EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 4, 2024, 4225-4237
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Weights of Codewords in Random Binary Linear Codes
and Their Correlation Properties

İbrahim Özen

Department of Matheamatics, Faculty of Science, Marmara University, Istanbul, Turkey.

Abstract. A classical problem in coding theory addresses moments of the weight spectrum dis-
tribution. The results in this work are on the weight moments of individual codewords rather
than the weight spectrum. The expectations of single and pairwise products of weights of nonzero
words in a random binary linear code are given. We show that the covariance between the weights
of any pair of distinct nonzero words is zero. Our main theorem has an application to sequence
correlations problem. We prove that the sums of out of phase self correlations, as well as sums of
cross-correlations, of nonzero words in a random binary linear code are equal to zero.

2020 Mathematics Subject Classifications: 94B05, 94A55, 94B65

Key Words and Phrases: Random binary linear codes, Weight moments of codewords, Binary
sequences, Auto-correlations, Cross-correlations

1. Introduction

A binary linear code C is a subspace of Fn
2 . We call n the length of the code. Dimension

of the code is its dimension as a linear space. A vector in the code is called a codeword.
Let c = (c1, c2, . . . , cn) and e = (e1, e2, . . . , en) be two binary vectors. The Hamming
distance between them is the number of coordinates that they differ and is defined by

dH(c, e) = |{i : ci ̸= ei}|.

For every vector c, the number of nonzero positions in c is called its weight and we denote
it by ∥c∥. The smallest weight of the nonzero codewords is called the minimum weight
of the code. Since C is a linear space, the minimum weight of the code coincides with
the minimum of the distances between distinct codewords of the code. If C ⊂ Fn

2 is a
linear code with dimension k and minimum weight d, we say that C is a binary [n, k, d]
code. If we let Ai denote the number of codewords with weight i, then the sequence
(A0, Ad, Ad+1, . . . , An) is called the weight distribution or the weight spectrum of the
code.

DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5600

Email address: iozen@marmara.edu.tr (İ. Özen)

https://www.ejpam.com 4225 Copyright: © 2024 The Author(s). (CC BY-NC 4.0)



İbrahim Özen / Eur. J. Pure Appl. Math, 17 (4) (2024), 4225-4237 4226

Every important feature of a code is encoded in its weight distribution. For exam-
ple, if the weight distribution of a linear code is known we can obtain estimations of its
decoding error probability. Gallager estimated the decoding error probability of a ran-
dom linear code in [4]. So we have a way to compare this property of a specific code as
well as information on what is possible for the same quality. This estimation is based on
the expectations of the weight distribution of random linear codes. The expectations of
weights are part of the code weight distribution problem, namely the evaluations of the
expectations

E (Ai1Ai2 · · ·Aik)

for random linear codes. Second and third joint moments of the weight distribution of
a random linear code were explored in [6] and [1] independently. Fourth moments were
given in [8]. In [9] and [7], big orders of the moments were studied.

In this work we focus on the weight moments of individual codewords in a random
binary linear code. Our results have applications on the expectations of correlation sums
of words.

Let us denote the words of an [n, k] random binary linear code by C = {c0 =
0, c1, . . . , c2k−1}. The weights of the words will be denoted by {w0 = 0, w1, . . . , w2k−1}
respectively. In Theorem 1 we obtain the expectations of the nonzero words’ weights in a
random [n, k] binary linear code and they are given by

E(wi) =
n

2
, for i ̸= 0.

Expectations of pairwise products of the weights are obtained in Theorem 2 and they can
be stated as

E(wiwj) =

{
n2+n

4 , for i = j, i, j ≥ 1,
n2

4 , for i ̸= j, i, j ≥ 1.

An immediate consequence is that the weights of distinct nonzero words in a random
binary linear code are statistically uncorrelated.

covariance(wi, wj) = E(wiwj)− E(wi)E(wj)

= 0, for all i, j ≥ 1 and i ̸= j

We apply these results to obtain correlation properties of words in a random code.
Let c = (c1, c2, . . . , cn) and e = (e1, e2, . . . , en) be two vectors in Fn

2 . The periodic
auto-correlation and cross-correlation functions on vectors of Fn

2 are defined respectively
by

rc,c(u) = rc(u) =

n∑
i=1

(−1)ci+ci+u and

rc,e(u) =

n∑
i=1

(−1)ci+ei+u ,



İbrahim Özen / Eur. J. Pure Appl. Math, 17 (4) (2024), 4225-4237 4227

where the indices are evaluated modulo n.
Let c and e be two distinct nonzero codewords in a random binary linear [n, k] code.

