EMAT 20205 Data Analysis WEEK 2

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EMAT 20205Data AnalysisWEEK -2

• Nello Cristianini

2
Axioms of Probability

• The probability law (assigning a number to each
  event E) must satisfy the following axioms
• Nonnegativity
• Additivity if E and F are two disjoint events,
  then the probability of their union satisfies
• Normalization the probability of the entire
  sample space is equal to 1

3
Some comments

• The maximum value for the probability of an event
  is 1 (probability of the entire sample space)
• This means that that event is CERTAIN
• P(W)1 means the outcome will be one of the
  possible outcomes (obviously)(e.g. the dice
  roll will certainly give outcome 1 or 2 or 3 or
  4 or 5 or 6)

4
Comments

• An event E is IMPOSSIBLE if it has zero
  probability P(E)0
• An event is CERTAIN if it has probability P(E)1
• The interesting things happen in between

5
Comments on Additivity

• Additivity if E and F are two disjoint events,
  then the probability of their union satisfies
• Probability of E or F is P(E) P(F)
• E.g. in dice roll probability of 1 or 2 is
  P(1)P(2)

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Consequences

• If we use a sample space WO1, O2, O3, O4,On
  the probabilities of the outcomes Oi must
  satisfyP(O1)P(O2)P(On)1
• We will write this sum as

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Consequences

• From this axiom we can see thatthe probability
  of the empty event is 0(so there MUST be
  an outcome, think of dice roll example)

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Probability Law

• We have seen 3 axioms that must be satisfied by
  the probability assignment to the outcomes
  (sample space) and some of their consequences
• BUT who gives us the probabilities ?
• They are largely an arbitrary design choice
  (although we will see practical methods)

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Example

• Think again of the case of the dice roll.
• Given our knowledge of physics, and the symmetry
  of a dice, we see no reason why a certain outcome
  should be more likely than another. So we
  wantP(1)P(2)P(3)P(4)P(5)P(6)
• The normalization axiom gives P()1/6 for each
  of them
• We can then use these probabilities and the
  axioms to compute probabilities of more complex
  events

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Example

• Coin toss.
• Again no reason to prefer one outcome over
  another, soP(H)P(T)1/2
• Unless

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Frequency information

• Unless we actually know that specific coin (or
  dice) and we know the exact frequency of the
  outcomes in the last 1000s experiments
• Possibly the coin is not fair, and we observe 80
  head, 20 tail outcomes
• We can incorporate this in the model, assigning
  P(H)0.8 P(T)0.2In the first case we have
  used our knowledge of the situation in the
  second case we have estimated the probabilities
  by using frequencies

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Probabilistic Model of Coin Toss

• Sample space is WH,T
• Possible events are all subsetsH,T, H, T,
  0 (empty)
• Fair coin ? P(H)P(T)0.5
• P(H,T)P(H)P(T)1
• P(0)0
• So we have assigned a probability to EACH
  possible event based on the probabilities on the
  outcomes, in a way to satisfy all axioms

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Model Toss of Three Coins

• Sample space (8 possible outcomes)WHHH,HHT,HTH
  , HTT, TTT, THH, THT, TTH
• We assume they are all equally likely, so we
  assign to each of them probability 1/8
• The probability law should assign probabilities
  to EVERY POSSIBLE EVENT

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Tossing Three Coins

• A possible event 2 heads occur
• How many outcomes are in this event ?HHT, HTH,
  THH
• 3 disjoint events, their union has probability
  equal to the sum of their probabilities
• P(HHT, HTH, THH) P(HHT)P(HTH)P(THH)
  1/81/81/83/8

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Tossing Three Coins

• We can calculate similarly the probability of all
  possible events, and this gives a probability law
  that satisfies the axioms.
• We can see that obtaining 3 heads has probability
  1/8, less than observing 2 heads (3/8), and so on
  

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Probability law for finite sample spaces

• For finite sample spaces, we specify the
  probability law by just assigning probabilities
  to the individual outcomes
• Often the outcomes are equiprobable,
  thenP(E)number of outcomes in E / total number
  of outcomes

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Continuous Sample Space

• In the case of the dart and target, things are
  different
• If each outcome is a point,its probability
  cannotbe bigger than zero,else the total
  probability will exceed one
• Solution outcomes must be(infinitesimally)
  small areas, not points
• Do not worry too much about this for now

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Properties of Probability Law
Assume area of set probability of event!
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Using Probabilistic Models

• Say we want to model an uncertain situation(e.g.
  an experiment)
• We first decide a sample space and a probability
  law. This step is somewhat arbitrary, and fully
  specifies the model.
• Then operating within the model we derive the
  probabilities of the events of interest, or other
  properties. This is fully unambiguous.

