in seperations for each step
j
, i.e.
Δ
X
(
t
) = Δ
X
(
n
Δ
t
) =
n
j
=1
Δ
X
j
(5)
So in order to find
Δ
X
2
(
t
) , one will have to square equation (5) and take
its average, i.e.
Δ
X
2
(
t
)
=
n
i
=1
Δ
X
i
n
j
=1
Δ
X
j
(6)
Now insert the expression for Δ
X
j
,
c.f.
equation (4)
Δ
X
2
(
t
)
=
n
i
=1
(Δ
x
2
,i

Δ
x
1
,i
)
n
j
=1
(Δ
x
2
,j

Δ
x
1
,j
)
(7)
Δ
X
2
(
t
)
=
n
i
=1
n
j
=1
(Δ
x
2
,i

Δ
x
1
,i
)(Δ
x
2
,j

Δ
x
1
,j
)
Now evaluate the sums
Δ
X
2
(
t
)
=
Δ
x
2
,
1
Δ
x
2
,
1

Δ
x
2
,
1
Δ
x
1
,
1

Δ
x
1
,
1
Δ
x
2
,
1
+
...
(9)
Δ
x
2
,
1
Δ
x
1
,
1
=
Δ
x
2
,
1
Δ
x
1
,
1
= 0 as the particles are uncorrelated and thus
Δ
x
i
= 0.
Similarly,
Δ
x
2
,
1
Δ
x
2
,
2
=
Δ
x
2
,
1
Δ
x
2
,
2
= 0 as the steps are uncorrelated.
Thus the only remaining terms in the double sum will be:
Δ
X
2
(
t
)
=
n
j
=1
(Δ
x
1
,j
)
2
+ (Δ
x
2
,j
)
2
(10)
Now realize that:
Δ
X
2
j
(
t
)
=
(Δ
x
2

Δ
x
1
)
2
(11)
Δ
X
2
j
(
t
)
=
(Δ
x
2
2

2Δ
x
1
Δ
x
2
+ Δ
x
2
1
)
(12)
2