<p>That implication is not actually true.
Consider $P$, $Q$, and the pseudodistribution given by:</p>

<p>$$P = x, \quad Q = x^3,$$
$$r = 4,$$
$$\mathbb{\tilde E} 1 = 1, \quad \mathbb{\tilde E} x = 0,$$
$$\mathbb{\tilde E} x^2 = 0, \quad \mathbb{\tilde E} x^3 = 0,$$
$$\mathbb{\tilde E} x^4 = 1$$</p>

<p>Then $\mathbb{\tilde E}P^2 = 0$ but $\mathbb{\tilde E} PQ = 1$.</p>

<p>This is a valid degree-4 pseudodistribution since its moment matrix is positive-semidefinite.</p>

<p>$$ \left[\begin{array}{ccc} \mathbb{\tilde E}1 &amp; \mathbb{\tilde E}x &amp; \mathbb{\tilde E}x^2 \\ \mathbb{\tilde E}x &amp; \mathbb{\tilde E}x^2 &amp; \mathbb{\tilde E}x^3 \\ \mathbb{\tilde E}x^2 &amp; \mathbb{\tilde E}x^3 &amp; \mathbb{\tilde E}x^4 \\\end{array}\right] = \left[\begin{array}{ccc} 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0  \\ 0 &amp; 0 &amp; 1 \\\end{array}\right] $$</p>

CC BY-SA 3.0
2015-12-01T20:23:17.700Z
2015-12-01T20:23:17.700Z
Jonathan Shi
<p>Let $\tilde{\mathbb{E}}$ be a degree-$r$ pseudo-distribution. If $\tilde{\mathbb{E}}P^2 = 0$, how to prove that $\tilde{\mathbb{E}}PQ=0$ for all polynomials $Q$ such that $\mathrm{deg}(PQ)\le r$?</p>

<p>We can prove the case $\mathrm{deg}(PQ) &lt; r$ by induction, but it fails when $\mathrm{deg}(PQ) = r$.</p>

<sum-of-squares>
How to prove $\tilde{\mathbb{E}}PQ=0$ when $\tilde{\mathbb{E}}P^2 = 0$?
Chao Liao
# How to prove $\tilde{\mathbb{E}}PQ=0$ when $\tilde{\mathbb{E}}P^2 = 0$?

(Asked by: username_0 on 2015-11-30)

Let $\tilde{\mathbb{E}}$ be a degree-$r$ pseudo-distribution. If $\tilde{\mathbb{E}}P^2 = 0$, how to prove that $\tilde{\mathbb{E}}PQ=0$ for all polynomials $Q$ such that $\mathrm{deg}(PQ)\le r$?

We can prove the case $\mathrm{deg}(PQ) < r$ by induction, but it fails when $\mathrm{deg}(PQ) = r$.

**Tags:** sum-of-squares

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## 1 answers

### Answer by username_1 on 2015-12-01. Score: 6 | Accepted answer

That implication is not actually true.
Consider $P$, $Q$, and the pseudodistribution given by:

$$P = x, \quad Q = x^3,$$
$$r = 4,$$
$$\mathbb{\tilde E} 1 = 1, \quad \mathbb{\tilde E} x = 0,$$
$$\mathbb{\tilde E} x^2 = 0, \quad \mathbb{\tilde E} x^3 = 0,$$
$$\mathbb{\tilde E} x^4 = 1$$

Then $\mathbb{\tilde E}P^2 = 0$ but $\mathbb{\tilde E} PQ = 1$.

This is a valid degree-4 pseudodistribution since its moment matrix is positive-semidefinite.

$$ \left[\begin{array}{ccc} \mathbb{\tilde E}1 & \mathbb{\tilde E}x & \mathbb{\tilde E}x^2 \\ \mathbb{\tilde E}x & \mathbb{\tilde E}x^2 & \mathbb{\tilde E}x^3 \\ \mathbb{\tilde E}x^2 & \mathbb{\tilde E}x^3 & \mathbb{\tilde E}x^4 \\\end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\\end{array}\right] $$


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