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Example

• We want to choose a day in 2009 when to organize
  a picnic
• We want to avoid rain, cold and traffic
• These are three possible events(dayrain
  daycold daytraffic)not mutually exclusive

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Assume this is a generic month. A random day will
havevalues for R,C,T we can compute the
probability forR (rain), or for nT (not
traffic) but also for R AND T
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Event RAIN
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Event COLD
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Event TRAFFIC
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Unions and Intersections of Events

• We may want to calculate the probability to
  randomly selecting a day that is both not-rainy
  and not-cold
• Today we talk of probabilities of COMBINATIONS of
  events

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Intersection of Events

• Probability that BOTH events occur simultaneously
• We DEFINE A NEW EVENT consisting of the outcomes
  that are in both events E and F and we calculate
  its probability
• New event
• The probability of both events occurring is

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Intersection of Events

• The probability of this event is the sum of the
  probabilities of the outcomes that are both in E
  and in F (e.g. fraction of days that are both R
  and T)
• Two events are mutually exclusive (or disjoint)
  if their intersection is empty(e.g. R and nR
  are disjoint)

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rain
cold
EventRain and cold
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Union of Events

• We want to calculate the probability that at
  least one of the events E and F occurs
• This is the probability of the union event
• The probability of G is the sum of the
  probability of the outcomes that are in either E
  or F(e.g. number of days that are either R or C)

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rain
cold
EventRain OR cold
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Other combinations

• We can consider the probability of being in E and
  not in F by considering the probability of being
  in E and in FC

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Dice Example

• Event E 1,2,3 outcome is small
  (less than 3)
• Event F 2,4,6 outcome is even number
• Probability of being either even OR small ?
• Probability of being even AND small ?

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R,C,T 6 6/30 nR,C,T 0 0/30 R,nC,T 16 16/30 nR,
NC,T 1 1/30 R,C,nT 0 0/30 nR,C,nT 6 6/30 R,nC,nT
1 1/30 nR,NC,nT 0 0/30
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Important

• Calculate the joint probabilities from the table
  
• P(R,C,nT)0/30
• P(R,C,T)6/30
• P(R,C)P(R,C,nT)P(R,C,T)6/30

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Conditional Probability

• What is the probability of rain in this
  month?(count all rainy days and divide by 30)
• P ( R )R / Days
• What is the probability of rain given that it is
  cold ?

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Conditional Probability

• Outcomes of experiment days
• Being a cold day is an event
• Being a rainy day is an event
• Probability of being cold AND rainy ?
• Cold AND NOT rainy ?
• NOW Is it more likely to be cold in rainy days ?
• What about COLD given that it is RAINY ?

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R,C,T 6 6/30 nR,C,T 0 0/30 R,nC,T 16 16/30 nR,
NC,T 1 1/30 R,C,nT 0 0/30 nR,C,nT 6 6/30 R,nC,nT
1 1/30 nR,NC,nT 0 0/30

• P(d is rainy d is cold)
• P(d is cold) 12/30
• P(d is rainy) 23/30
• P(d is rainy and cold)6/30

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Is it more likely to have rain in cold days ?

• P(rain)23/30
• What is the rain probability IN THE COLD DAYS ?
• Probability of rain given cold is
  P(raincold) P(rain AND cold)/P(cold)
• P(raincold) 6/120.5

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Definition

• We define conditional probability of E given F
• Given that F is true, what is the probability of
  E ?
• In a way, restrict to the case when only F
  exists, F is the universe here

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Conditional probability

• We can consider the conditional probability
  P(EF) as a new probability law defined on a new
  universe, F
• P(FF)1
• All other axioms also remain valid

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Properties of Conditional Probability

• It satisfies all the axioms to be a probability
  law

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Properties of Conditional Probability

• Definition
• This can be seen as a new probability law in the
  restricted universe F
• For finite sample spaces

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Independent Events

• We define 2 independent events as follows

Independent eventsP(EF)P(E)
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Independent Events

• 2 independent eventsrain and monday
• 2 dependent eventsrain and january
• 2 dependent (?) eventstraffic and Monday
• 2 independent eventsjanuary and monday

In theory(not sure aboutour finite dataset)
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Bayes Theorem

• Calculation
• P(cold)12/302/5
• P(traffic)6/301/5
• P(cold AND traffic)6/301/5
• P(coldtraffic)1
• P(trafficcold)1/2

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Bayes Theorem
P(coldtraffic)P(traffic)P(cold AND traffic)

• Calculation
• P(cold)12/302/5
• P(traffic)6/301/5
• P(cold AND traffic)6/301/5
• P(coldtraffic)1
• P(trafficcold)1/2

11/51/5
P(trafficcold)P(cold)P(traffic AND cold)
½2/51/5
P(coldtraffic)P(traffic) P(trafficcold)P(cold)
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Bayes Theorem

• P(coldtraffic)P(traffic) P(trafficcold)P(cold)

P(coldtraffic) P(trafficcold)P(cold)/P(traffic)
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Independent Events

• P(EF)P(E)
• E independent of F

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Independent Events

• Since it was
• And we are assuming
• it follows that for independent events

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Independent Events

• If E and F are independent, so are E and FC

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Independence of 3 events

• E,F,G are independent if every subset of these 3
  events is independent
• E,F are independent
• E,G are independent
• F,G are independent
• And P(E,F,G)P(E)P(F)P(G)

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Independent Events

• We can decompose joint probabilitiesP(E,F,G)P(E
  )P(F)P(G)if they are independent
• Otherwise, we should writeP(E,F,G)P(EF,G)P(FG
  )P(G)

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Bernoulli Trials

• Toss a coin N times
• Probability of starting with H ½
• Probability of starting with HH ½ ½
• Probability of N consecutive H (½)N

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MATLAB INTERLUDE

• INTERSECT Set intersection.
• INTERSECT(A,B) when A and B are vectors returns
  the values common to both A and B. The result
  will be sorted. A and B can be cell arrays of
  strings.

