Entropy of random variable is defined as:

$$H(X)= \sum_{i=1}^n p_i \log_2(p_i)$$

Which as far as I understand can be interpreted as how many yes/no questions one would have to ask on average, to find out the value of the random variable $X$.

But what if the log base is changed to for example e? What would the interpretation be then? Is there an intuitive explanation?

  • $\begingroup$ it will multiply $H(X)$ by $\ln 2$.. $\endgroup$ – reuns Jul 27 '16 at 19:57
  • $\begingroup$ To those calculating the new value but not explaining what it "means" conceptually, you seem to be missing the point of the question. The OP wants an interpretation of the concept of information when the log base isn't 2. $\endgroup$ – Chill2Macht Jul 27 '16 at 21:36
  • $\begingroup$ Perhaps this is a little disingenuous, but I see no reason why there should be a nice interpretation. One could argue that the reason we use base two to define entropy is precisely because one can then couch it in the intuitive and pleasing terms of 'number of yes/no questions' (although this itself has issues - what are $1.5$ yes/no questions?) $\endgroup$ – stochasticboy321 Jul 27 '16 at 23:15
  • $\begingroup$ @stochasticboy321 well you could take the ensemble average over all recursive trees whose average branching is 1.5 -- in any case there is a clear interpretation for integer bases $\endgroup$ – Chill2Macht Jul 27 '16 at 23:46
  • $\begingroup$ @William right, so we've gone from the simple yes/no question to ensemble averages over recursive trees, which is already a sharp decrease in simplicity/niceness of interpretation. Far as I see, that supports my point. (That said, I do agree with you, interpretations for integer bases seem well within reach). $\endgroup$ – stochasticboy321 Jul 27 '16 at 23:55

Obviously this interpretation breaks down (at least somewhat) for non-integer bases, but for any logarithm base $b$, not just base $b=2$, we can interpret the information with respect to that base, $$H_b(X) = \sum_{i=1}^n p_i \log_b(p_i)$$ to be the average number of $b-$ary questions one would have to ask on average in order to find out the value of the random variable $X$.

What do I mean by a $b-$ary question? One that has at most $b$ possible answers. Any yes/no question is a $2-$ary question, and if one can reword it cleverly enough, any $2-$ary question can be stated as a yes/no question (since yes and no are both exhaustive and mutually exclusive).

More precisely, a $b$-ary question is one that has $b$ possible answers, which together exhaust all possibilities and each of which is mutually exclusive. In general, for $b \not=2$, it might be difficult to think of truly $b-$ary questions which don't involve artificially restricting the possible answers.

Consider, for example, trying to determine the value of a dice roll. If you are limited to $2-$ary questions, (e.g. "did it roll a $1$?") then on average it will take $\log_2 6$ questions to ascertain what value it rolled. However, if you are allowed to ask the $6-$ary question "did it roll a 1, 2, 3, 4, 5, or 6"? Then it will only take you $\log_6 6 =1$ question on average to ascertain the value.

If you consider the question tree consisting of all possible questions to derive the answer, then $b$ is just the number of branches emanating from each node (since each node is a question). If you increase $b$, you will decrease the number of questions necessary on average to ascertain the answer because each node will have more branches emanating from it, and thus you can traverse a larger portion of the set of possible answers by hitting fewer nodes.

This is in fact very similar to the idea of recursion equations and recursion trees in computer science; in particular, I encourage you to look at Section 4.4 of Cormen et al, Introduction to Algorithms, for an explanation of how the logarithm enters naturally into these types of situations.

We can think of asking a $b-$ary question as dividing the problem of finding the value of the random variable $X$ into $b$ subproblems of equal size -- then the analogy with recursion should become more clear (hopefully).

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    $\begingroup$ I think we can extend this to say that base $2.5$ entropy counts the total number of 2-ary and 3-ary questions required on average (given that we ask the same number of each type of question on average). $\endgroup$ – Curtis Bechtel Sep 14 '18 at 22:29
  • $\begingroup$ @CurtisBechtel That sounds about right, although it might be something like the geometric mean of $2$ and $3$, $\sqrt{6}$, and not the arithmetic mean $2.5$, if both $2$-ary and $3$-ary questions are asked on average. (Since of course as you know logarithms translate multiplicative features into additive ones.) To be honest though I never really bothered to put in enough effort into thinking about it. $\endgroup$ – Chill2Macht Sep 14 '18 at 23:58

One has $$\log_a(x)=\frac{\log b}{\log a}\:\log_b(x)$$ thus $$H(X)= \sum_{i=1}^n p_i \log_a(p_i)=\frac{\log b}{\log a}\sum_{i=1}^n p_i \log_b(p_i)$$ or $$ H_a(X)=\frac{\log b}{\log a}H_b(X). $$


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