1 the gaussian approximation to the binomial we start with the probability of ending up j steps from the origin when taking a total of N steps, given by
P j
=
2 N
N + j
2
N !
!
N − j
2
!
(1) taking the logarithm of both sides, we have ln P j
= ln N !
− N ln 2 − ln
N + j
2
!
− ln
N − j
2
!
now we apply stirling’s approximation, which reads ln N !
≈ N ln N − N +
1
2 ln 2 πN for large N . this gives ln P j
≈ N ln N − N +
1
2 ln 2 πN − N ln 2 −
N + j N + j N + j
N
2
−
2 j ln ln
N
2
−
2 j
−
−
N
2
−
2 j
+
1
2 ln 2 π
+
1
2 ln 2 π
N + j
2
N − j
2
−
(2)
(3) the second term in each of the square brackets cancel each other. regrouping the first term in each of the square brackets together, ln P j
≈ N ln N −
N + j
2
+
1
2 ln 2 πN −
1
2 ln 2 π ln
N
N + j
2
+
2 j
−
N
−
1
2 ln 2 π
−
2 j ln
N − j
2
N − j
2
− N ln 2
(4)
(5) looking at only the terms in square brackets and rearranging a bit,
[ ] = N ln N −
N
2 ln
N +
2 j N − j
2
− j
2 ln
N
2
(1 −
4 j
2
N 2
) − j
2 ln
N
2
N
(1 +
+
2 j
+ j
N
) + j
2 j
2 ln ln
N −
2
= N ln N −
N
2 ln
= N ln N −
N
2 ln
N 2
4
−
N
2 ln(1 − j 2
N 2
) − j
2 j ln(1 +
N
) + j
2 ln(1 − j
N
) j
N
(1 −
2 j
N
) now we use the taylor expansion ln(1 ± x ) ≈ ± x for x 1 and work to “second order in j
N
”:
N j
2
[ ] ≈ N ln 2 +
2 j 2
N 2
= N ln 2 −
2 N
− j j
2 N j
−
2 j
N simplifying Eqn. (4) a bit and plugging the above in gives ln P j j 2
≈ N ln 2 −
2 N
−
1
2 ln
2 π
N
N + j N − j
2 2
− N ln 2 (6) now we approximate the second term in the same manner as before: ln P j j 2
≈ −
2 N j
2
≈ −
2 N
−
1
2 ln
π
2 N
−
1
2 ln
πN
2
N
2
(1 − j 2
N 2
)
1

we exponentiate to get P j back:
P j
≈ r
2
πN e
− j
2
2 N to get P ( x ) =
P j
2 L
, we use j = x/L and L = p
2 Dt/N :
P ( x ) =
=
1
2 r
N
2 Dt
1
√
4 πDt r
2
πN x 2 exp( −
2 N x 2 exp( −
4 Dt
)
N
2 Dt
) which is, at last, a gaussian distribution.
lesson: the largeN limit of a “fair” binomial distribution is a gaussian distribution.
c 2006 jake hofman.
(7)
(8)
(9)
2