Properties of the Sample Mean

Transcription

Consider X1, . . . , Xn independent and identically distributed (iid)
with mean µ and variance σ 2.
n
1P
¯
X=
Xi
n i=1
(sample mean)
Then
n
1P
¯
(X) =
µ=µ
n i=1
n
σ2
1 P
2
¯
σ =
var(X) = 2
n i=1
n
Remarks:
◦ The sample mean is an unbiased estimate of the true mean.
◦ The variance of the sample mean decreases as the sample size
increases.
◦ Law of Large Numbers: It can be shown that for n → ∞
n
1P
¯
X=
Xi → µ.
n i=1
Question:
◦ How close to µ is the sample mean for finite n?
◦ Can we answer this without knowing the distribution of X?
Central Limit Theorem, Feb 4, 2003
-1-
Chebyshev’s inequality
Let X be a random variable with mean µ and variance σ 2.
Then for any ε > 0
¡
¢ σ2
|X − µ| > ε ≤ 2 .
ε
Proof: Let
1{|xi − µ| > ε} =
½
1
0
if |xi − µ| > ε
otherwise
Then
o
n n (x − µ)2
P
i
> 1 p(xi )
1{|xi − µ| > ε} p(xi ) =
1
ε2
i=1
i=1
n (x − µ)2
P
σ2
i
≤
p(x
)
=
i
ε2
ε2
i=1
n
P
Application to the sample mean:
³
´
3σ
3σ
¯ ≤ µ + √ ≥ 1 − 1 ≈ 0.889
µ− √ ≤X
n
n
9
However: Known to be not very precise
iid
Example: Xi ∼ N (0, 1)
n
1P
¯
X=
Xi ∼ N (0, n1 )
n i=1
Therefore
³
´
3
3
¯
− √ ≤ X ≤ √ = 0.997
n
n
-2-
Central Limit Theorem
Let X1, X2, . . . be a sequence of random variables
◦ independent and identically distributed
◦ with mean µ and variance σ 2.
For n ∈
define
n
√ X¯ − µ
Xi − µ
1 P
=√
.
Zn = n
σ
n i=1
σ
Zn has mean 0 and variance 1.
For large n, the distribution of Zn can be approximated by the
standard normal distribution N (0, 1). More precisely,
³
´
√ X¯ − µ
lim
a≤ n
≤ b = Φ(b) − Φ(a),
σ
n→∞
where Φ(x) is the standard normal probability
Φ(z) =
Z
z
f (x) dx,
−∞
that is, the area under the standard normal curve to left of z.
Example:
◦ U1, . . . , U12 uniformly distributed on [ 0, 12).
◦ What is the probability that the sample mean exceeds 9?
³√ ¯
´
U −6
¯
(U > 9) =
12 √ > 3 ≈ 1 − Φ(3) = 0.0013
12
-3-
0.4
1.0
U[0,1],n=1
0.8
density f(x)
0.3
density f(x)
Exp(1),n=1
0.2
0.1
0.6
0.4
0.2
0.0
0.0
−3
−2
−1
0
1
2
3
−3
−2
−1
0
1
U[0,1],n=2
0.4
2
3
Exp(1),n=2
0.5
0.4
density f(x)
density f(x)
0.3
0.2
0.3
0.2
0.1
0.1
0.0
0.0
−3
−2
−1
0
1
2
3
−3
−1
0
1
0.5
U[0,1],n=6
0.4
−2
2
3
Exp(1),n=6
0.4
density f(x)
density f(x)
0.3
0.2
0.1
0.2
0.1
0.0
0.0
−3
−2
−1
0
1
2
3
−3
U[0,1],n=12
0.4
−2
−1
0
1
2
3
Exp(1),n=12
0.4
0.3
0.3
density f(x)
density f(x)
0.3
0.2
0.1
0.2
0.1
0.0
0.0
−3
−2
−1
0
1
2
3
−3
−2
−1
0
1
U[0,1],n=100
0.4
2
3
Exp(1),n=100
0.4
density f(x)
density f(x)
0.3
0.2
0.1
0.3
0.2
0.1
0.0
0.0
−3
−2
−1
0
1
2
3
−3
−2
−1
0
1
2
3
-4-
Example: Shipping packages
Suppose a company ships packages that vary in weight:
◦ Packages have mean 15 lb and standard deviation 10 lb.
◦ They come from a arge number of customurs, i.e. packages are
independent.
Question: What is the probability that 100 packages will have a
total weight exceeding 1700 lb?
Let Xi be the weight of the ith package and
T =
100
P
Xi .
i=1
Then
T − 1500 lb
1700 lb − 1500 lb
√
> √
(T > 1700 lb) =
100 · 10 lb
100 · 10 lb