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MATLAB INTERLUDE

• UNION Set union.
• UNION(A,B) when A and B are vectors returns the
  combined values from A and B but with no
  repetitions. The result will be sorted.

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MATLAB INTERLUDE

• FIND Find indices of nonzero elements.
• I FIND(X) returns the linear indices
  corresponding to the nonzero entries of the array
  X.
• X may be a logical expression.
• So you can find elements in a set with a given
  property, and make a new set

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MATLAB INTERLUDE

• LENGTH Length of vector.
• LENGTH(X) returns the length of vector X. It is
  equivalent to MAX(SIZE(X)) for non-empty arrays
  and 0 for empty ones.

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MATLAB INTERLUDE

• You can use these set commands to count the
  elements in various sets, and hence to compute
  probabilities

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Topics

• Modeling with Random Variables
• Discrete Random Variables
• Events and
• Probability Mass Function
• Examples of RV
• Bernoulli
• Binomial
• Geometric
• The concept of Expectation

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RANDOM VARIABLES

• We have studied Probabilistic Models in general,
  the notions of outcome, sample space and event.
• Now an important special casein many
  probabilistic models the outcomes are NUMBERS, or
  can be associated to numbers

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RANDOM VARIABLES

• Examples of numerical outcome
• How many people showed up today ?
• How many are sitting next to a statistics major?
• How many days of rain in january ?
• Temperature on a given day ?
• OR we can ASSOCIATE numerical values to
  non-numerical outcomes

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RANDOM VARIABLES

• Associating numerical values to non-numerical
  outcomes HOMEWORK EXPERIMENT
• Outcome the homework
• Sample space set of all possible answers you
  COULD have given
• Associated numerical value the GRADE

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RANDOM VARIABLES

• Easier model multiple choice quiz
• 10 questions, 3 choices each (A,B,C)
• Experiment give the test to a student
• Outcome a string of 10 symbols
• Sample space set of all possible 10 symbols
  strings
• Numeric value the grade assigned to each string
  (some form of distance to correct string)

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RANDOM VARIABLES

• We call RANDOM VARIABLE a real-valued function of
  the outcome of an experiment
• Given an experiment, and the corresponding set of
  possible outcomes, a random variable associates a
  particular number with each outcome

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RANDOM VARIABLES

• Example
• Sample space AAA, AAB, AAC, .
• Random variableAAA?3AAB?2AAC?3
• This could be a model of grading a test

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RANDOM VARIABLES

• Why are RANDOM VARIABLES important ?
• They allow us to model uncertain situations in a
  quantitative way, we will talk aboutthe
  EXPECTED temperature on january 25, or the
  EXPECTED number of students that will pass the
  test, etc.
• We can also talk about expected deviations from
  this estimate

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RANDOM VARIABLES(continuous vs discrete)

• A random variable is called discrete if its range
  (the set of values it can take) is finite or
  COUNTABLY infinite
• It is called continuous for example - if its
  range is the real axis (but we will not deal with
  this case today)

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RANDOM VARIABLES

• Examples of discrete random variables
• Number of things (number of tails in1000 coin
  tosses)
• Number of minutes this class will last
• Roll of 2 dice, sum or product of the outputs is
  a discrete random variable

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RANDOM VARIABLES

• The 2- dice example
• Let us call A B C D E
  F

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RANDOM VARIABLES

• Let us consider the following random variable N
  associated to one diceN(A)1N(B)2N(
  C)3N(D)4N(E)5N(F)6

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RANDOM VARIABLES

• Sample space of the 2 dice experiment
  AA,AB,AC,AD,AE,AF, BA,BB,BC,BD,BE,BF,
  CA,CB,CC,CD,CE,CF, DA,DB,DC,DD,DE,DF,
  EA,EB,EC,ED,EE,EF, FA,FB,FC,FD,FE,FF,

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RANDOM VARIABLES

• Sum random variableAA?112 S(AA)AB?123
  S(AB)FF?6612 S(FF)
• Range of random variable2,3,4,5,6,7,8,9,10,11,1
  2

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RANDOM VARIABLES

• Similarly we can define the random variable
  PRODUCT, etc
• So after the same experiment (rolling 2 dice) we
  may define different random variables (sum,
  absolute difference, product, max, min, etc of
  the two individual outcomes )
• Whatever attaches a numeric value to the OUTCOME
  of the experiment is a RANDOM VARIABLE

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RANDOM VARIABLESimportant concepts

• A discrete random variable is a real valued
  function of the outcome of the experiment that
  can take a finite or countably infinite number of
  values
• A function of a discrete random variable defines
  another random variable
• We will define MEAN and VARIANCE of a random
  variable
• We will define independence and all other
  concepts we defined in the previous classes

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RANDOM VARIABLES

• For discrete random variables we will define
  PROBABILITY MASS FUNCTIONS, that are probability
  laws that assign a probability to each possible
  numerical value the random variable can assume
• It will be analogous to what done so far

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RANDOM VARIABLESnotation

• We will denote by uppercase letters (X) the
  random variable, by lowercase letters (x) the
  actual value it assumes in a given experiment
• So we will talk about the probability that Xx,
  for example and we will write it P(Xx)

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RANDOM VARIABLES

• Look at the website of the course, where we
  publish the statistics about the past homeworks
• Random variable GRADE, G
• A particular grade g
• For example we can talk about P(G27)