We obtain in Theorem 3 that

n−1∑
u=1

E(rc(u)) = 0 and

n−1∑
u=0

E(rc,e(u)) = 0.

2. Characteristic Vectors of Codes and Weights of Nonzero Codewords

Our main theorem is based on the characterization of weights of nonzero words in
terms of the characteristic vector of a code, given in [2].

We will review this characterization in the following paragraphs.
Let k ≥ 2 and i be integers with 1 ≤ i ≤ 2k−1. We denote by i = (i0, i1, . . . , ik−1) ∈ Fk

2,
the binary expansion of i =

∑k−1
j=0 ij2

j . We form the k × (2k − 1) matrix Gk, whose ith

column is iT .
Let C be a binary linear [n, k] code and let G be a generator matrix of C. We will

show below that with high probability a random matrix G has no columns of zeros. We
will assume from now on that C is a code with a generator matrix without zero columns.
For such C and G, we define the characteristic vector χC of C with respect to G as the
vector χC = (h1, h2, . . . , h2k−1)

T , where hi is the number of columns in G equal to the ith
column iT of Gk. Clearly we have

∑
j hj = n (recall that G has no zero columns).

Now since any nonzero codeword is obtained by a nontrivial linear combination of rows
of G, all nonzero codewords of C are obtained as the rows of the matrix product GT

kG. So

we identify the nonzero words of C with the rows of the product GT
kG; C = {0}∪{ci}2

k−1
i=1 ,

where ci is the ith row of GT
kG.

If we denote the matrix product GT
kGk by Mk, we can express the relation between the

weights of the nonzero codewords of C and the characteristic vector χC as in the following
lemma.

Lemma 1. [2, Lemma 1] Let C be a binary [n, k] linear code, let G be a generator matrix
of C and let χC be the characteristic vector of C with respect to G. Then the weight of
ith row ci of G

T
kG equals to the ith entry of the column vector MkχC

MkχC =


w1

w2
...

w2k−1

 ,

where {w1, w2, . . . , w2k−1} is the set of weights of all nonzero words in C.



İbrahim Özen / Eur. J. Pure Appl. Math, 17 (4) (2024), 4225-4237 4228

Let f be a sequence of real numbers of length 2k, whose terms are indexed from 0 to
2k−1 and let us denote by f(i) its ith coordinate. Then we define its Walsh transformation
f̂ as a sequence of real numbers with the same length, whose ith entry is given by

f̂(i) =
2k−1∑
j=0

f(j)(−1)⟨j,i⟩, 0 ≤ i ≤ 2k − 1,

where ⟨ , ⟩ is the standard inner product on Fk
2, i and j are the binary expansions of i and

j respectively.
From χC = (h1, h2, . . . , h2k−1)

T we obtain a sequence of integers of length 2k by χC =
(h0 = 0, h1, h2, . . . , h2k−1)

T . Then the Walsh transform of χC is the vector whose ith entry
is given by

χ̂C(i) =

2k−1∑
j=0

χC(j)(−1)⟨j,i⟩ =

2k−1∑
j=0

hj(−1)⟨j,i⟩.

We will also need the matrix

Mk =

 0 0 . . . 0
... Mk

0

 .

The Walsh transform of the characteristic vector is expressed in terms of the Hadamard
matrix obtained by Sylvester construction. These matrices are constructed recursively
with the help of the Hadamard product as follows:

H1 =

[
1 1
1 −1

]
, Hi = H1 ⊗Hi−1.

If we index the rows and columns of Hk by 0 ≤ i ≤ 2k − 1 and 0 ≤ j ≤ 2k − 1 respectively,
then the (i, j) entry of Hk is given by [5, Page 11]

Hk[i, j] = (−1)⟨i,j⟩. (1)

The lemma below is taken from [2], we will prove it for completeness.