µ
¶
T − 1500 lb
√
=
>2
100 · 10 lb
µ
¶
≈ 1 − Φ(2) = 0.023
-5-
Remarks
• How fast approximation becomes good depends on distribution
of Xi’s:
◦ If it is symmetric and has tails that die off rapidly, n can
be relatively small.
iid
Example: If Xi ∼ U [0, 1], the approximation is good for
n = 12.
◦ If it is very skewed or if its tails die down very slowly, a
larger value of n is needed.
Example: Exponential distribution.
• Central limit theorems are very important in statistics.
• There are many central limit theorems covering many situations, e.g.
◦ for not identically distributed random variables or
◦ for dependent, but not “too” dependent random variables.
-6-
The Normal Approximation to the Binomial
Let X be binomially distributed with parameters n and p.
Recall that X is the sum of n iid Bernoulli random variables,
X=
n
P
Xi ,
i=1
iid
Xi ∼ Bin(1, p).
Therefore we can apply the Central Limit Theorem:
Normal Approximation to the Binomial Distribution
¡
¢
For n large enough, X is approximately N np, np(1 − p)
distributed:
¢
¡
¡
1
1
a ≤ X ≤ b) ≈ a − 2 ≤ Z ≤ b + 2
where
¡
¢
Z ∼ N np, np(1 − p) .
Rule of thumb for n: np > 5 and n(1 − p) > 5.
In terms of the standard normal distribution we get
µ
¶
1
¡
−
np
a − 12 − np
b
+
p
a ≤ X ≤ b) =
≤ Z0 ≤ p 2
np(1 − p)
np(1 − p)
µ
¶
µ
¶
b + 21 − np
a − 12 − np
=Φ p
−Φ p
np(1 − p)
np(1 − p)
where Z 0 ∼ N (0, 1).
-7-
Bin(1,0.5)
1.0
0.8
0.6
0.6
p(x)
0.8
p(x)
Bin(1,0.1)
1.0
0.4
0.4
0.2
0.2
0.0
0
1
2
3
4
5
6
7
8
9
10
12
14
16
18
0.0
20
0
1
2
3
4
5
6
7
8
9
x
10
12
14
16
18
20
x
Bin(2,0.5)
1.0
0.8
0.6
0.6
p(x)
0.8
p(x)
Bin(5,0.1)
1.0
0.4
0.4
0.2
0.2
0.0
0
1
2
3
4
5
6
7
8
9
10
12
14
16
18
0.0
20
0
1
2
3
4
5
6
7
8
9
x
10
12
14
16
18
20
x
Bin(5,0.5)
0.5
0.4
0.3
0.3
p(x)
0.4
p(x)
Bin(10,0.1)
0.5
0.2
0.2
0.1
0.1
0.0
0
1
2
3
4
5
6
7
8
9
10
12
14
16
18
0.0
20
0
1
2
3
4
5
6
7
8
9
x
10
12
14
16
18
20
x
Bin(10,0.5)
0.3
Bin(20,0.1)
0.3
p(x)
0.2
p(x)
0.2
0.1
0.0
0.1
0
1
2
3
4
5
6
7
8
9
10
12
14
16
18
0.0
20
0
1
2
3
4
5
6
7
8
9
x
10
12
14
16
18
20
x
Bin(20,0.5)
0.3
Bin(50,0.1)
0.3
p(x)
0.2
p(x)
0.2
0.1
0.0
0.1
0
1
2
3
4
5
6
7
8
9
10
12
x
14
16
18
20
0.0
0
1
2
3
4
5
6
7
8
9
10
12
14
16
18
20
x
-8-
Example: The random walk of a drunkard
Suppose a drunkard executes a “random” walk in the following
way:
◦ Each minute he takes a step north or south, with probability 21
each.
◦ His successive step directions are independent.
◦ His step length is 50 cm.
How likely is he to have advanced 10 m north after one hour?
◦ Position after one hour: X · 1 m − 30 m
◦ X binomially distributed with parameters n = 60 and p =
1
2
◦ X is approximately normal with mean 30 and variance 15:
(X · 1 m − 30 m > 10 m)
= (X > 40)
≈ (Z > 39.5)
µ
¶
Z − 30
9.5
√
=
>√
15
15
= 1 − Φ(2.452) = 0.007
Z ∼ N (30, 15)
How does the probability change if he has same idea of where he
wants to go and steps north with probability p = 23 and south with
probability 31 ?
-9-

Properties of the Sample Mean

Transcription

Similar documents