2.7 Proofs of theorems in Chapter 2 83

2.7.1 Symmetry

Lemma 6. If W is a symmetric LDR r**and**om vector, then its extension W ×A , for any

linear map A selected from E 1,2 , is also symmetric. The truncation of W by the inverse

of A, denoted by W ×A−1 , is also symmetric.

Proof: We first prove that any q 2 -sized extension of a q 1 -sized symmetric r**and**om

vector remains symmetric. We want to show that

Case b /∈ Im(A):

∀b ∈ [0, q 2 − 1], P(W ×A = w) = e w b

P(W ×A = w +b ) (2.26)

• In the case when w b ≠ −∞:

We have to show that

e −w b

P(W ×A = w) = P(W ×A = w +b )

If w b ≠ ∞, then P(W ×A = w) = 0. If w b = ∞, then e −w b = 0. Thus, we have to

show that

∀b /∈ Im(A), P(W ×A = w +b ) = 0 (2.27)

This is equivalent to show that ∃i /∈ Im(A) such that w i +b ≠ ∞. We have w i +b =

w b+i −w b . It is sufficient to choose i = b, then w +b

b

= −w b . Since w +b

b

= −w b ≠ ∞

by hypothesis, P(W ×A = w +b ) = 0.

• In the case w b = −∞, to prove that equation (2.26) is fulfilled we have to prove

that P(W ×A = w) = 0, which is straight forward because b /∈ Im(A), **and** hence

P(W ×A = w) ≠ 0 ⇒ w b = ∞. By taking the contraposition, we end on the

wanted result.

Hence we have proved equation (2.26) in the case where b /∈ Im(A).

Case b ∈ Im(A):

We have

P(W ×A = w) = P(W = w ×A−1 )Π i/∈Im(A) δ w i ,∞

Since b belongs to Im(A), we denote by a the element in [0, q 1 − 1] such that b = Aa.

The input message W is symmetric, hence we have

P(W = w ×A−1 ) = e w Aa

P(W = (w ×A−1 ) +a )

∀i ∈ [0, q 1 − 1], (w ×A−1 ) +a

i

= w ×A−1

i+a

− w ×A−1

a

= w A(i+a) − w Aa

= w +Aa

Ai

= (w +Aa ) ×A−1

i