Transcript Random Variable

Stat 155, Section 2, Last Time
• Big Rules of Probability:
– Not Rule ( 1 – P{opposite})
– Or Rule (glasses – football)
– And rule (multiply conditional prob’s)
– Use in combination for real power
• Bayes Rule
– Turn around conditional probabilities
– Write hard ones in terms of easy ones
– Recall surprising disease testing result
Reading In Textbook
Approximate Reading for Today’s Material:
Pages 266-271, 311-323, 277-286
Approximate Reading for Next Class:
Pages 291-305, 334-351
Midterm I
Coming up: Tuesday, Feb. 27
Material:
HW Assignments 1 – 6
Extra Office Hours:
Mon. Feb. 26, 8:30 – 12:00, 2:00 – 3:30
(Instead of Review Session)
Bring Along:
1 8.5” x 11” sheet of paper with formulas
Recall Pepsi Challenge
In class taste test:
•
Removed bias with randomization
•
Double blind approach
•
Asked which was:
–
Better
–
Sweeter
–
which
Recall Pepsi Challenge
Results summarized in spreadsheet
Eyeball impressions:
a. Perhaps no consensus preference
between Pepsi and Coke?
–
Is 54% "significantly different from 50%? (will
develop methods to understand this)
–
Result of "marketing research"???
Recall Pepsi Challenge
b. Perhaps no consensus as to which is
sweeter?
•
Very different from the past, when Pepsi was
noticeably sweeter
•
This may have driven old Pepsi challenge
phenomenon
•
Coke figured this out, and matched Pepsi in
sweetness
Recall Pepsi Challenge
c. Most people believe they know
–
Serious cola drinkers, because now flavor driven
–
In past, was sweetness driven, and there were many
advertising caused misperceptions!
d. People tend to get it right or not??? (less clear)
–
Overall 71% right. Seems like it, but again is that
significantly different from 50%?
Recall Pepsi Challenge
e. Those who think they know tend to be right???
–
People who thought they knew: right 71% of the
time
f. Those who don't think they know seem to right as
well. Wonder why?
–
People who didn't: also right 70% of time? Why?
"Natural sampling variation"???
–
Any difference between people who thought they
knew, and those who did not think so?
Recall Pepsi Challenge
g. Coin toss was fair (or is 57% heads significantly
different from %50?)
How accurate are those ideas?
•
Will build tools to assess this
•
Called “hypo tests” and “P-values”
•
Revisit this example later
Independence
(Need one more major concept at this level)
An event A does not depend on B, when:
Knowledge of B does not change
chances of A:
P{A | B} = P{A}
Independence
E.g. I Toss a Coin, and somebody on South
Pole does too.
P{H(me) | T(SP)} = P{H(me)} = ½.
(no way that can matter, i.e. independent)
Independence
E.g. I Toss a Coin twice:
PH 2 | H1  ???
(toss number indicated with subscript)
•
Is it < ½?
•
What if have 5 Heads in a row?
(isn’t it more likely to get a Tail?)
(Wanna bet?!?)
Independence
E.g. I Toss a Coin twice, …
Rational approach:
Look at Sample Space
Model all as equally likely
Then:
H1H 2
H 1T2
T1 H 2
T1T2
PH 2 & H1 1 / 4 1
PH 2 | H1 

  PH 2 
PH1
1/ 2 2
So independence is good model for coin tosses
New Ball & Urn Example
H 
RRRRGG
T

RRG
Again toss coin, and draw ball:
2
PR | H  
3
PR  PR | H PH  PR | T PT  0
 2  1   2  1  2
        
 3  2   3  2  3
Same, so R & H are independent events
Not true above, but works here, since proportions of
R & G are same
Independence
Note, when A is independent of B:
PA & B
PA  PA | B 
PB
so
And thus
PAP{B}  PA & B
PA & B
P{B} 
 PB | A
PA
i.e. B is independent of A
Independence
Note, when A in independent of B:
It follows that: B is independent of A
I.e. “independence” is symmetric in A and B
(as expected)
More formal treatments use symmetric version
as definition
(to avoid hassles with 0 probabilities)
Independence
HW:
4.31
Special Case of “And” Rule
For A and B independent:
P{A & B} = P{A | B} P{B} = P{B | A} P{A} =
= P{A} P{B}
i.e. When independent, just multiply
probabilities…
Textbook: Call this another rule
Me:
Only learn one, this is a special case
Independent “And” Rule
E.g. Toss a coin until the 1st Head appears,
find P{3 tosses}:
Model: tosses are independent
(saw this was reasonable last time, using
equally likely sample space ideas)
P{3 tosses} = PT1 & T2 & H 3 
When have 3: group with parentheses
Independent “And” Rule
E.g. Toss a coin until the 1st Head appears,
 PT1 & T2 & H 3 
 PT1 & T2  & H 3 
find P{3 tosses}
 PH 3 | T1 & T2 PT1 & T2 
(by indep:)
 PH 3 PT2 | T1PT1
 PH 3 PT2 PT1
I.e. “just multiply”
Independent “And” Rule
E.g. Toss a coin until the 1st Head appears,
P{3 tosses}  PH 3 PT2 PT1
•
Multiplication idea holds in general
•
So from now on will just say:
“Since Independent, multiply probabilities”
•
Similarly for Exclusive Or rule,
Will just “add probabilities”
Independent “And” Rule
HW:
4.29 (hint: Calculate
P{G1&G2&G3&G4&G5&G6&G7})
4.33
Overview of Special Cases
Careful:
OR
these can be tricky to keep separate
works like adding,
for mutually exclusive
AND
works like multiplying,
for independent
Overview of Special Cases
Caution:
special cases are different
Mutually exclusive
 independent
For A and B mutually exclusive:
P{A | B} = 0
Thus not independent

