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A section of lecture notes from math331, fall 2008, covering the central limit theorem. The theorem states that the distribution of the sum of a large number of independent and identically distributed random variables approaches a normal distribution. Examples, calculations, and proof references.
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Section 11.5 lecture notes
Math331, Fall 2008
Instructor: David Anderson
HW: pg. 506, #’s 2, 6 (uniform), 8.
Consider X 1 , X 2 , X 3 ,... , which are independent and identically distributed with mean μ
and variance σ^2. In nature, it is observable that no matter what the distribution of Xi is,
X 1 + X 2 + · · · + Xn
looks like a normal distribution (if you back far enough away).
Example 1: Consider rolling a die 100 times (each Xi is the output from one roll) and
adding outcomes. You will get a value around 350, plus or minus some. Do this experiment
10000 times and plot the number of times you get each outcome. Will look like a bell curve.
Example 2: Go to a library and go to the stacks. Each row of books is divided into n >> 1
pieces. Let Xi be number of books on piece i. Then
∑n i=1 Xi^ is the number of books on
a given row. Do this for all rows of same length. You will get a plot that looks like a bell
curve. Do these bell curves have anything in common?
Consider Wn = X 1 + Xn + · · · + Xn. How far “back” should we go out to view this? What
does this mean? (scaling) Why not standardize it? Recall, to standardize we do
Wn − E[Wn]
σWn
This has mean zero (shift over to zero) and variance 1. This seems like right way to “back
off.” So,
E[Wn] = E[X 1 + Xn + · · · + Xn] = nμ.
σWn =
V ar(X 1 + · · · + Xn) =
V ar(X 1 ) + · · · + V ar(Xn) =
nσ^2 = σ
n.
So, what does
Wn − nμ
σ
n
X 1 + · · · + Xn − nμ
σ
n
look like for large n?
Theorem 1 (Central Limit Theorem). Let X 1 , X 2 ,... be a sequence of independent and
identically distributed random variables, each with expectation μ and variance σ^2. Then the
distribution of
Zn =
X 1 + · · · + Xn − nμ
σ
n
converges to the distribution of a standard normal random variable. That is,
lim n→∞
P (Zn ≤ t) = lim n→∞
X 1 + · · · + Xn − nμ
σ
n
≤ t
2 π
∫ (^) t
−∞
e
−x^2 / 2 dx.
Proof in book is based on moment generating functions. But I won’t spend time on it.
Example 1. Let X 1 , X 2 ,... be independent and identically distributed RVs with mean μ
and standard deviation σ. Set Sn = X 1 +X 2 +· · ·+Xn. For large n, what is the approximate
probability that Sn is between E[Sn] − kσSn and E[Sn] + kσSn (the probability of being k
deviates from mean).
Solution. We have that
E[Sn] = E[X 1 + · · · + Xn] = nμ
σSn =
V ar(Sn) =
V ar(X 1 + · · · Xn) =
V ar(X 1 ) + · · · V ar(Xn) = σ
n.
Thus, letting Z ∼ N(0, 1),
P (E[Sn] − kσSn ≤ Sn ≤ E[Sn] + kσSn ) = P (nμ − kσ
n ≤ Sn ≤ nμ + kσ
n)
= P (−kσ
n ≤ Sn − nμ ≤ kσ
n)
= P (−k ≤
Sn − nμ
σ
n
≤ k)
≈ P (−k ≤ Z ≤ k)
2 π
∫ (^) k
−k
e
−x^2 / 2 dx
6826 k = 1
9545 k = 2
9973 k = 3
9999366 k = 4
Recall that Chebyschev’s inequality gave the following:
P (|Sn − nμ| < kσ
n) = 1 − P (|Sn − nμ| ≥ kσ
n)
σ^2 n
k^2 σ^2 n
k^2
0 k = 1
75 k = 2
8889 k = 3
9375 k = 4
Example 2. At a party each person will independently eat 1, 2, or 3 appetizers with a
probability of 1/ 4 , 1 / 2 , 1 /4 respectively. You know there will be 80 people at this party. You
want to buy enough appetizers so that with probability .95 you do not run out. How many
should you buy?
Solution: Let X be the number of appetizers eaten and Xi be the number eaten by the ith
person. Then
i=
Xi.