Download random variable

Review
• Probability
• Basic definitions:
Randomization experiment
Sample spaces
Elementary outcomes
Event
• Basic operations—conditional probability
• Bayes Theorem
Objectives
• Random Variable
Discrete random variable
Continuous random variable
• Two probability distributions
Binomial distribution
Normal distribution
Random variables
4
• A random variable is a function that assigns numeric
values to different events in a sample space. Usually we
denote a random variable using a capital letter X, Y or
Z…
•
NOTE: (1) Randomness; (2) Numeric values
•
Example 1: Randomly select a student from a class.
X=student’s number of siblings. X could be 0, 1, 2 …
•
Example 2: Randomly select a student from a class.
X=student’s height. X could be any value bigger than 0
Two types of random variables
5
1.
Discrete random variable: their outcomes are set of
discrete (isolated) values.
Eg. X=number of siblings
2.
Continuous random variable: its possible values
cannot be enumerated; infinite number of values, all
outcomes have probability zero. p(x)=0 for every x.
Eg. X=the student’ height
EG1. Tossing two coins
6
let X=number of heads
Outcome
TT
x
0
Notation:
X: variable
x: observed values
HT
TH
1
HH
2
Probability distribution function
7
• A probability distribution function (pdf) is a
mathematical relationship, or rule, that assigns to any
possible value x of a discrete random variable X the
probability Pr(X=x).
8
Probability Distribution of the
Random Variable
X=number of heads.
Probability
histogram
Outcome
TT
WT
TW
WW
x
0
1
2
P(X=x)
1/4
1/2
1/4
EG2. Tossing two dice
9
Y: the sum of the dots on the two
Dice.
What’s the possible values of Y?
10
Probability Distribution of the
Random Variable
Y: the sum of the dots on the two Dice.
Relative frequency
In practice, the probability can be estimated by
the relative frequency of an event “in a long
run”.
frequency of occurrences
Probability =
frequency of all possible occurrences
0 ≤ Probability ≤ 1
Relative frequency histogram should look very
much like the probability histogram, if the
experiment is repeated many times.
Data set vs. Probability distributions
12

Sample properties—based
on
data
set
n
x = ∑ i =1 xi / n
Sample mean:
1
n
2
2
Sample variance:
=
−
s
x
x
(
)
∑ i
n −1

i =1
Model or population properties—based on
probability distribution.
R
=
µ ∑
=
xi Pr( X xi )
Population mean:
i =1
R
Population variance: σ 2 =
( x − µ ) 2 Pr( X =
x)
∑
i =1
i
i
Mean of Random Variable
13

Mean or expected value of X, denoted as E(X)
or µ, is defined as
R
E ( X ) = µ = ∑ xi Pr( X = xi )
i =1
It is the sum of the possible values, each
weighted by its probability
 Expectation represents “average” value of the
random variable

Mean of X
14
X=number of heads.
Outcome
TT
x
0
1
2
P(X=x)
1/4
1/2
1/4
xP(x)
0
1/2
1/2
E( X =
) µ=
WT
3
∑ x Pr( X=
i =1
i
TW
xi =
) 1
WW
Variance of Random Variable
15

The variance of X is the expected squared
distance from the population mean.
R
Var ( X ) = σ = ∑ ( xi − µ ) 2 Pr( X = xi )
2
i =1

The standard deviation σ is the square root of
variance
sd ( X ) = σ = Var ( X )
Variance of X
16
X=number of heads.
x
P(x)
(X-µ)2 P(x)
0
0.25
(0-1)2*0.25=0.25
1
0.5
(1-1)2*0.25=0
2
0.25
(2-1)2*0.25=0.25
Total
0.50
σ 2 = 0.5
Thus,
Summary, µ and σ are computed from probability
distribution. They are population properties.
Two types of random variables
17
1.
Discrete random variable: their outcomes are set of
discrete (isolated) values.
2.
Continuous random variable: its possible values
cannot be enumerated; infinite number of values, all
outcomes have probability zero. p(x)=0 for every x.
Continuous random variables
18