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RANDOM VARIABLES

• Easier model multiple choice quiz
• 10 questions, 3 choices each (A,B,C)
• Experiment give the test to a student
• Outcome a string of 10 symbols
• Sample space set of all possible 10 symbols
  strings
• Numeric value the grade assigned to each string
  (some form of distance to correct string)

79
RANDOM VARIABLES

• We call RANDOM VARIABLE a real-valued function of
  the outcome of an experiment
• Given an experiment, and the corresponding set of
  possible outcomes, a random variable associates a
  particular number with each outcome

80
RANDOM VARIABLESimportant concepts

• A discrete random variable is a real valued
  function of the outcome of the experiment that
  can take a finite or countably infinite number of
  values
• A function of a discrete random variable defines
  another random variable
• We will define MEAN and VARIANCE of a random
  variable
• We will define independence and all other
  concepts we defined in the previous classes

81
RANDOM VARIABLES

• For discrete random variables we will define
  PROBABILITY MASS FUNCTIONS, that are probability
  laws that assign a probability to each possible
  numerical value the random variable can assume
• It will be analogous to what done so far

82
RANDOM VARIABLESnotation

• We will denote by uppercase letters (X) the
  random variable, by lowercase letters (x) the
  actual value it assumes in a given experiment
• So we will talk about the probability that Xx,
  for example and we will write it P(Xx)

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Probability Mass Function (PMF)

• The most important way to characterize a random
  variable is through the probabilities of the
  values that it can take
• For the random variable X, these are given by the
  PMF of X, denoted pX.
• If x is any possible value of X, the probability
  mass of x, pX(x) is the probability of the event
  Xx, consisting of all outcomes that give rise
  to a value of X equal to x
• pX(x)P(Xx)

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PMF

• Example experiment tossing 2 fair coins
• Random Variable X number of heads obtained
  (range 0,1,2)
• Compute the PMF of X
• pX(x)

¼ if x0 ½ if x1 ¼ if x2 0 otherwise
(impossible)
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PMF

• Event x0 ? corr. Outcome TT
• Event x1 ? corr. Outcomes HT or TH
• Event x2 ? corr. Outcome HH
• Each outcome has probability ¼
• ? hence the probabilities given before
• (grouping outcomes based on value of random
  variable a way to define events )

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PMF

• Some propertiessince the events corresponding
  to each value of the random variable must be
  disjoint, and form a partition of the sample
  space,
• From probability axioms we obtain

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PMF

• By a similar argument, we have for any set S of
  possible values of X In coin example
  before, we can say probability of at least 1
  head is ¾ (sum of prob 1 heat prob 2 heads)

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PMF
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PMF

• Some propertiessince the events corresponding
  to each value of the random variable must be
  disjoint, and form a partition of the sample
  space,
• From probability axioms we obtain

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PMF

• By a similar argument, we have for any set S of
  possible values of XIn coin example before,
  we can say probability of at least 1 head is ¾
  (sum of prob 1 heat prob 2 heads)

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Functions of Random Variables

• One can generate new random variables as
  functions of random variables

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CALCULATION OF PMF OF A RANDOM VARIABLE X

• For each possible value x of X
• Collect all the possible outcomes that give rise
  to the event Xx
• Add their probabilities to obtain pX(x)
• THIS IS IMPORTANT !!

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Example

• Probability of having HW grade larger than 30 ?
• Prob G30 prob G31 prob G40
• Each probability count number of outcomes,
  divide by total sample space size

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Expectation

• The PMF of a random variable provides us with
  several numbers the probabilities of all
  possible values of X
• We would like to summarize this in few numbers
  that represent the PMF
• One such number is the EXPECTATION

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Expectation

• Expected value of Xweighted average of all
  possible values of X (using probabilities as
  weights)

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Expectation

• Suppose you roll a dice many times, and each time
  you receive as many dollars as the outcome of the
  dice-roll
• How much money would you expect for each roll ?
• We need to specify these terms

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Expectation

• Suppose you roll the dice K times, and Ki is the
  number of times the outcome is i
• Sample space 1,2,3,4,5,6
• The total amount of money you receive is

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Expectation

• The total amount in K rolls is
• So the amount per roll is

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Expectation

• If we have been rolling the dice many times (K
  is v. large), we can approximate the probability
  of an outcome with its frequency piKi/K
• Then we can write the expected amount of money
  as

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Expectation

• We define the expected value (expectation, or
  mean) of a random variable X, with PMF pX, by

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Expectation

• Remark we can consider this as the center of
  gravity of the distribution

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Variance

• Other important quantity to describe PMF.
• Expectation we know the average behavior of
  the random variable
• But how often does the random variable deviate
  from the average behavior ?

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Variance

• Let us create a NEW random variable describing
  the deviation of X from its mean EX, and let us
  study it
• What is the expected value of the random variable
  (X-EX)2 ?

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Variance

• New random variable (X-EX)2
• Its expectation E(X-EX)2Var(X) is called
  the variance of X
• It is always nonnegative
• Provides a measure of dispersion of X around its
  mean

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Variance
106
Variance

• Another related measure of dispersion is the
  standard deviation of X, defined as the square
  root of the variance

• From a practical viewpoint, the STD is easier to
  use because its has the same units as X(I.e. if
  X is in meters, STD will be in meters, Var(X) in
  square meters)

107
Calculation of Variance

• Can just study expectation of R.V. Z(X-EX)2
• X
• Z
• Var(X)EZ

108
Expected Value of Functions of Random Values

• Let X be a random variable with PMF p(x), and let
  g(X) be a function of X
• ?The expected value of the random variable g(X)
  is

109
Variance

• So the variance can be calculated as

110
Properties of Mean and Variance

• Let X be a random variable and let us consider
  the linear function YaXb where a,b are given
  scalars. Then
• EYaEXb
• Var(Y)a2Var(X)
• THIS ONLY if g(X) is linear !!

