Download COM2023 Mathematics Methods for Computing II

COM2023
Mathematics Methods for Computing II
Lecture 13& 14
Gianne Derks
Department of Mathematics (36AA04)
http://www.maths.surrey.ac.uk/Modules/COM2023
Autumn 2010
Use channel 04 on your EVS handset
Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Uniform distribution
3
The Uniform Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
The Uniform Distribution and Matlab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Normal distribution
Normal distribution . . . . . . . . . . . . . . .
Normal distribution: changing µ and σ . .
The normal distribution and Matlab . .
Spread of the normal distribution. . . . . .
Samples and normal distribution . . . . . .
Properties of normal variables . . . . . . . .
Examples for combined normal variables .
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6
. 7
. 8
. 9
10
11
12
13
Central limit theorem
Averages of a normal distribution . . . . .
Central limit theorem. . . . . . . . . . . . . .
Example Central limit theorem: n = 200.
Example Central limit theorem: n = 400.
Example Central limit theorem: n = 600.
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14
15
16
17
18
19
1
Overview
●
Distributions for continuous random variables:
◆ The uniform distribution
◆ The normal distribution
●
The central limit theorem
The deadline for the assignment has been extended to
Monday 22 November, 2pm
Uniform distribution (§4.7)
The Uniform Distribution
The random variable X is uniformly distributed over the interval [a, b], in short X ∼ U [a, b], if it has a
constant probability density function over the interval [a, b]:
f(x)
f (x) =
(
1
b−a ,
1
b−a
a≤x≤b
0, otherwise
a
The expectation is E(X) =
b+a
2
and the variance is Var(X) =
b
x
(b−a)2
12 .
Example
The random variable X has a uniform distribution over the interval [1, 6], i.e., X ∼ U [1, 6]. What is
the probability that X takes a value between 2 and 4? The probability is the area of the graph for x
between 2 and 4, hence
P (2 ≤ X ≤ 4) = (4 − 2) ·
1
6−1
= 25 .
2
The Uniform Distribution and Matlab
The package Matlab has the uniform distribution build in. For X ∼ U [a, b]:
●
unifpdf(x,a,b) gives the pdf f (x).
●
unifcdf(x,a,b) gives P (X ≤ x).
●
unifrnd(a,b,1,y) gives a random sample in a 1×y matrix.
Examples
●
f (4) for X ∼ U [2, 7] is obtained by unifpdf(4,2,7) and is 0.2;
●
P (X ≤ 3) for X ∼ U [−1, 9] follows from unifcdf(3,-1,9) and is 0.4;
●
P (−1 ≤ X ≤ 3) for X ∼ U [−6, 8] is obtained by
unifcdf(3,-6,8)-unifcdf(-1,-6,8) and is 0.2857143;
●
P (X > 5) for X ∼ U [4, 9] follows from 1-unifcdf(5,4,9) and is 0.8;
●
P (X ≤ 8 | X > 6) for X ∼ U [5, 10] is obtained by using that
P (X ≤ 8 | X > 6) = PP(6<X≤8)
(X>6) , hence in Matlab we find it is 0.5 by
(unifcdf(8,5,10)-unifcdf(6,5,10))/(1-unifcdf(6,5,10)).
Normal distribution (§4.8)
Normal distribution
The normal distribution or Gaussian distribution is the most important distribution in statistics. It has
two parameters: the location, controlled by µ and the dispersion, controlled by σ. The probability
density function is
!
1
1 x−µ 2
f (x) = √
exp −
2
σ
2πσ 2
(you don’t have to remember this).
A random variable X satisfying a normal distribution is denoted by X ∼ N (µ, σ 2 ).
For a random variable X ∼ N (µ, σ 2 ) the expectation and variance are
E(X) = µ
and
3
Var(X) = σ 2 .
Normal distribution: changing µ and σ
Note the effect of changing µ and σ and note that the graph is symmetric about X = µ.
1.6
N(2,0.25)
N(2,1)
N(0,1)
1.4
pdf f(x) of normal distribution
1.2
1
0.8
0.6
0.4
0.2
0
-2
-1
0
1
x
2
3
4
The normal distribution and Matlab
Matlab has the uniform distribution build in: For X ∼ N (µ, σ 2 ).
●
normpdf(x,µ,σ) gives f (x);
●
normcdf(x,µ,σ) gives P (X ≤ x);
●
normrnd(µ,σ,1,y) gives a random sample of y values in a 1×y matrix.
Note: Matlab uses the standard deviation σ, while N uses the variance σ 2 .
Examples for X ∼ N (15, 11):
●
●
P (X ≤ 18) follows from normcdf(18,15,sqrt(11)) and is 0.90928;
P (X > 15) follows from 1-normcdf(15,15,sqrt(11)) and is 0.5;
●
P (18 < X ≤ 25) is 0.90928, which follows from
normcdf(25,15,sqrt(11))-normcdf(18,15,sqrt(11));
●
P (15 < X < 25) is 0.4987, which follows from
normcdf(25,15,sqrt(11))-normcdf(15,15,sqrt(11));
●
P (X ≤ 22 | X ≤ 26) follows from P (X ≤ 22 | X ≤ 26) = PP (X≤22)
(X≤26) , i.e.,
normcdf(22,15,sqrt(11))/normcdf(26,15,sqrt(11)) is 0.9830.
4
Spread of the normal distribution
The area of the normal distribution is such that:
●
95% is in the range [µ − 1.96σ, µ + 1.96σ];
●
99% is in the range [µ − 2.58σ, µ + 2.58σ].
