Upper bound for entropy-like term

For a probability vector $$P = (p_1,p_2,.. p_n)$$ i.e. $$p_i\geq 0$$ and $$\sum_{i=1}^{n} p_i = 1$$, define an entropy-like quantity

$$X(P) = \sum_{i=1}^n p_i\left(\log \frac{1}{p_i}\right)^2$$

Is there any upper bound (tighter than the one pointed out by Rammus) to $$X(P)$$ in terms of $$n$$? For $$H(P) = \sum_{i=1}^n p_i\left(\log \frac{1}{p_i}\right)$$, it is known that $$H(P)\leq \log n$$.

• Sure, here's a simple upper bound (not tight). A single term in the sum is maximized when $p = 1/\mathrm{e}^2$ and for that we get each term is no larger than $\frac{4}{e^2}$ where $e$ is the exponent related to your logarithm. Thus the sum is bounded above by $\frac{4 n}{e^2}$. Apr 4, 2021 at 14:11

Ok, here are the tight bounds. For $$n=1$$ we get $$X(P)=0$$. For the case of $$n=2$$ we just have a single parameter $$p$$ (as $$p_2 = 1-p$$) and we can directly optimize the quantity. We find that $$X(\{p,1-p\})$$ has two maxima at $$p^* = \frac{1}{2} \pm \frac{\sqrt{e^2 - 4}}{2 e}.$$ This gives a value of $$X(\{p^*,1-p^*\}) \approx 0.563$$.
Things change however for $$n>2$$. It turns out, see Proposition 8 part 3 in this paper for details, that for $$n>2$$, $$X(P)$$ has a unique maximum which is attained by the uniform distribution. Therefore, for $$n>2$$ we have $$X(P) \leq (\log n)^2.$$