111
A useful relation(variance as a function of
moments)

• Var(X)E(X-EX) 2
• Var(X)EX2-(EX) 2
• Proof SEE IN LATER SLIDES FOR FULL PROOF Use
  the relation

112
Variance

• The variance can be calculated as

113
Properties of Mean and Variance

• Let X be a random variable and let us consider
  the linear function YaXb where a,b are given
  scalars. Then
• EYaEXb
• Var(Y)a2Var(X)
• THIS ONLY if g(X) is linear !!

114
A useful relation

• Var(X)E(X-EX 2)
• Var(X)EX2-(EX) 2
• Proof either as HW or with TasUse the relation

115
Variance Calculation

• Var(X)E(X-EX 2) EX2-(EX) 2

We will use this a lot
116
Covariance of 2 RVs

• In probability theory and statistics, covariance
  is a measure of how much two variables change
  together (variance is a special case of the
  covariance when the two variables are identical).
• If two variables tend to vary together (that is,
  when one of them is above its expected value,
  then the other variable tends to be above its
  expected value too), then the covariance between
  the two variables will be positive. On the other
  hand, when one of them is above its expected
  value the other variable tends to be below its
  expected value, then the covariance between the
  two variables will be negative.
• from wikipedia

117
Covariance of 2 RVs

• The covariance between two real-valued random
  variables X and Y, with expected values E(X)m
  E(Y)n is defined as
• Cov(X, Y) E(X - m) (Y - n)

118
In Matlab

• COV Covariance matrix.
• COV(X), if X is a vector, returns the
  variance. For matrices,
• where each row is an observation, and each
  column a variable,
• COV(X) is the covariance matrix.
  DIAG(COV(X)) is a vector of
• variances for each column, and
  SQRT(DIAG(COV(X))) is a vector
• of standard deviations. COV(X,Y), where X and
  Y are matrices with
• the same number of elements, is equivalent to
  COV(X() Y()).

119
Correlation Coefficient
From wikipedia
120
Correlation CoefficientBetween 2 Random Variables

• CORRCOEF Correlation coefficients.
• RCORRCOEF(X) calculates a matrix R of
  correlation coefficients for
• an array X, in which each row is an
  observation and each column is a
• variable.
• RCORRCOEF(X,Y), where X and Y are column
  vectors, is the same as
• RCORRCOEF(X Y).
• If C is the covariance matrix, C COV(X),
  then CORRCOEF(X) is
• the matrix whose (i,j)'th element is
• C(i,j)/SQRT(C(i,i)C(j,j)).

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EXTRA MATERIALBELOW THIS POINT

• WHAT FOLLOWS IS EXTRA MATERIAL FOR REFERENCE
• Not covered in class 1 of week 2 (refers to
  class 2 of week 2)

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Bernoulli Random Variable

• Consider the toss of a (generally not fair) coin,
  probability H p prob T 1-p
• The BERNOULLI random variable is a RV that takes
  the two values 0 or 1 depending on whether the
  outcome is H or T(remember RV is a function of
  the outcome)
• X1 if outcome is H X0 if outcome is T

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Bernoulli Random Variable

• The PMF of this Bernoulli RV is
• PX(x)
• Very important RV in modeling any generic
  situation with just 2 outcomes, e.g. outcome of
  the football match on Sunday,

P if x1 1-p if x0
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Binomial Random Variable

• Experiment N coin tosses, each one with
  prob(H)p prob(T)1-p
• The random variable X is the number of heads in
  the n-toss sequence
• We refer to X as a BINOMIAL RANDOM VARIABLE WITH
  PARAMETERS n AND p

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Binomial Random Variable

• The PMF of X consists of the binomial
  probabilities we have seen some time ago
• Two parts
• probability of a sequence with k heads and n-k
  tails
• Number of sequences with k heads and n-k tails

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Binomial Random Variable

• The normalization property can be written as
• We will study this more in the future

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Geometric Random Variable

• We repeatedly toss the same coin as before.
• RV number of tosses before the first head comes
  up
• TTTTTTTH
• TTH
• H
• TTTTTTTTTTTTTTTTTTTTTTTTTTTTH

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Geometric Random Variable

• PMFtwo parts probability of the prefix of
  k1 tails, and probability of the end H
• Normalization

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Geometric Random Variable

• This can model the process of you trying to
  connect with the modem to an internet service
  provider (how many fails before 1 success ?)

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Poisson Random Variable
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Functions of Random Variables

• One can generate new random variables as
  functions of random variables

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Expectation

• The PMF of a random variable provides us with
  several numbers the probabilities of all
  possible values of X
• We would like to summarize this in few numbers
  that represent the PMF
• One such number is the EXPECTATION

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Expectation

• Expected value of Xweighted average of all
  possible values of X (using probabilities as
  weights)
• Next time we will develop this and other
  concepts

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Conclusion

• Random Variables
• Probability Mass Functions
• How to calculate PMFs
• Bernoulli
• Binomial
• Geometric
• Poisson ?