P{A}
Overview of Special Cases
HW: C15 Suppose events A, B, C all have
probability 0.4, A & B are independent,
and A & C are mutually exclusive.
(a) Find P{A or B}
(0.64)
(b) Find P{A or C}
(0.8)
(c) Find P{A and B}
(0.16)
(d) Find P{A and C}
(0)
Random Variables
Text, Section 4.3 (we are currently jumping)
Idea: take probability to next level
Needed for probability structure of political
polls, etc.
Random Variables
Definition:
A random variable, usually denoted as X,
is a quantity that
“takes on values at random”
Random Variables
Two main types
(that require different mathematical models)
•
Discrete, i.e. counting
(so look only at “counting numbers”, 1,2,3,…)
•
Continuous, i.e. measuring
(harder math, since need all fractions, etc.)
Random Variables
E.g:
X = # for Candidate A in a randomly
selected political poll: discrete
(recall all that means)
Power of the random variable idea:
•
Gives something to “get a hold of…”
•
Similar in spirit to high school algebra…
High School Algebra
Recall Main Idea?
Rules for solving equations???
No, major breakthrough is:
•
Give unknown(s) a name
•
Find equation(s) with unknown
•
Solve equation(s) to find unknown(s)
Random Variables
E.g:
X = # that comes up, in die rolling:
Discrete
•
But not very interesting
•
Since can study by simple methods
•
As done above
•
Don’t really need random variable concept
Random Variables
E.g: Measurement error:
Let X = measurement:
Continuous
•
How to model probabilities???
Random Variables
HW on discrete vs. continuous:
4.40
((b) discrete, (c) continuous, (d)
could be either, but discrete is more
common)
And now for something
completely different
My idea about “visualization” last time:
•
30% really liked it
•
70% less enthusiastic…
•
Depends on mode of thinking
•
–
“Visual thinkers” loved it
–
But didn’t connect with others
So hadn’t planned to continue that…
And now for something
completely different
But here was another viewpoint:
Professor Marron,
Could you focus on something more intelligent in
your "And now for something completely
different" section once every two weeks,
perhaps, instead of completely abolishing it? I
really enjoyed your discussion of how to view
three dimensions in 2-D today.
And now for something
completely different
A fun example:
•
Faces as data
•
Each data point is a digital image
•
Data from U. Carlos, III in Madrid
(hard to do here for confidentiality reasons)
Q: What distinguishes men from women?
And now for something
completely different
And now for something
completely different
Context: statistical problem of “classification”,
i.e. “discrimination”
Basically “automatic disease diagnosis”:
•
Have measurm’ts on sick & healthy cases
•
Given new person, make measm’ts
•
Closest to sick or healthy populations?
And now for something
completely different
Approach: Distance Weight Discrimination
(Marron & Todd)
Idea: find “best separating direction” in high
dimensional data space
Here:
•
Data are images
•
Classes: Male & Females
•
Given new image: classify make - female
And now for something
completely different
Fun visualization:
•
March through point clouds
•
Along separating direction
•
Captures “Femaleness” & “Maleness”
•
Note relation to “training data”
And now for something
completely different
Random Variables
A die rolling example
(where random variable concept is useful)
Win $9 if 5 or 6, Pay $4, if 1, 2 or 3,
otherwise (4) break even
Notes:
• Don’t care about number that comes up
• Random Variable abstraction allows
focusing on important points
• Are you keen to play? (will calculate…)
Random Variables
Die rolling example
Win $9 if 5 or 6, Pay $4, if 1, 2 or 4
Let X = “net winnings”
Note: X takes on values 9, -4
and 0
Probability Structure of X is summarized by:
P{X = 9} = 1/3 P{X = -4} = 1/2 P{X = 0} = 1/6
(should you want to play?, study later)
Random Variables
Die rolling example,
for X = “net winnings”:
Win $9 if 5 or 6, Pay $4, if 1, 2 or 4
Probability Structure of X is summarized by:
P{X = 9} = 1/3 P{X = -4} = 1/2 P{X = 0} = 1/6
Convenient form:
Winning
Prob.
a table
9
-4
0
1/3
1/2
1/6
Summary of Prob. Structure
In general: for discrete X, summarize
“distribution” (i.e. full prob. Structure) by a
table:
Values
x1
x2
…
xk
Prob.
p1
p2
…
pk
Where:
i.
All pi are between 0 and 1
k
ii.  pi  1 (so get a prob. funct’n as above)
i 1
Summary of Prob. Structure
Summarize distribution, for discrete X,
by a table:
Values
x1
x2
…
xk
Prob.
p1
p2
…
pk
Power of this idea:
•
Get probs by summing table values
•
Special case of disjoint OR rule
Summary of Prob. Structure
E.g.
Die Rolling game above:
P{X = 9} = 1/3
Winning 9 -4 0
Prob.
1/3 1/2 1/6
P{X < 2} = P{X = 0} + P{X = -4} =1/6+1/2 = 2/3
P{X = 5} = 0
(not in table!)
Summary of Prob. Structure
E.g.
Die Rolling game above:
Winning 9 -4 0
Prob.
1/3 1/2 1/6
P X  9  &  X  0
PX  9 | X  0 
PX  0
1
3
1
PX  9
2
3