A balanced spinning pointer.
Can stop anywhere in the
circle
X—the proportion of the
total circumference it lands
on.
X can be any value between
0 and 1. Infinite values.
p(0.25≤x ≤0.75)=0.5
p(x=0.5)=0, for x can take on
an infinite number of values.
Probability density function(pdf) of X
19
• The curve
y = f ( x)
is
the probability density
function (pdf) of the random
variable X
• Pr(a≤X ≤b)= is the area
under the curve between
the x value a and b.
y = f ( x)
b
P ( a ≤ X ≤ b) =
∫ f ( x)dx
a
• The total area under the
density function curve over
the entire range of possible
values for the random
variable is 1
P (−∞ =
≤ X ≤ ∞)
∞
f ( x)dx
∫=
−∞
1
Probability density function(pdf) of X
20
• The pdf has large values in
regions of high probability and
small values in regions of low
probability
• Pr(X=x)=0 for any specific
value x
• Generally, a distinction is not
made between probabilities
such as Pr(X<x) and Pr(X≤x),
Pr(a≤X≤b) and Pr(a<X<b)
when X is a continuous
y = f ( x)
21
Expectation and variance of a continuous
random variable
• Mean
µ:
E (X)= µ=
∫
∞
−∞
xf ( x)dx
Center of the probability density
• Variance σ :
2
Var (X)
= σ=
2
∫
∞
−∞
( x − µ ) f ( x)dx
2
Spread of the probability density
• The standard deviation, or σ, is the square root of
the variance, that is,
σ = Var ( X )
Two distributions
22
Binomial --discrete
 Normal -- continuous

Bernoulli trial
23
Examples:
 A heads-or-tails Coin toss
 A win-or-lose football game
 A pass-or-fail automotive smog inspection
Properties:
 Two outcomes: success or failure
 Success probability(p) is the same in each
trial
 Trials are independent.
Binomial random variable
24
---X is the number of success in n repeated
Bernoulli trial with probability p of success.
 Success
probability(p) is the same in each
trial
 Trials
are independent.
Binomial random variable
25
Probability Distribution: the probability of
obtaining k successes in n trial, with success
probability p:
n k
n−k
P( X= k=
)   p (1 − p )
k 
n
n!
:
=
 
 k  k !(n − k )!
counts all possible ways of getting k
success and n-k failures
where n ! = n × (n − 1) × ... ×1
p (1 − p )
k
n−k
: probability for getting k success and
n-k failures
26
Mean and Variance of the Binomial
Distribution
µ = np
np (1 − p )
=
σ
2
Exercise
27
Newborns were screened for HIV in a Massachusetts
hospital. The positive rate for inner-city baby is p=0.01.
If 500 newborns are screened,
1. what is the exact binomial probability of 5 HIV
positive test results?
Exercise
28
Newborns were screened for HIV in a Massachusetts
hospital. The positive rate for inner-city baby is p=0.01.
If 500 newborns are screened,
1. what is the exact binomial probability of 5 HIV
positive test results?
Answer:
 500 
5
495
P( X= 5)= 
0.01
(1
−
0.01)

5


= 0.176
EXCEL: BINOMDIST(5,500,0.01,FALSE)
Exercise
29
Newborns were screened for HIV in a Massachusetts
hospital. The positive rate for inner-city baby is p=0.01.
If 500 newborns are screened,
2. What is the exact binomial probability of at least 5
HIV positive test results?
Exercise
30
Newborns were screened for HIV in a Massachusetts
hospital. The positive rate for inner-city baby is p=0.01.
If 500 newborns are screened,
2. What is the exact binomial probability of at least 5
HIV positive test results?
Answer:
P( X ≥ 5) =1 − P( X ≤ 4)
= 1 − F (4)
= 1 − 0.44
= 0.56
EXCEL: F(4)= BINOMDIST(4,500,0.01,TRUE)
Normal distribution
31
• Normal distribution is also called Gaussian
distribution, after the well-known
mathematician Karl Gauss (1777-1855,
“the Prince of Mathematicians“)
Normal distribution
32
• Normal distribution is very useful
• Many things closely follow a normal distribution
• Heights of people
• Errors in measurement
• Blood pressure
• Scores on a test
• Many other distributions can be made approximately
normal by transformation—Binomial et al.
• Most statistical methods considered in this text are
based on normal distribution
The pdf of normal distribution
33
• The normal distribution is defined by its pdf, which is
given as for some parameters µ and σ
f ( x) =
1
e
2π σ
 ( x − µ )2
−
2