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Probability Mass Function (PMF)

• The most important way to characterize a random
  variable is through the probabilities of the
  values that it can take
• For the random variable X, these are given by the
  PMF of X, denoted pX.
• If x is any possible value of X, the probability
  mass of x, pX(x) is the probability of the event
  Xx, consisting of all outcomes that give rise
  to a value of X equal to x
• pX(x)P(Xx)

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PMF

• Example experiment tossing 2 fair coins
• Random Variable X number of heads obtained
  (range 0,1,2)
• Each outcome has probability ¼
• ? hence the probabilities are
• Event x0 ? corr. Outcome TT
• Event x1 ? corr. Outcomes HT or TH
• Event x2 ? corr. Outcome HH

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PMF

• Compute the PMF of X
• pX(x)

¼ if x0 ½ if x1 ¼ if x2 0 otherwise
(impossible)
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PMF
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PMF

• Some propertiessince the events corresponding
  to each value of the random variable must be
  disjoint, and form a partition of the sample
  space,
• From probability axioms we obtain

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PMF

• By a similar argument, we have for any set S of
  possible values of XIn coin example before,
  we can say probability of at least 1 head is ¾
  (sum of prob 1 heat prob 2 heads)

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Functions of Random Variables

• One can generate new random variables as
  functions of random variables

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Bernoulli Random Variable

• Consider the toss of a (generally not fair) coin,
  probability H p prob T 1-p
• The BERNOULLI random variable is a RV that takes
  the two values 0 or 1 depending on whether the
  outcome is H or T(remember RV is a function of
  the outcome)
• X1 if outcome is H X0 if outcome is T

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Bernoulli Random Variable

• The PMF of this Bernoulli RV is
• PX(x)
• Very important RV in modeling any generic
  situation with just 2 outcomes, e.g. outcome of
  the football match on Sunday,

P if x1 1-p if x0
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Mean and Variance

• EX1p 0(1-p)p
• EX2 12p 02(1-p)p
• Var(X)EX2-(EX) 2p-p2p(1-p)

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Uniform Distribution dice roll

• see later slides

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Binomial Random Variable

• Experiment N coin tosses, each one with
  prob(H)p prob(T)1-p
• The random variable X is the number of heads in
  the n-toss sequence
• We refer to X as a BINOMIAL RANDOM VARIABLE WITH
  PARAMETERS n AND p

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Binomial Random Variable

• The PMF of X consists of the binomial
  probabilities we have seen some time ago
• Two parts
• probability of a sequence with k heads and n-k
  tails
• Number of sequences with k heads and n-k tails

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Binomial Random Variable

• The normalization property can be written as
• We will study this more in the future

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Geometric Random Variable

• We repeatedly toss the same coin as before.
• RV number of tosses before the first head comes
  up
• TTTTTTTH
• TTH
• H
• TTTTTTTTTTTTTTTTTTTTTTTTTTTTH

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Geometric Random Variable

• PMFtwo parts probability of the prefix of
  k1 tails, and probability of the end H
• Normalization

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Bernoulli Random Variable

• Consider the toss of a (generally not fair) coin,
  probability H p prob T 1-p
• The BERNOULLI random variable is a RV that takes
  the two values 0 or 1 depending on whether the
  outcome is H or T(remember RV is a function of
  the outcome)
• X1 if outcome is H X0 if outcome is T

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Bernoulli Random Variable

• The PMF of this Bernoulli RV is
• PX(x)
• Very important RV in modeling any generic
  situation with just 2 outcomes, e.g. outcome of
  the football match on Sunday,

P if x1 1-p if x0
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Mean and Variance

• EX1p 0(1-p)p
• EX2 12p 02(1-p)p
• Var(X)EX2-(EX) 2p-p2p(1-p)

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Two Important Series

• We do not derive them here.We will apply these
  to calculations of variance

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Uniform Distribution dice roll

• Discrete Uniform PMF over a,b(case of the dice
  rolls)

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Uniform
The expectation is This can be seen
directly, since the PMF is symmetric around
(ab/2). Or use the series given before... Dice
example 12345621Direct Computation of
Expectation 21/63.5Formula says (16)/23.5

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Variance of Discrete Uniform

• We first study case where a1 bn the general
  case will reduce to this
• We will use relation Var(X)EX2-(EX)2

Can verify this by inductionof just believe it
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Variance of Discrete Uniform
Notice we are still working with special case
a1 bn
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Variance of Discrete Uniform

• Now we can study the general case by SHIFTING a
  distribution, its variance does not change (so we
  can study a,b case by studying variance of
  1,b-a1 case)
• So setting nb-a1 in the previous equation
  gives the general case

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Variance of Discrete Uniform

• Example I get 1 for each point on the dice, I
  can expect 3.5 dollars at each roll, and a
  Standard Deviation of sqrt(35/12)1.7

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Binomial Random Variable

• Experiment N coin tosses, each one with
  prob(H)p prob(T)1-p
• The random variable X is the number of heads in
  the n-toss sequence
• We refer to X as a BINOMIAL RANDOM VARIABLE WITH
  PARAMETERS n AND p

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Binomial Random Variable

• The PMF of X consists of the binomial
  probabilities we have seen some time ago
• Two parts
• probability of a sequence with k heads and n-k
  tails
• Number of sequences with k heads and n-k tails

