3
$\begingroup$

I need to show that $-\sum_\limits i{p_i \log{p_i}}$ is maximal iff $p_i=p_j$ for all $i\neq j$ using the convexity inequality:

$$\phi \left(\frac{\sum{a_i}}{N}\right)\leq \frac{\sum{\phi (a_i)}}{N}$$

I tried expanding with cross-entropy and KL-divergence

$$-\sum_i{\frac{1}{N} \log{\frac{1}{N}}}=-\sum_i{\frac{1}{N} \log{p_i}}-\sum_i{\frac{1}{N} \log{\frac{N}{p_i}}}$$

But I got only trivial answers like $\log N \leq \log N$

$\endgroup$
4
$\begingroup$

Let $\phi(x) = x\log x$. This function is convex in the set $[0, 1]$.

The entropy $S$ is defined as:

$$S(p) = -\sum \phi(p_i),$$

where $p=[p_1, p_2, \ldots, p_N]$.

Then, using the Jensen's inequality, you get that:

$$ \phi \left(\frac{\sum{p_i}}{N}\right)\leq \frac{\sum{\phi (p_i)}}{N} \Rightarrow \phi \left(\frac{\sum{p_i}}{N}\right)\leq -\frac{S(p)}{N} \Rightarrow S(p) \leq -N \phi \left(\frac{\sum{p_i}}{N}\right).$$

Notice that, whichever is $p$, then by definition $\sum p_i = 1$, and hence:

$$S(p) \leq -N\phi\left(\frac{1}{N}\right) = -N\left(\frac{1}{N}\log\frac{1}{N}\right) = \log N.$$

This means that the entropy is at most equal to $\log N$.

Now, notice that $S(p) = \log(N)$ for $p = \left[\frac{1}{N}, \ldots, \frac{1}{N}\right]$.

Then: $$p_i = \frac{1}{N} \forall i \implies S(p) ~\text{is maximum}.$$

We conclude the proof by observing that the maximum $p_i = \frac{1}{N} ~\forall i$ is unique since $S(p)$ is strictly concave.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.