 2σ




Other properties of Normal pdf
34
•Mean=median=mode
•Symmetry about the center
•50% of values less than the mean
Location is measured by µ
35
• In the graph, µ2>µ1
Spread is measured by σ2
36
• In the graph, σ2>σ1
Standard normal distribution N(0, 1)
37
• A normal distribution with mean 0 and variance 1
is called a standard normal distribution. Denoted
as N(0, 1)
• In the following, we will examine the standard
normal distribution N(0, 1) in details.
• We will see that any information concerning a
general normal distribution N(µ, σ2) can be
obtained from appropriate manipulations of an
N(0,1) distribution
38
Density of standard normal N(0,1)
µ =0
σ =1
f ( x) =
1
e
2π
x2
−
2
Properties of the standard normal N(0, 1)
39
• It can be shown that about 68% of the area under the standard
normal density lies between -1 and +1, about 95% of the area lies
between -2 and +2, and about 99% lies between -2.5 and +2.5
NOTE: You will see that, more precisely,
Pr(-1<x<1)=0.6827, Pr(-1.96<X<1.96)=0.95, Pr(-2.576<X<2.576)=0.99
Cumulative probability
40
• The cumulative distribution function (cdf) for a
standard normal distribution is denoted by
F=
( a )Φ(x)=Pr(X≤x),
P ( Z ≤ a )where Z~N(0,1)
Excel: F(a): NORMSDIST(a);
P ( a ≤ Z ≤ b=
)
F (b) − F ( a )
41
P ( −1 ≤ Z ≤ 1)
= F (1) − F ( −1)
=0.8413-0.1587
=0.6826
Excel: F(1): NORMSDIST(1); F(-1): NORMSDIST(-1);
P( Z ≥ a) =
1 − F (a)
42
• Eg.
P ( Z ≥ 1) =1 − F (1)
=1-0.8413
=0.1587
Excel: F(1):
NORMSDIST(1);
(1) NORMSDIST(1)
How to standardize the normal distribution?
43
How to standardize the normal distribution?
X −µ
Z=
σ
44
Then Z has a standard normal distribution, Z ~ N(0, 1)
Standardization
45
• IF X~
N(µ,
σ 2)
and Z =
X −µ
σ
then Z~N(0,1)
Then
a−µ
b−µ
b−µ
a−µ
< b) P(
< Z < =) F (
P(a < X =
) − F(
)
σ
σ
σ
σ
Use standardization for many problems
46
• Example:If X~N(80, 12^2), what is Pr(90<X<100)?
• Solution:
90 − 80 X − 80 100 − 80
<
<
)
12
12
12
= Pr(0.83 < Z < 1.67)
=F(1.67)-F(0.83)
=0.9522-0.7977
Pr(90 < X =
< 100) Pr(
=0.155
Always draw a graph…
47
Exercise
48
• Suppose we know that among men aged 30-34 who have
ever smoked, the mean number of years they smoked is
12.8 with a standard deviation of 5.1 years. Assuming
that the duration of smoking is normally distributed, what
proportion of men in this age group have smoked for
more than 20 years?
Exercise
49
Suppose we know that among men aged 30-34 who have
ever smoked, the mean number of years they smoked is
12.8 with a standard deviation of 5.1 years. Assuming
that the duration of smoking is normally distributed, what
proportion of men in this age group have smoked for
more than 20 years?
Answer: We have X ~ N (12.8, 5.12 )
And we need to compute P( X > 20)
P( X > 20) =
1 − P( X ≤ 20)
20 − 12.8
=1-P(Z ≤
)
5.1
= 1 − F (1.412)
=1-0.9210=0.079
EXCEL:
NORMDIST(20,12.8,5.1,TRUE)
Or NORMSDIST(1.412)