Download Properties of the Sample Mean

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Properties of the Sample Mean
Consider X1, . . . , Xn independent and identically distributed (iid)
with mean µ and variance σ 2.
n
1P
X̄ =
Xi
n i=1
(sample mean)
Then
n
1P
(X̄) =
µ=µ
n i=1
n
σ2
1 P
2
σ =
var(X̄) = 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 → ∞
X̄ =
n
1P
Xi
n i=1
→ µ.
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-
Properties of the Sample Mean
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
´
1
3σ
3σ
µ − √ ≤ X̄ ≤ µ + √ ≥ 1 − ≈ 0.889
n
n
9
However: Known to be not very precise
iid
Example: Xi ∼ N (0, 1)
X̄ =
n
1P
Xi
n i=1
∼ N (0, n1 )
Therefore
³
´
3
3
− √ ≤ X̄ ≤ √ = 0.997
n
Central Limit Theorem, Feb 4, 2003
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
√ X̄ − µ
Xi − µ
1 P
=√
.
Zn = n
σ
n i=1
σ
Zn has mean 0 and variance 1.
Central Limit Theorem
For large n, the distribution of Zn can be approximated by the
standard normal distribution N (0, 1). More precisely,
³
´
√ X̄ − 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√
´
Ū − 6
(Ū > 9) =
12 √ > 3 ≈ 1 − Φ(3) = 0.0013
12
Central Limit Theorem, Feb 4, 2003
-3-
Central Limit Theorem
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
Central Limit Theorem, Feb 4, 2003
1
2
3
−3
−2
−1
0
1
2
3
-4-
Central Limit Theorem
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
Central Limit Theorem, Feb 4, 2003
-5-
Central Limit Theorem
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.
Central Limit Theorem, Feb 4, 2003
-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).
Central Limit Theorem, Feb 4, 2003
-7-
The Normal Approximation to the Binomial
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
Central Limit Theorem, Feb 4, 2003
14
16
18
20
0.0
0
1
2
3
4
5
6
7
8
9
10
12
14
16
18
20
x
-8-
The Normal Approximation to the Binomial
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 ?
Central Limit Theorem, Feb 4, 2003
-9-
Related documents