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RANDOM VARIABLES, EXPECTATIONS, VARIANCES ETC. - THEORY. Variable. Recall: Variable: A characteristic of population or sample that is of interest for us. Random variable: A function defined on the sample space S that associates a real number with each outcome in S.

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Variable
Variable

  • Recall:

  • Variable: A characteristic of population or sample that is of interest for us.

  • Random variable: A function defined on the sample space S that associates a real number with each outcome in S.


Discrete random variables
DISCRETE RANDOM VARIABLES

  • If the set of all possible values of a r.v. X is a countable set, then X is called discrete r.v.

  • The function f(x)=P(X=x) for x=x1,x2, … that assigns the probability to each value x is called probability density function (p.d.f.) or probability mass function (p.m.f.)


Example
Example

  • Discrete Uniform distribution:

  • Example: throw a fair die. P(X=1)=…=P(X=6)=1/6


Continuous random variables
CONTINUOUS RANDOM VARIABLES

  • When sample space is uncountable (continuous)

  • Example: Continuous Uniform(a,b)


Cumulative density function c d f
CUMULATIVE DENSITY FUNCTION (C.D.F.)

  • CDF of a r.v. X is defined as F(x)=P(X≤x).

  • Note that, P(a<X ≤b)=F(b)-F(a).

  • A function F(x) is a CDF for some r.v. X iff it satisfies

F(x) is continuous from right

F(x) is non-decreasing.


Example1
Example

  • Consider tossing three fair coins.

  • Let X=number of heads observed.

  • S={TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

  • P(X=0)=P(X=3)=1/8; P(X=1)=P(X=2)=3/8


Example2
Example

  • Let


Joint distributions
JOINT DISTRIBUTIONS

  • In many applications there are more than one random variables of interest, say X1, X2,…,Xk.

    JOINT DISCRETE DISTRIBUTIONS

  • The joint probability mass function (joint pmf) of the k-dimensional discrete rv

    X=(X1, X2,…,Xk) is


Joint discrete distributions
JOINT DISCRETE DISTRIBUTIONS

  • A function f(x1, x2,…, xk) is the joint pmf for some vector valued rv X=(X1, X2,…,Xk) iff the following properties are satisfied:

    f(x1, x2,…, xk) 0 for all (x1, x2,…, xk)

    and


Example3
Example

  • Tossing two fair dice  36 possible sample points

  • Let X: sum of the two dice;

    Y: |difference of the two dice|

  • For e.g.:

    • For (3,3), X=6 and Y=0.

    • For both (4,1) and (1,4), X=5, Y=3.


Example4
Example

  • Joint pmf of (x,y)

Empty cells are equal to 0.

e.g. P(X=7,Y≤4)=f(7,0)+f(7,1)+f(7,2)+f(7,3)+f(7,4)=0+1/18+0+1/18+0=1/9


Marginal discrete distributions
MARGINAL DISCRETE DISTRIBUTIONS

  • If the pair (X1,X2) of discrete random variables has the joint pmf f(x1,x2), then the marginal pmfs of X1 and X2 are


Example5
Example

  • In the previous example,


Joint discrete distributions1
JOINT DISCRETE DISTRIBUTIONS

  • JOINT CDF:

  • F(x1,x2) is a cdf iff


Joint continuous distributions
JOINT CONTINUOUS DISTRIBUTIONS

  • A k-dimensional vector valued rvX=(X1, X2,…,Xk) is said to be continuous if there is a function f(x1, x2,…, xk), called the joint probability density function (joint pdf), of X, such that the joint cdf can be given as


Joint continuous distributions1
JOINT CONTINUOUS DISTRIBUTIONS

  • A function f(x1, x2,…, xk) is the joint pdf for some vector valued rv X=(X1, X2,…,Xk) iff the following properties are satisfied:

    f(x1, x2,…, xk) 0 for all (x1, x2,…, xk)

    and


Joint continuous distributions2
JOINT CONTINUOUS DISTRIBUTIONS

  • If the pair (X1,X2) of discrete random variables has the joint pdf f(x1,x2), then the marginal pdfs of X1 and X2 are


Joint distributions1
JOINT DISTRIBUTIONS

  • If X1, X2,…,Xk are independent from each other, then the joint pdf can be given as

    And the joint cdf can be written as


Conditional distributions
CONDITIONAL DISTRIBUTIONS

  • If X1 and X2 are discrete or continuous random variables with joint pdf f(x1,x2), then the conditional pdf of X2 given X1=x1 is defined by

  • For independent rvs,


Example6
Example

Statistical Analysis of Employment Discrimination Data (Example from Dudewicz & Mishra, 1988; data from Dawson, Hankey and Myers, 1982)

Affected class might be a minority group or e.g. women


Example cont
Example, cont.

