Multidistance
We continue to let \(G\) be an abelian group.
DefinitionMultidistance#
Let \(m\) be a positive integer, and let \(X_{[m]} = (X_i)_{1 \leq i \leq m}\) be an \(m\)-tuple of \(G\)-valued random variables \(X_i\). Then we define
where the \(\tilde X_i\) are independent copies of the \(X_i\).
Used by 14
LemmaMultidistance of copy#
If \(X_{[m]} = (X_i)_{1 \leq i \leq m}\) and \(Y_{[m]} = (Y_i)_{1 \leq i \leq m}\) are such that \(X_i\) and \(Y_i\) have the same distribution for each \(i\), then \(D[X_{[m]}] = D[Y_{[m]}]\).
Proof
Clear from Lemma Copy preserves entropy.
Lean
Formalized as multiDist_copy.
Sources
Uses 1 · Used by 2
LemmaMultidistance of independent variables#
If \(X_{[m]} = (X_i)_{1 \leq i \leq m}\) are jointly independent, then \(D[X_{[m]}] = \mathbb{H}[\sum_{i=1}^m X_i] - \frac{1}{m} \sum_{i=1}^m \mathbb{H}[X_i]\).
Uses 1 · Used by 3
LemmaNonnegativity#
For any such tuple, we have \(D[X_{[m]}] \geq 0\).
Proof
From Independent lower bound on sumset one has
for each \(1 \leq i \leq m\). Averaging over \(i\), we obtain the claim.
Lean
Formalized as multiDist_nonneg.
Sources
Uses 2 · Used by 3
LemmaRelabeling#
If \(\phi: \{1,\dots,m\} \to \{1,\dots,m\}\) is a bijection, then \(D[X_{[m]}] = D[(X_{\phi(j)})_{1 \leq j \leq m}]\).
Uses 1 · Used by 2
LemmaMultidistance and Ruzsa distance, I#
Let \(m \ge 2\), and let \(X_{[m]}\) be a tuple of \(G\)-valued random variables. Then
Proof
By Copy preserves Ruzsa distance, Multidistance of copy we may take the \(X_i\) to be jointly independent. From Independent lower bound on sumset, we see that for any distinct \(1 \leq j,k \leq m\), we have
and hence by Ruzsa distance in independent case
Summing this over all pairs \((j,k)\), \(j \neq k\) and using Lemma Multidistance of independent variables, we obtain the claim.
Lean
Formalized as multidist_ruzsa_I.
Sources
Uses 6 · Used by 2
LemmaMultidistance and Ruzsa distance, II#
Let \(m \ge 2\), and let \(X_{[m]}\) be a tuple of \(G\)-valued random variables. Then
Proof
From Ruzsa triangle inequality we have \(d[X_j;X_j] \leq 2 d[X_j;-X_k]\), and applying this to every summand in Multidistance and Ruzsa distance, I, we obtain the claim.
Lean
Formalized as multidist_ruzsa_II.
Sources
Uses 3 · Used by 2
LemmaMultidistance and Ruzsa distance, III#
Let \(m \ge 2\), and let \(X_{[m]}\) be a tuple of \(G\)-valued random variables. If the \(X_i\) all have the same distribution, then \(D[X_{[m]}] \leq m d[X_i;X_i]\) for any \(1 \leq i \leq m\).
Proof
By Copy preserves Ruzsa distance, Multidistance of copy we may take the \(X_i\) to be jointly independent. Let \(X_0\) be a further independent copy of the \(X_i\). From Kaimonovich–Vershik–Madiman inequality, we have
and hence by Negation preserves entropy and Ruzsa distance in independent case
On the other hand, by Independent lower bound on sumset we have
and the claim follows.
Lean
Formalized as multidist_ruzsa_III.
Sources
Uses 8 · Used by 2
LemmaMultidistance and Ruzsa distance, IV#
Let \(m \ge 2\), and let \(X_{[m]}\) be a tuple of independent \(G\)-valued random variables. Let \(W := \sum_{i=1}^m X_i\). Then
Proof
Take \((X'_i)_{1 \leq i \leq m}\) to be further independent copies of \((X_i)_{1 \leq i \leq m}\) (which exist by Existence of independent copies), and write \(W' := \sum_{i=1}^m X'_i\). Fix any distinct \(a,b \in I\).
From Kaimanovich-Vershik-Madiman inequality one has
and also
Combining this with (1) and then applying Independent lower bound on sumset we have
Averaging this over all choices of \((a,b)\) gives \(\mathbb{H}[W] + 2 D[X_{[m]}]\), and the claim follows from Ruzsa distance in independent case.
Lean
Formalized as multidist_ruzsa_IV.
Sources
Uses 6 · Used by 1
PropositionVanishing#
If \(D[X_{[m]}]=0\), then for each \(1 \leq i \leq m\) there is a finite subgroup \(H_i \leq G\) such that \(d[X_i; U_{H_i}] = 0\).
Proof
From Multidistance and Ruzsa distance, II and Distance nonnegative we have \(d[X_j; X_j]=0\) for all \(1 \leq j \leq m\). The claim now follows from General 100% inverse theorem.
Lean
Formalized as multidist_eq_zero.
Sources
Uses 4 · Used by 1
With more effort one can show that \(H_i\) is independent of \(i\), but we will not need to do so here.
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