To check for N (3, 4), use Matlab:
●
●
normcdf(3+1.96*2,3,2)normcdf(3-1.96*2,3,2)
0.9500042;
gives
normcdf(3+2.58*2,3,2)normcdf(3-2.58*2,3,2)
0.99012.
gives
Samples and normal distribution
With Matlab, we can get a sample from a normal distribution N (µ, σ 2 ) by using
normrnd(µ,σ,1,n) to get a 1×n matrix.
N(12,8): scaled plot and histogram of a sample of size 200
25
On the right is the histogram of a sample of size 200 from the normal distribution
N (12, 8), together with a scaled plot of the
normal distribution.
15
10
5
0
Frequency
20
The sample follows from
normrnd(12,sqrt(8),1,200)
5
10
15
Sample value
5
20
Properties of normal variables
Recall that for any two independent random variables X and Y :
E(aX + bY + c) = aE(X) + bE(Y ) + c;
Var(aX + bY + c) = a2 Var(X) + b2 Var(Y ).
For random variables with normal distributions, we have
1. If X ∼ N (µ, σ 2 ), then
aX + b ∼ N (aµ + b, a2 σ 2 ),
as E(aX + b) = aµ + b and Var(aX + b) = a2 σ 2 .
2. If X ∼ N (µx , σx2 ) and Y ∼ N (µy , σy2 ) and X and Y are independent, then
aX + bY ∼ N (aµx + bµy , a2 σx2 + b2 σy2 ).
as E(aX + bY ) = aµx + bµy and Var(aX + bY ) = a2 σx2 + b2 σy2 .
Examples for combined normal variables
For X ∼ N (µx , σx2 ) and Y ∼ N (µy , σy2 ), we have
aX + bY + c ∼ N (aµx + bµy + c, a2 σx2 + b2 σy2 .
Examples
2X − Y − 8:
Answer: 2X − Y − 8 ∼ N (1, 9).
80
●
1
● 4
0
20
40
(X + X + X + X):
Answer:
8
1
4 (X + X + X + X) ∼ N (3, 16 )6= X!
60
X +Y:
Answer: X + Y ∼ N (0, 3)
Frequency
●
100
120
Let X ∼ N (3, 2) and Y ∼ N (−3, 1) (X, Y indep.). Find the distributions of
● −X + 2:
2X−Y−8: histogram of sample and scaled N(1,9) plot
Answer: −X + 2 ∼ N (−1, 2);
−10
−5
0
sample value
6
5
10
Central limit theorem (§4.8.3)
Averages of a normal distribution
For independent normal variables X1 , X2 , . . . , Xn , each with distribution N (µ, σ 2 ), the average
X = n1 (X1 + . . . + Xn ) has a N (µ, σ 2 /n) distribution.
This follows immediately from property 2 and is refeered to as the sampling distribution of the sample
mean
0.15
0.10
0.00
0.05
probility density function
0.20
0.25
N(25,20/10)
N(25,20/3)
N(25,20)
10
15
20
25
30
35
40
x
Central limit theorem
Central limit theorem
For any population distribution with mean µ and variance σ:
If is n sufficiently large, the sampling distribution of the sample mean X =
N (µ, σ 2 /n) distribution.
1
n
(X + . . . + X) has a
Example Consider the Poisson distribution with average 8, hence µ = 8 = σ 2 .
●
n = 200: Using poissrnd, we take a sample of size 200, register its average and repeat this
500 times. This average of these 500 numbers is µ = 8.014 and the variance is
σ 2 = 0.03802665, while 8/200 = 0.04.
●
n = 400: Using poissrnd, we take a sample of size 400, register its average and repeat this
500 times. This average of these 500 numbers is µ = 8.0068 and the variance is σ 2 = 0.0206,
while 8/400 = 0.02.
●
n = 600: Using poissrnd, we take a sample of size 600, register its average and repeat this
500 times. This average of these 500 numbers is µ = 7.9963 and the variance is σ 2 = 0.01377,
while 8/600 = 0.01333.
7
Example Central limit theorem: n = 200
Histogram of the 500 observations of
means of samples of size 200, together
with the scaled graph of N (8, 8/200).
QQplot of quantiles of N (8.8/200)
against the quantiles of the 500 observations of means of samples of size 200.
QQ plot for means of samples of size 200
0
7.6
20
7.8
8.0
Quantiles of means
60
40
Frequency
8.2
80
8.4
100
8.6
Means of samples of size 200 of Po(8)
7.4
7.6
7.8
8.0
8.2
8.4
8.6
−3
−2
−1
mean of sample
0
1
2
3
Quantiles of N(8,8/200)
Example Central limit theorem: n = 400
QQplot of quantiles of N (8.8/400)
against the quantiles of the 500 observations of means of samples of size 400.
Histogram of the 500 observations of
means of samples of size 400, together
with the scaled graph of N (8, 8/400).
QQ plot for means of samples of size 400
60
8.0
7.8
40
7.6
20
0
Frequency
80
Quantiles of means
8.2
100
120
8.4
Means of samples of size 400 of Po(8)
7.4
7.6
7.8
8.0
8.2
8.4
8.6
−3
mean of sample
−2
−1
0
Quantiles of N(8,8/400)
8
1
2
3
Example Central limit theorem: n = 600
Histogram of the 500 observations of
means of samples of size 600, together
with the scaled graph of N (8, 8/600).
QQplot of quantiles of N (8.8/600)
against the quantiles of the 500 observations of means of samples of size 600.
QQ plot for means of samples of size 600
8.0
Quantiles of means
7.9
7.8
40
7.7
20
0
Frequency
60
8.1
8.2
80
8.3
Means of samples of size 600 of Po(8)
7.4
7.6
7.8
8.0
8.2
8.4
8.6
−3
mean of sample
−2
−1
0
Quantiles of N(8,8/600)
9
1
2
3