<p>(Comment --> answer)</p>

<p>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 $x=2$.</p>

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

CC BY-SA 3.0
2015-07-15T00:55:15.740Z
2015-07-15T00:55:15.740Z
usul
<p>I want to show the following inequality, which seems like it should have an elementary proof (or even a well-known name).</p>

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

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

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

CC BY-SA 3.0
2015-07-14T15:23:01.223Z
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.
usul
<pr.probability>
Inequality for p-norm of probability distributions
David Harris
# Inequality for p-norm of probability distributions

(Asked by: username_0 on 2015-07-14)

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 $\sum p\_i^r \leq x^r \sum q\_i^r$.

**Tags:** pr.probability
> **Comments:**
> **username_1**: 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.  

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## 1 answers

### Answer by username_1 on 2015-07-15. Score: 2 | Accepted answer

(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 $x^{r-1}\sum q\_i^r \to \frac{2^{r-1}}{2^r} = \frac{1}{2}$.


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