Lemma 2. [2, Page 267] Let C be a binary [n, k] linear code, let G be a generator matrix
of C, and let χC be the characteristic vector of C with respect to G. Then we have

HkχC = (J − 2Mk)χC =


n

n− 2w1

n− 2w2
...

n− 2w2k−1

 = χ̂C ,

where J is the all one matrix, {w1, w2, . . . , w2k−1} is the set of weights of all nonzero words
in C.



İbrahim Özen / Eur. J. Pure Appl. Math, 17 (4) (2024), 4225-4237 4229

Proof. We will index all entries of the matrices and vectors starting from 0. All
assertions for the terms indexed by zero are trivial. Let i be an integer with 1 ≤ i ≤ 2k−1.
Then by (1) the term indexed by i in the product HkχC equals

2k−1∑
j=0

(−1)⟨i,j⟩χC(j) =
2k−1∑
j=0

(−1)⟨i,j⟩hj , (2)

with i and j being the binary representations of i and j respectively.

The same term in the product (J − 2Mk)χC is
∑2k−1

j=0 (1− 2⟨si, sj⟩)hj , where sl’s are
the column vectors of Gk. One can easily confirm that we have

1− 2⟨si, sj⟩ = 1− 2⟨i, j⟩ = (−1)⟨i,j⟩.

So we have
∑2k−1

j=0 (1− 2⟨si, sj⟩)hj =
∑2k−1

j=0 (−1)⟨i,j⟩hj and by (2), the first equality of the
lemma follows. The second equality follows from Lemma 1.

If we compare the ith coordinates of HkχC and χ̂C we obtain the last equality of the
lemma

(HkχC)(i) =

2k−1∑
j=0

hj(−1)⟨i,j⟩ = χ̂C(i).

In the next section we will make use of the identity in the following lemma. The
assertion of the first line is trivial and the second line is an easy consequence of [3, Lemma
3, Page 58].

Lemma 3. For any integer 0 ≤ i ≤ 2k − 1, we have

2k−1∑
j=0

(−1)⟨j,i⟩ =

{
2k, for i = 0,
0, for i ̸= 0.

3. First and Second Moments of the Codeword Weights

A random binary linear code is the row-space of a random k×n matrix, whose entries
are identically and independently distributed in F2. A well known property of a random
k×n matrix over a finite field is that when k/n < 1, the matrix has rank k with probability
tending to 1, as n tends to infinity. So we will assume that a random binary k × n
matrix whose terms are distributed identically and independently in F2 has rank k and
its row space C is an [n, k] linear code. The words of such a code will be denoted by
C = {c0 = 0, c1, c2, . . . , c2k−1}, with the same relation to GT

kG as above. The weights of
the nonzero words will be denoted by {w1, w2, . . . , w2k−1} respectively.

Another fact about this ensemble of matrices is that, the probability that a random
k × n matrix has zero columns tends to 0 as n tends to infinity with k/n > 0. This can
easily be observed by

lim
n→∞

1− (2k − 1)n

2kn
= 0,



İbrahim Özen / Eur. J. Pure Appl. Math, 17 (4) (2024), 4225-4237 4230

with the condition that R = k/n ̸= 0.
By Lemma 2, the expectations of weights of the nonzero codewords in a random code

can be calculated by

E(n− 2wi) = E (χ̂C(i))

=
2k−1∑
j=0

E(hj)(−1)⟨j,i⟩.

For a positive integer n, we denote by [n] the set {0, 1, . . . , n − 1}. We can calculate
the expectations of the pairwise products of nonzero words’ weights as follows.

E[(n− 2wi)(n− 2wj)] = E (χ̂C(i)χ̂C(j))

=
∑

s,m∈[2k]

E(hshm)(−1)⟨s,i⟩+⟨m,j⟩.

Let us denote by P (hs), the probability that a random binary k×n matrix has exactly
hs columns that are equal to sT , for 0 ≤ s ≤ 2k − 1. Similarly, let P (hs, hm) denote the
probability that a random matrix has exactly hs columns equal to sT and hm columns
equal to mT , for 0 ≤ s,m ≤ 2k − 1 and s ̸= m.

