I want to show the following inequality, which seems like it should have an elementary proof (or even a well-known name).

Suppose $p, q$ are discrete probability distributions. And suppose that $p_i \leq x q_i$ for some $x > 1$. Then I want to show that for $r > 1$ we have the inequality: $$ \sum p_i^r \leq x^{r-1} \sum q_i^r $$

If we take away the requirement that $p$ is a probability distribution, then of course we have instead $\sum p_i^r \leq x^r \sum q_i^r$.

EDIT: This is not true. The best inequality one can show is the obvious one $\sum p_i^r \leq x^r \sum q_i^r$.

  • 1
    $\begingroup$ Hmm, is this a counterexample to the inequality, or am I missing something? The support size is $n+1$. Let $$p = (1,0,\dots,0)$$ and $$q = (\frac{1}{2},\frac{1}{2n},\dots,\frac{1}{2n}).$$ Here $x = 2$ and think of $n \to \infty$. Then $\sum p_i^r = 1$ and $\sum q_i^r = \frac{1}{2^r} + n\left(\frac{1}{2n}\right)^r$. So we get $1 \leq 2^{r-1} \cdot \frac{1}{2^r} + O(\frac{1}{n^{r-1}})$, or $$1 \leq \frac{1}{2} + O(\frac{1}{n^{r-1}}),$$ a contradiction. $\endgroup$
    – usul
    Commented Jul 14, 2015 at 15:23

1 Answer 1


(Comment --> answer)

The inequality unfortunately fails to hold, a counterexample is $$p = (1,0,\dots,0)$$ and $$q = \left(\frac{1}{2},\frac{1}{2n},\dots,\frac{1}{2n}\right),$$ where the support size is $n+1$ and $x=2$.

Then $\sum p_i^r = 1$, but $x^{r-1}\sum q_i^r \approx \frac{1}{2}$: \begin{align} \sum q_i^r &= \frac{1}{2^r} + n\left(\frac{1}{2n}\right)^r \\ &= \frac{1}{2^r}\left(1 + \frac{1}{n^{r-1}}\right) \\ &\to \frac{1}{2^r} \end{align} as $n \to \infty$, so $x^{r-1}\sum q_i^r \to \frac{2^{r-1}}{2^r} = \frac{1}{2}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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