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Binomial Random Variable

• The normalization property can be written as
• We will study this more in the future

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QUESTION

• There are 94 students
• Each has probability 1/3 to get an A
• The number of students that get an A is a random
  variable
• What is its mean ? (how many are expected to get
  an A)

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Mean of the Binomial

• If we want the mean of the binomial, we first
  need to learn how to handle JOINT PMFs of
  MULTIPLE RANDOM VARIABLES

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JOINT PMFs of MULTIPLE RANDOM VARIABLES

• Consider 2 discrete random variables, X and Y
  associated with the same experiment
• The probabilities of the values that X and Y can
  take, are captured by the JOINT PMF of X and Y,
  written pX,Y
• pX,Y(x,y)P(Xx,Yy)

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JOINT PMF of 2 RV

• (if we consider the pair X,Y as a random
  variable, all ideas transfer )
• If A is an event (set of pairs (x,y) that have a
  certain property) then

P((X,Y) in A)S(x,y in A)pX,Y(x,y)
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students

• Consider the random variable Xi that is 1 if
  student i gets an A, and 0 otherwise
• If n students, probability p, this is np

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Conclusion

• Mean of Random Variables
• Variance of Random Variables
• Properties, relations for variance and moments
• Bernoulli
• Discrete Uniform,
• General Methods for variance calculation

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topics

• Some probability distributions
• Some real applicationsdecision making modeling
  clashes between ants
• Modeling the distribution of ping times

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Marginalization

• For a fixed value y,
• Using the definition of conditional probability,
  we have

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Random Variables

• Joint probability
• Conditional probability
• Independence

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Joint Probability

• It is common for several random variables to be
  defined on the same sample space. If X and Y are
  random variables, the functionf(x,y) PrX x
  and Y y is the joint probability mass function
  of X and Y.

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Independent Random Variables

• We define two random variables X and Y to be
  independent if for all x and y, the events X x
  and Y y are independent or, equivalently, if
  for all x and y, we have PrX x and Y y
  PrX x PrY y.

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Functions of Random Variables

• Given a set of random variables defined over the
  same sample space, one can define new random
  variables as sums, products, or other functions
  of the original variables.

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Expected value of a random variable

• The simplest and most useful summary of the
  distribution of a random variable is the
  "average" of the values it takes on. The expected
  value (or, synonymously, expectation or mean) of
  a discrete random variable X is

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Expectation of joint RVs

• Given random variables X and Y, and given their
  PMF PXx and Yy, what is their joint
  expectation ? EX,Y
• Easy if they are independent

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Expectation of Joint Independent RVs
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In general

• In general, when n random variables X1, X2, . . .
  , Xn are mutually independent,EX1X2 Xn
  EX1EX2 EXn .

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More about independent RVs

• When X and Y are independent random variables,
• VarX Y VarX VarY.
• (whereas for ANY random variablesthe expectation
  of the sum is the sum of their expectations, that
  is,
• EX Y EX EY , )

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The Geometric Distribution

• A coin flip is an instance of a Bernoulli trial,
  which is defined as an experiment with only two
  possible outcomes success, which occurs with
  probability p, and failure, which occurs with
  probability q 1 - p.
• When we speak of Bernoulli trials collectively,
  we mean that the trials are mutually independent
  and that each has the same probability p for
  success.
• Two important distributions arise from Bernoulli
  trials the geometric distribution and the
  binomial distribution.

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Geometric Distribution

• Take a sequence of Bernoulli trials, each with a
  probability p of success and a probability q 1
  - p of failure.
• How many trials occur before we obtain a success?
  

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Geometric Distribution

• Let the random variable X be the number of trials
  needed to obtain a success. Then X has values in
  the range 1, 2, . . ., and PrX k qk-1p ,
  (for k larger than 0)since
  we have k - 1 failures before the one success.
• A probability distribution satisfying this
  equation is said to be a geometric distribution.

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Geometric Distribution

• This is the geometric dictribution (picture
  taken from Cormen, Leiserson and Rivests book on
  Algorithms)
• In this case, the coin has probability p 1/3 of
  success and a probability q 1 - p of failure

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Geometric distribution

• Expectationwe can use the relation
• That holds when the summation is infinite and x
  lt 1

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Geometric Distribution
The expectation of the distribution is 1/p 3.
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Geometric Distribution

• The variance, which can be calculated similarly,
  isVarX q/p2
• Example repeatedly roll two dice until we
  obtain either a seven or an eleven.Of the 36
  possible outcomes, 6 yield a seven and 2 yield an
  eleven. Thus, the probability of success is p
  8/36 2/9, and we must roll 1/p 9/2 4.5
  times on average to obtain a seven or eleven.
• NEXT WEEK we will implement things like this .

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BINOMIAL DISTRIBUTION

• How many successes occur during n Bernoulli
  trials, where a success occurs with probability p
  and a failure with probability q 1 - p?

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Binomial Distribution

• Define the random variable X to be the number of
  successes in n trials. Then X has values in the
  range 0, 1, . . . , n, and for k 0, . . . ,
  n,
• since there are ways to pick which k of
  the n trials are successes, and the probability
  that each occurs is pkqn-k. A probability
  distribution satisfying this equation is said to
  be a binomial distribution.