  • Does this data indicate discrimination against the affected class in promotions in this company?

  • Let X=(X1,X2,X3) where X1 is pay grade of an employee; X2 is an indicator of whether the employee is in the affected class or not; X3 is an indicator of whether the employee was promoted or not

  • x1={5,7,9,10,11,12,13,14}; x2={0,1}; x3={0,1}


Example cont1
Example, cont.

  • E.g., in pay grade 10 of this occupation (X1=10) there were 102 members of the affected class and 695 members of the other classes. Seven percent of the affected class in pay grade 10 had been promoted, that is (102)(0.07)=7 individuals out of 102 had been promoted.

  • Out of 1950 employees, only 173 are in the affected class; this is not atypical in such studies.


Example cont2
Example, cont.

  • E.g. probability of a randomly selected employee being in pay grade 10, being in the affected class, and promoted: P(X1=10,X2=1,X3=1)=7/1950=0.0036 (Probability function of a discrete 3 dimensional r.v.)

  • E.g. probability of a randomly selected employee being in pay grade 10 and promoted:

    P(X1=10, X3=1)= (7+56)/1950=0.0323 (Note: 8% of 695 -> 56) (marginal probability function of X1 and X3)


Example cont3
Example, cont.

  • E.g. probability that an employee is in the other class (X2=0) given that the employee is in pay grade 10 (X1=10) and was promoted (X3=1):

    P(X2=0| X1=10, X3=1)= P(X1=10,X2=0,X3=1)/P(X1=10, X3=1)

    =(56/1950)/(63/1950)=0.89 (conditional probability)

  • probability that an employee is in the affected class (X2=1) given that the employee is in pay grade 10 (X1=10) and was promoted (X3=1):

    P(X2=1| X1=10, X3=1)=(7/1950)/(63/1950)=0.11


Production problem
Production problem

  • Two companies manufacture a certain type of sophisticated electronic equipment for the government; to avoid the lawsuits lets call them C and company D. In the past, company C has had 5% good output, whereas D had 50% good output (i.e., 95% of C’s output and 50% of D’s output is not of acceptable quality). The government has just ordered 10,100 of these devices from company D and 11,000 from C (maybe political reasons, maybe company D does not have a large enough capacity for more orders). Before the production of these devices start, government scientists develop a new manufacturing method that they believe will almost double the % of good devices received. Companies C and D are given this info, but its use is optional: they must each use this new method for at least 100 of their devices, but its use beyond that point is left to their discretion.


Production problem cont
Production problem, cont.

  • When the devices are received and tested, the following table is observed:

  • Officials blame scientists and companies for producing with the lousy new method which is clearly inferior.

  • Scientists still claim that the new method has almost doubled the % of good items.

  • Which one is right?


Production problem cont1
Production problem, cont.

  • Answer: the scientists rule!

  • The new method nearly doubled the % of good items for both companies.

  • Company D knew their production under standard method is already good, so they used the new item for only minimum allowed.

  • This is called Simpson’s paradox. Do not combine the results for 2 companies in such cases.


Describing the population
Describing the Population

  • We’re interested in describing the population by computing various parameters.

  • For instance, we calculate the population mean and population variance.


Expected values
EXPECTED VALUES

Let X be a rv with pdf fX(x) and g(X) be a function of X. Then, the expected value (or the mean or the mathematical expectation) of g(X)

providing the sum or the integral exists, i.e.,

<E[g(X)]<.


Expected values1
EXPECTED VALUES

  • E[g(X)] is finite if E[| g(X) |]is finite.