Proposition 1. For a random binary k × n matrix, we have

P (hs) =

(
n

hs

)
(2k − 1)(n−hs)

2kn

and

P (hs, hm) =

(
n

hs + hm

)(
hs + hm

hm

)
(2k − 2)(n−(hs+hm))

2kn
·

Proof. The number of matrices with exactly hs columns equal to sT is given by(
n

hs

)
(2k − 1)(n−hs).

The binomial coefficient gives the number of different choices for placing sT and the factor
(2k − 1)(n−hs) counts the number of ways to fill the other columns with the remaining
(2k − 1) vectors arbitrarily. The number of k × n binary matrices is 2kn. So we have

P (hs) =

(
n

hs

)
(2k − 1)(n−hs)

2kn
·

Now let 0 ≤ s,m ≤ 2k − 1 and s ̸= m. For the number of k × n binary matrices with
exactly hs columns equal to sT and hm columns equal to mT , we fix (hs + hm) positions
for sT and mT . Hence we have the binomial coefficient

(
n

hs+hm

)
. Then we determine the

hm positions among the (hs + hm) columns for the vector mT . The number of choices



İbrahim Özen / Eur. J. Pure Appl. Math, 17 (4) (2024), 4225-4237 4231

for that is
(
hs+hm

hm

)
. The choice for columns sT is determined by that of mT uniquely.

The columns other than these (hs + hm) positions can be filled by the remaining (2k − 2)
vectors arbitrarily. So we have

P (hs, hm) =

(
n

hs + hm

)(
hs + hm

hm

)
(2k − 2)(n−(hs+hm))

2kn
·

Theorem 1. Let C be a random binary linear [n, k] code with words and weights denoted
as above. Then the expectations of the weights of nonzero codewords are given by

E(wi) =
n

2
for i ≥ 1. (3)

Proof. By Lemma 2, we have

E(n− 2wi) = E (χ̂C(i)) =
∑
s∈[2k]

E(hs)(−1)⟨s,i⟩.

By Proposition 1 we proceed as follows:

E(n− 2wi) =
∑
s∈[2k]

E(hs)(−1)⟨s,i⟩

=
∑
s∈[2k]

∑
hs∈[n+1]

P (hs)hs(−1)⟨s,i⟩

=
∑
s∈[2k]

∑
hs∈[n+1]

(
n
hs

)
(2k − 1)n−hs

2kn
hs(−1)⟨s,i⟩

=
∑
s∈[2k]

∑
hs∈[n+1]

(
n
hs

)
(2k − 1)hs

2kn
(n− hs)(−1)⟨s,i⟩

In the last equality we performed a change of variables and put (n−hs) for hs, by the
symmetry of the binomial coefficients. Let K1 = 2k−1. In the following equation we have
derivative with respect to K1.

E(n− 2wi) =
∑
s∈[2k]

(
n− K1

2kn
[(K1 + 1)n]′

)
(−1)⟨s,i⟩

=

(
n− K1

2kn

[
n(2k)n−1

]) ∑
s∈[2k]

(−1)⟨s,i⟩

=
( n

2k

) ∑
s∈[2k]

(−1)⟨s,i⟩

= 0 for i ≥ 1.



İbrahim Özen / Eur. J. Pure Appl. Math, 17 (4) (2024), 4225-4237 4232

The last equality is obtained by Lemma 3. So the expectations of the nonzero weights
follow

E(wi) =
n

2
for i ≥ 1.

We will explore the expectations of the products of nonzero weights in the next theo-
rem.

Theorem 2. Expectations of the pairwise products of nonzero weights in a random binary
linear [n, k] code are given by

E(wiwj) =

{
n2+n

4 , for i = j, i, j ≥ 1,
n2

4 , for i ̸= j, i, j ≥ 1.
(4)

Proof. By Lemma 2, the expectations of the products of nonzero weights are given by

E((n− 2wi)(n− 2wj)) = E (χ̂C(i)χ̂C(j))

=
∑

s,m∈[2k]

E(hshm)(−1)⟨s,i⟩+⟨m,j⟩

=
∑
s∈[2k]

E(h2s)(−1)⟨s,(i+j)⟩

+
∑

s,m∈[2k]
s ̸=m

E(hshm)(−1)⟨s,i⟩+⟨m,j⟩.

