.

$$ H(X) = - \sum\limits_{i = 1}^{k} p(x_i) \log p(x_i) $$

If I further solve $H(X)$ as follows, I will get $H(X)$ in terms of N(X).

$$ H(X) = - \sum\limits_{i = 1}^{k} p(x_i) \log p(x_i) = - \sum\limits_{i = 1}^{k} \log p(x_i)^{p(x_i)} $$ $$\implies H(X)= - \log \prod \limits_{i = 1}^{k} p(x_i)^{p(x_i)} = - \log N(X)$$

Entropy is used to characterize the unpredictability of a random variable.

A logarithm is generally applied to many quantities in AI in order to bring them into the desirable range where overflow and underflow won't happen. Hence I am thinking that $\dfrac{1}{N(X)}$ is the actual quantity one has to measure (the entropy?). Hence I am guessing that $N(X)$ can be treated as reciprocal of entropy. So, does $N(X)$ is a quantity that has quantified the predictability of a random variable?

$$N(X) = \dfrac{1}{2^{H(X)}} = \dfrac{1}{2^{entropy}}$$

So, I am wondering whether there is any quantity that $N(X)$ quantify.

  • 2
    $\begingroup$ I don't know if it signifies anything or has a term, but it looks likely to underflow on a computer if you use floating point. $\endgroup$ Jun 13, 2021 at 7:52

1 Answer 1


I don't know if $N(X)$ has a name or has any applicability in AI, but I can comment on how this function varies as the $H(X)$ based on your equation

$$N(X) = \dfrac{1}{2^{H(X)}}$$

which looks correct to me (just apply the $\log_2$ to both sides).

In the case of a Bernoulli random variable (which is a categorical r.v. that can take 2 values, $0$ or $1$, which is a special case of your categorical r.v., if you set $k=2$), then this is the relationship between the probability that this random variable $X = 1$ and the entropy of this r.v.

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So, the entropy is $1$ when the probability is $0.5$ and decreases as the probability tends to $0$ or $1$, which makes sense, because the entropy quantifies the uncertainty about the r.v. Here, the entropy is computed in bits because we use the logarithm $\log_2$.

So, for $k=2$ (in your example), then, if $H(X) = 1$,

$$N(X) = \dfrac{1}{2^{H(X)}} = \frac{1}{2} = \frac{1}{2}^{\frac{1}{2}} * \frac{1}{2}^{\frac{1}{2}}$$

which is equal to $P(X = 1) = 1 - P(X = 0)$.

Now, as $H(X)$ decreases to $0$, then $\dfrac{1}{2^{H(X)}}$ increases, because the denominator $2^{H(X)}$ becomes smaller, where the smallest value is $H(X) = 0$

$$N(X) = \dfrac{1}{2^{H(X)}} = 1 = 1^1 * ? $$

This is already problematic because $0^0$ is not well-defined. So, I already see a problem with $N(X)$.

For $k > 2$, I think the same reasoning applies, but now the maximum value of $H(X)$ should be $\log_2 k$.

So, I don't think that $N(X)$ is of any practical value as it can lead to expressions like $0^0$.

However, this type of problem also arises in the original formulation of the entropy, because $\log_2 0$ is not defined. The convention for when $P(X) = 0$ is to set $P(X) \log P(X)$ to zero [1].

Anyway, it seems to me that one way to interpret $N(X)$ is as the (average?) probability that $X$ takes one of the values, and it could be that the information content is what you are looking for. The definition can be found in [1].


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