Population mean expected value
Population Mean (Expected Value)

  • Given a discrete random variable X with values xi, that occur with probabilities p(xi), the population mean of X is


Population variance
Population Variance

  • Let X be a discrete random variable with possible values xi that occur with probabilities p(xi), and let E(xi) =. The variance of X is defined by

Unit*Unit

Unit


Expected value
EXPECTED VALUE

  • The expected value or mean value of a continuous random variable X with pdf f(x) is

  • The variance of a continuous random

  • variable X with pdf f(x) is


Example7
EXAMPLE

  • The pmf for the number of defective items in a lot is as follows

Find the expected number and the variance of

defective items.


Example8
EXAMPLE

  • Let X be a random variable. Its pdf is

    f(x)=2(1-x), 0< x < 1

    Find E(X) and Var(X).


Laws of expected value
Laws of Expected Value

  • Let X be a rv and a, b, and c be constants. Then, for any two functions g1(x) and g2(x) whose expectations exist,


Laws of expected value and variance

Laws of Expected Value

E(c) = c

E(X + c) = E(X) + c

E(cX) = cE(X)

Laws of Variance

V(c) = 0

V(X + c) = V(X)

V(cX) = c2V(X)

Laws of Expected Value and Variance

Let X be a rv and c be a constant.


Expected value1
EXPECTED VALUE

If X and Y are independent,

The covariance of X and Y is defined as


Expected value2
EXPECTED VALUE

If X and Y are independent,

The reverse is usually not correct! It is only correct under normal distribution.

If (X,Y)~Normal, then X and Y are independent iff

Cov(X,Y)=0


Expected value3
EXPECTED VALUE

If X1 and X2 are independent,



Conditional expectation and variance1
CONDITIONAL EXPECTATION AND VARIANCE

(EVVE rule)

Proofs available in Casella & Berger (1990), pgs. 154 & 158


Example advanced
Example - Advanced

  • An insect lays a large number of eggs, each surviving with probability p. Consider a large number of mothers. X: number of survivors in a litter; Y: number of eggs laid

  • Assume:

  • Find: expected number of survivors, i.e. E(X)


Example solution
Example - solution

EX=E(E(X|Y))

=E(Yp)

=p E(Y)

=p E(E(Y|Λ))

=p E(Λ)

=pβ


Some mathematical expectations
SOME MATHEMATICAL EXPECTATIONS

  • Population Mean:  = E(X)

  • Population Variance:

(measure of the deviation from the population mean)

  • Population Standard Deviation:

  • Moments:


Skewness
SKEWNESS

  • Measure of lack of symmetry in the pdf.

If the distribution of X is symmetric around its mean ,

3=0  Skewness=0


Kurtosis
KURTOSIS

  • Measure of the peakedness of the pdf. Describes the shape of the distribution.

Kurtosis=3  Normal

Kurtosis >3  Leptokurtic

(peaked and fat tails)

Kurtosis<3  Platykurtic

(less peaked and thinner tails)


Kurtosis1
KURTOSIS

  • What is the range of kurtosis?

  • Claim: Kurtosis ≥ 1. Why?

  • Proof:


Problems
Problems

  • True or false: The mean, median and mode of a normal distribution with mean µ and std deviation σ coincide.


Problems1
Problems

2. True or false: In a symmetrical population, mean, median, and mode coincide. (Kendall & Stuart, 1969, p. 85)


Problems2
Problems

3. True or False: “The mean, median and mode occur in the same order (or reverse order) as in the dictionary; and that the median is nearer to the mean than that to the mode, just as the corresponding words are nearer together in the dictionary. “ (Kendall & Stuart, 1969, p. 39)


Problems3
Problems

4. If X, Y, Z and W are random variables, then find (show the derivations):

a) Cov(X+Y,Z+W)

b) Cov(X-Y,Z)


Problems4
Problems

5. Calculate

a)the skewness for . Comment.

b) the kurtosis for the following pdf and comment:


Problems5
Problems

5. c) Consider the discrete random variable X with pdf given below:

i) Is the distribution of X symmetric around mean?

ii) Show that the 3rd central moment, and hence skewness, are 0. What does this imply?


Problem
Problem

6. Let X1, X2, X3 be three independent r.v.s each with variance . Define new r.v.s W1, W2, W3 by W1=X1; W2=X1+X2; W3=X2+X3.

Find Cor(W1,W2), Cor(W2,W3), Cor(W1,W3)


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