86 Chapitre 2 : HybridLDPC Codes Therefore ∀(i, j) ∈ [1, q k − 1] × [1, q l − 1], 2.7.3 LM-invariance Card(A ∈ E k,l : Ai = j) Card(E k,l ) = 1 q l − 1 Lemma 7. If a probability-vector random variable Y of size q 2 is LM-invariant, then for all (i, j) ∈ [0, q 2 − 1] × [0, q 2 − 1], the random variables Y i and Y j are identically distributed. Proof: For any (q 1 , q 2 ), q 1 < q 2 , T 1,2 denotes the set of all truncations from G(q 2 ) to G(q 1 ). We assume Y LM-invariant. A −1 and B −1 denote two truncations independently arbitrary chosen in T 1,2 . For any l and k in [0, q 2 − 1], we can choose extension A such that l ∈ Im(A) and A −1 l is denoted by i. Also, we choose B such that Bi = k. Y LM-invariant implies ∀(i, A −1 , B −1 ) ∈ [0, q 1 − 1] × T 1,2 × T 1,2 , P(Y ×A−1 i = x) = P(Y ×B−1 i = x) This is equivalent to P(Y Ai = x) = P(Y Bi = x) and hence P(Y l = x) = P(Y k = x), ∀(l, k) ∈ [0, q 2 − 1] × [0, q 2 − 1] Lemma 8. A probability-vector random variable Y of size q 2 is LM-invariant if and only if there exist q 1 and a probability-vector random variable X of size q 1 such that Y = ˜X. Proof: Let us first assume Y = ˜X and prove that Y is LM-invariant. This means that we want to prove that for any (B, C) ∈ E 1,2 × E 1,2 , Y ×B−1 and Y ×C−1 are identically distributed. By hypothesis Y = X ×A , with A uniformly chosen in E 1,2 . We define the matrix α A of size q 2 × q 1 . This matrix is such that Y = α A X and is defined by ∀j = 0 . . .q 1 − 1, ∀i = 0 . . .q 2 − 1, α A (i, j) = 1 if i=Aj Thus, vector Y truncated by any linear map B is expressed by: The same holds for linear map C: Y ×B−1 = α T B α AX Y ×C−1 = α T Cα A X = 0 otherwise □

2.7 Proofs of theorems in Chapter 2 87 αB T α A and αC T α A correspond to a selection of q 1 rows of α A . Thus, showing that Y ×B−1 and Y ×C−1 are identically distributed is equivalent to show that both matrices αB T α A and αC T α A are identically distributed, for any B and C in E 1,2 and for A uniformly chosen in E 1,2 . The number of elements of X, whose indexes are in Im(A) and which are selected by αB T , is equal to the cardinality of Im(A) ∩ Im(B). The same holds for C. E A (f(A, B)) denotes the expectation of the function f applied to random variables A and B, over all the realizations of A. Let us first show that E A (Card(Im(B) ∩ Im(A))) = E A (Card(Im(C) ∩ Im(A))) , ∀(B, C) ∈ E 1,2 ×E 1,2 , A ∼ U E1,2 (2.32) 1 ∑ E A (Card(Im(B) ∩ Im(A))) = Card(Im(B) ∩ Im(A)) Card(E 1,2 ) A∈E 1,2 = 1 Card(E 1,2 ) q 1 ∑ r=1 r · = 1 Card(E 1,2 ) q 1 ∑ r=1 r · Card(A ∈ E 1,2 : Card(Im(B) ∩ Im(A)) = r) ( ) q1 ∑ Card(A ∈ E 1,2 : Ai 1 = j 1 ,... ,Ai r = j r ) r i 1 ≠···̸=i r ∈G(q 1 ) where j 1 . . .j r are subsets of Im(B). In the same way as for lemma 10, we can show that Card (A ∈ E 1,2 : Ai 1 = j 1 , . . .,Ai r = j r ) is independent of j 1 . . .j r . Hence we conclude on equality (2.32). Let us now consider a given subset j 1 . . .j r of size r, taken from the image of any linear map in E 1,2 (hence with r ≤ q 1 ), and a given subset i 1 . . .i r of G(q 1 ) of size r. In the same way as lemma 10, we can prove that Card (A ∈ E 1,2 : Ai 1 = j 1 , . . .,Ai r = j r ) is independent of j 1 . . .j r . The first part of the proof ensures each row, of both matrices α T B α A and α T C α A, to have the same probability to contain a 1 (they have at most one 1). The second part of the proof ensures that, given r rows of α A of indexes j 1 , . . .,j r , the combination of locations of ones in the matrix α T B α A is independent of which rows j 1 , . . .,j r of α A have been selected by α T B . Hence, this combination is independent of αT B . For any (B, C) ∈ E 1,2 ×E 1,2 , for A uniformly distributed in E 1,2 , both matrices α T B α A and α T C α A are therefore identically distributed. Since Y ×B−1 = α T B α AX and Y ×C−1 = α T C α AX, Y ×B−1 and Y ×C−1 are identically distributed for any (B, C) ∈ E 1,2 × E 1,2 , that means that Y is LM-invariant. Let us now assume Y LM-invariant, and define X by X = Y ×A−1 with A uniformly chosen in E 1,2 and independent of Y. We have to show that X is independent of A. P(X = x|A) = P(Y ×A−1 = x|A) We can write, thanks to definition 8, for all B arbitrary selected from E 1,2 independently on A, P(Y ×A−1 = x|A) = P(Y ×B−1 = x|A) = P(Y ×B−1 = x)