The first sum in the last equality is the contribution of cases where s = m and the second
sum is the contribution of the cases where s ̸= m. Now let K1 and K2 denote 2k − 1 and
2k − 2 respectively.

E((n− 2wi)(n− 2wj))=
1

2kn

∑
s∈[2k]

hs∈[n+1]

(
n

hs

)
K

(n−hs)
1 h2s(−1)⟨s,(i+j)⟩

+
∑

s,m∈[2k]
s ̸=m

hs,hm∈[n+1]

(
n

hm hs

)
K

(n−hs−hm)
2 hshm(−1)⟨s,i⟩+⟨m,j⟩

2kn
,

where (
n

hm hs

)
=

n!

(hm)!(hs)!(n− (hm + hs))!

is the multinomial coefficient. We continue with a change of variables T = hs + hm in the
second sum and we get

E((n− 2wi)(n− 2wj))=
1

2kn

∑
s∈[2k]

hs∈[n+1]

(
n

hs

)
K

(n−hs)
1 h2s(−1)⟨s,(i+j)⟩



İbrahim Özen / Eur. J. Pure Appl. Math, 17 (4) (2024), 4225-4237 4233

+
∑

s,m∈[2k]
s ̸=m

T∈[n+1]
hm∈[T+1]

(
n
T

)(
T
hm

)
K

(n−T )
2 (T − hm)hm(−1)⟨s,i⟩+⟨m,j⟩

2kn

=
1

2kn

∑
s∈[2k]

hs∈[n+1]

(
n

hs

)
K

(n−hs)
1 (h2s − hs)(−1)⟨s,(i+j)⟩

+
1

2kn

∑
s∈[2k]

hs∈[n+1]

(
n

hs

)
K

(n−hs)
1 hs(−1)⟨s,(i+j)⟩

−
∑

s,m∈[2k]
s ̸=m

T∈[n+1]
hm∈[T+1]

(
n
T

)(
T
hm

)
K

(n−T )
2 (h2m − hm)(−1)⟨s,i⟩+⟨m,j⟩

2kn

+
∑

s,m∈[2k]
s ̸=m

T∈[n+1]
hm∈[T+1]

(
n
T

)(
T
hm

)
K

(n−T )
2 (T − 1)hm(−1)⟨s,i⟩+⟨m,j⟩

2kn

(5)

After summation over hs, hm and T we obtain the following, where the derivatives are
taken with respect to the variables u and v. After derivatives we substitute u = 1 and
v = 2.

E((n− 2wi)(n− 2wj)) =
1

2kn

∑
s∈[2k]

[(K1 + u)n]′′(−1)⟨s,(i+j)⟩

+
1

2kn

∑
s∈[2k]

[(K1 + u)n]′(−1)⟨s,(i+j)⟩

+
1

2kn

∑
s,m∈[2k]
s̸=m

[(K2 + v)n]′′(−1)⟨s,i⟩+⟨m,j)⟩

=
n(n− 1)

22k

∑
s∈[2k]

(−1)⟨s,(i+j)⟩



İbrahim Özen / Eur. J. Pure Appl. Math, 17 (4) (2024), 4225-4237 4234

+
n

2k

∑
s∈[2k]

(−1)⟨s,(i+j)⟩

+
n(n− 1)

22k

∑
s,m∈[2k]
s ̸=m

(−1)⟨s,i⟩+⟨m,j)⟩

By Lemma 3 the result is

E((n− 2wi)(n− 2wj)) =

{
n, for i = j, i, j ≥ 1,
0, for i ̸= j, i, j ≥ 1.

Finally we get the expectations of products of nonzero codeword weights as

E(wiwj) =

{
n2+n

4 , for i = j, i, j ≥ 1,
n2

4 , for i ̸= j, i, j ≥ 1.

Two random variables X and Y are called statistically uncorrelated if the covariance
between them is 0,

cov(X,Y ) = E(XY )− E(X)E(Y ) = 0.

We observe that this is the case for the weights of distinct nonzero codewords in a random
code.