 
PX  0 1  1 1 3
6 3 2
Summary of Prob. Structure
HW:
4.41 & (c) Find P{X = 3 | X >= 2}
4.52
(0.144, …, 0.352)
(3/7)
Probability Histogram
Idea: Visualize probability distribution using a
bar graph
E.g.
Die Rolling game above:
Winning 9 -4 0
Prob.
1/3 1/2 1/6
probability
Toy example probability histogram
0.6
0.5
0.4
0.3
0.2
0.1
0
-4 -3 -2 -1
0
1
2
3
X values
4
5
6
7
8
9
Probability Histogram
Construction in Excel:
•
Very similar to bar graphs (done before)
•
Bar heights = probabilities
•
Example:
Class Example 18
Probability Histogram
HW:
4.43
Random Variables
Now consider continuous random variables
Recall: for measurements (not counting)
Model for continuous random variables:
Calculate probabilities as areas,
under “probability density curve”, f(x)
Continuous Random Variables
Model probabilities for continuous random
variables, as areas under “probability
density curve”, f(x):
Pa  X  b = Area(
b
  f ( x )dx
a
)
a
b
(calculus notation)
Continuous Random Variables
Note:
Same idea as “idealized distributions” above
Recall discussion from:
Page 8, of Class Notes, Jan. 23
Continuous Random Variables
e.g. Uniform Distribution
Idea:
choose random number from [0,1]
Use constant density:
f(x) = C
Models “equally likely”
To choose C, want:
Area
1 = P{X in [0,1]} = C
So want C = 1.
0
1
Uniform Random Variable
HW:
4.54
(0.73, 0, 0.73, 0.2, 0.5)
4.56
(1, ½, 1/8)
Continuous Random Variables
e.g. Normal Distribution
Idea: Draw at random from a normal
population
f(x) is the normal curve (studied above)
Review some earlier concepts:
Normal Curve Mathematics
The “normal
 , density curve” is:
1
f ( x) 
e
2 
1 x 
 

2  
2
usual “function” of x
circle constant = 3.14…
natural number = 2.7…
Normal Curve Mathematics
Main Ideas:
•
•
Basic shape is:
e
“Shifted to mu”:
1
 x2
2
e

1
 x   2
2
1 x 
 

2  
2
•
“Scaled by sigma”:
•
Make Total Area = 1: divide by
•
f ( x )  0 as x   , but never  0
e
2 
Computation of Normal Areas
EXCEL
Computation:
works in terms of
“lower areas”
E.g. for N (1,0.5)
Area < 1.3
Computation of Normal Probs
EXCEL Computation:
probs given by “lower
areas”
E.g. for X ~ N(1,0.5)
P{X < 1.3} = 0.73
Normal Random Variables
As above, compute probabilities as areas,
In EXCEL, use NORMDIST & NORMINV
E.g. above:
X ~ N(1,0.5)
P{X < 1.3} =NORMDIST(1.3,1,0.5,TRUE)
= 0.73
(as in pic above)
Normal Random Variables
HW:
4.57, 4.58 (0.965, ~0)