Hybrid LDPC codes and iterative decoding methods - i3s

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Hybrid LDPC codes and iterative decoding methods - i3s

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)

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