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Binomial Distribution

• Let Xi be the random variable describing the
  number of successes in the ith trial. Then EXi
  p1 q0 p, and by linearity of expectation,
  the expected number of successes for n trials is

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Binomial Distribution

• Similarly we can do for the variance, exploiting
  the relation VarXEX2 - E2X
• Since Xi only takes on the values 0 and 1, we
  have EX2 EXp
• And hence VarXi p - p2 pq .

• Then we can use independence, to move from
  VarXi to the variance of the binomial

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Binomial Distribution

• The binomial distribution increases as k runs
  from 0 to n until it reaches the mean np, and
  then it decreases.
• Picture from cormen, leiserson, rivests book

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Binomial Distribution
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Conclusion

• Conditional PMF in RVs
• Independence
• Expectation and Variance for RVs
• Geometric distribution
• Binomial distribution
• ? next we will implement all of these ideas

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• EXTRA MATERIAL (NOT COVERED IN CLASS)

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Cards
? ? ? ?
Ace 2 3 4 5 6 7 8 9 10 Jack Queen King
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Counting

• Probability of Generating a growing sequence of
  cards (1,2,3,4,5,6,7,8,9,)
• Probability of starting with a 1 probability of
  having a 2 probability of having a king

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COUNTING METHODS

• How many ways to obtain K heads and N-K tails in
  N coin tosses ?
• How many ways to have a 4-of-a-kind ?

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Basic Counting

• Two experiments are performed. The first one can
  have any one of N possible outcomes, the second
  one any of M possible outcomes.
• ? there are MN possible outcomes for the two
  experiments considered together

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Basic Counting

• How many different arrangements of the letters
  A,B,C are possible ?
• ABCACBBACBCACABCBA
• Each arrangement known as a PERMUTATION.
• There are 6 possible permutations of a set of 3
  objects
• There are N! permutations of a set of N
  objectsN!N(N-1)(N-2)321

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Combinations

• How many different groups of M objects can I form
  from a total of N objects ?(e.g. how many groups
  of 5 cards can I form from a deck of 52 ?)
• (there are 52 ways to select the first 51 to
  select the second but we are counting each
  group each time we see one of its possible
  orderings we need to correct for this )
• (5251504948)/(54321)

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Combinations

• Ways of choosing k elements out of a set of n
  elements

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Combinations and Permutations

• How many ways to put N balls in K boxes ?
• OOO11O1O1OOO11 ? example
• 1 is the boundary of the box ? will use G(K-1)
  1s
• O is the ball ? N will use Os
• (NG)!
• Correct for permutations of the 1s and of the 0s
• (NG)!/(N!G!)
• If create and MNG
• M!/(M-G)!G! Same as before

In exampleG6 K7N8 M14
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COUNTING METHODS

• Combinations VS permutations
• How many sets of 3 numbers out of 10 ?
• How many ordered sets of 3 numbers ?

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Pascals Triangle
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Pascals Triangle
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Binomial Coefficient and Pascals Triangle

• A number in the triangle can be found by nCr (n
  Choose r) where n is the number of the row and r
  is the element in that row. For example, in row
  3, 1 is the zeroth element, 3 is element number
  1, the next three is the 2nd element, and the
  last 1 is the 3rd element. The formula for nCr
  is
• n!--------r!(n-r)!

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Examples

• How many ways to select 5 cards from the deck ?
• How many ways to have 4 equal cards in a set of 5
  ?
• Probability of selecting 5 cards containing a
  poker ?

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Poker Probabilities

• Deck of 52 cards, rankedace, king, queen, jack,
  10,9,8,7,6,5,4,3,2 (and ace again it can be
  either high or low)
• 4 suits spades, hearts, diamonds and clubs
• 5 card draw 5 cards make up a poker hand
• The highest hand wins
• Hands are ranked as follows

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Poker Probabilities

• Royal flush ? 10, J, Q, K, A of the same suit
• Four of a kind ? 4 cards of the same RANK
• Full house ?3 cards of the same rank 2 cards of
  the same rank
• Flush ? 5 cards of the same suit

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Poker Probabilities

• How many poker hands ?2,598,960

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Poker Probabilities

• How many combinations of royal flush?4
  (probability 0.00000154)
• How many combinations of 4-of-a-kind ?624

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• Consider a number of experiments with poker cards
• Write down SAMPLE SPACE
• Count possible outcomes for each experiment (see
  book, or handouts)

? ? ? ?
Ace 2 3 4 5 6 7 8 9 10 Jack Queen King
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Kind of questions

• Probability of having King of ? at first draw ?
• Probability of having 4 kings ?
• Probability of having any set of 4 equal cards ?
• When we ask to write sample space for 5-cards
  experiment, we do not mean to list all of the
  outcomes (they are about 2.5 million), just to
  show you know what the sample space is e.g.all
  hands of 5 cards, or 2S, 2C,2D,
  2H,3S,KS,KC,KD,KH, AC,

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How to do the homework

• Always write down probabilistic model
• Use one of the 3 formulae we have for COUNTING
  number of events of a certain type, or of
  outcomes
• Use definitions like P(event) outcomes in
  event / possible outcomes

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Combinations

• Ways of choosing k elements out of a set of n
  elements
• HOW MANY COMMITTEES OF 5 PEOPLE CAN WE MAKE OUT
  OF A CLASS OF 10 PEOPLE ?

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Poker Probabilities

• How many poker hands ?2,598,960

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Poker Probabilities

• How many combinations of royal flush?4
  (probability 0.00000154)
• How many combinations of 4-of-a-kind ?624