Corollary 1. Weights of distinct nonzero words of a random binary linear code are un-
correlated

cov(wi, wj) = 0, for all i, j ≥ 1 and i ̸= j.

4. Correlation Properties of Nonzero Words in a Random Binary Liner
Code

Let c = (c1, c2, . . . , cn) and e = (e1, e2, . . . , en) be two vectors in Fn
2 . We define the

periodic auto-correlation and cross-correlation functions on vectors of Fn
2 as follows

rc,c(u) = rc(u) =
n∑

i=1

(−1)ci+ci+u and

rc,e(u) =
n∑

i=1

(−1)ci+ei+u

respectively. The indices are evaluated modulo n.
Given e ∈ Fn

2 let us denote by eu, the vector obtained by shifting each coordinate of e
by u to the left

eu = (e1+u, e2+u, . . . , en, e1, . . . , eu).



İbrahim Özen / Eur. J. Pure Appl. Math, 17 (4) (2024), 4225-4237 4235

The correlation functions rc and rc,e evaluates the following: If in a coordinate the
vectors agree, then this coordinates contributes a (+1) and if the the entries in the same co-
ordinate are different this coordinate contributes a (−1). So we can rewrite the correlation
functions as follows

rc(u) =
n∑

i=1

(−1)ci+ci+u = n− 2dH(c, cu) and (6)

rc,e(u) =
n∑

i=1

(−1)ci+ei+u = n− 2dH(c, eu), (7)

where dH is the Hamming distance. The following formulation of Hamming distance on
Fn
2 will be useful

dH(c, e) = ∥c∥+ ∥e∥ − 2⟨c, e⟩, (8)

where ∥ · ∥ denotes the weight of the vector and the inner product is evaluated over R.
The following theorem gives the expectations of sums of auto-correlations and sums of

cross-correlations between the nonzero words in a random binary linear code.

Theorem 3. Let C be a random binary linear [n, k] code and let c and e be two distinct
nonzero codewords. Then we have

n−1∑
u=1

E(rc(u)) = 0 and

n−1∑
u=0

E(rc,e(u)) = 0.

Proof. By (6) and (8) we have

rc(u) = n− 2dH(c, cu)

= n− 2(∥c∥+ ∥cu∥ − 2⟨c, cu⟩).

Summation over u and taking expectations of both sides gives

n−1∑
u=1

E(rc(u)) =
n−1∑
u=1

[n− 2(E(∥c∥) + E(∥cu∥)− 2E(⟨c, cu⟩))]

= −n(n− 1) + 4E

(
n−1∑
u=1

⟨c, cu⟩

)
.

In the last equation we made use of (3). Now we will utilize the following simple fact: For
any c = (c1, c2, . . . , cn) and e = (e1, e2, . . . , en) in Fn

2 we have

∥c∥∥e∥ = (c1 + c2 + . . .+ cn)(e1 + e2 + . . .+ en)



REFERENCES 4236

=
n−1∑
u=0

⟨c, eu⟩. (9)

With this fact, (3) and (4), we have

n−1∑
u=1

E(rc(u)) = −n(n− 1) + 4 (E(∥c∥∥c∥)− E(∥c∥))

= −n(n− 1) + 4

(
n2 + n

4
− n

2

)
= 0. (10)

Now let c and e be two distinct nonzero codewords in C. For the cross-correlations,
following the same steps as above we get

n−1∑
u=0

E(rc,e(u))) =

n−1∑
u=0

E[n− 2dH(c, eu)]

=

n−1∑
u=0

[n− 2(E(∥c∥) + E(∥eu∥)− 2E(⟨c, eu⟩))]

= −n2 + 4E

(
n−1∑
u=0

⟨c, eu⟩

)
= −n2 + 4E(∥c∥∥e∥).

Finally by (4) we get the result

n−1∑
u=0

E(rc,e(u)) = 0.

5. Conclusion

We obtained the expectations of single and pairwise products of weights of codewords
in random binary linear codes. We showed that the weights of two nonzero words are
statistically uncorrelated. We used the expectations of the products of nonzero weights to
show that the expectations of the sums of out of phase auto-correlations and the expec-
tations of the sums of cross-correlations of nonzero words are zero.

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