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Discrete Random Variables 2

Discrete Random Variables 2

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Discrete Random Variables 2. Random Variable Numerical attribute of an experimental outcome. Probability Mass Function (PMF). Functions of Random Variables Y = 4*H 3 + 75 Y = H – E(H) Y = 1 if H = 0 0 if H >= 1. Expectation - PowerPoint PPT Presentation

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Page 1: Discrete Random Variables 2

Discrete Random Variables 2

Page 2: Discrete Random Variables 2

Random Variable Numerical attribute of an experimental outcome.

Page 3: Discrete Random Variables 2

Probability Mass Function (PMF)

0

1/8

2/8

3/8

4/8

0 1 2 3

# of heads

pro

bab

ilit

y

p(h)

Page 4: Discrete Random Variables 2

Functions of Random Variables Y = 4*H3 + 75 Y = H – E(H) Y = 1 if H = 0 0 if H >= 1

Page 5: Discrete Random Variables 2

Expectation Weighted average of all possible outcomes.

E[x] = ∑ [ x px(x) ]

Variance Measures the spread of the PMF around the expected value. or Y = (X – E(X))2

σx2 = E(Y)

Page 6: Discrete Random Variables 2

Functions of Random Variables (cont) Y = 1 if h <= 1 0 if h >= 2

E(H) = 1.5 E(Y) = ?

In general, for any var. X

and func. g(X): if Y = g(X): E(Y) = ∑ [ g(X) px(X) ]

Page 7: Discrete Random Variables 2

Bernoulli Random Variable

Experiment: Toss coin once

0 (T) 1 (H)

XT

H

Examples of experiments with 2 possible outcomes: - is a person healthy or sick? - do you like a song on pandora.com? - will event A occur or not?

0.5 0.51-p p

P(X=1) = p

P(X=0) = 1-p

Page 8: Discrete Random Variables 2

Bernoulli Random Variable (cont)

0 (T) 1 (H)

X

Experiment: Toss coin oncep

1-pPMF:

E(X) =

variance(X) =

p

p(1-p)

CDF:

0 (T) 1 (H)

X

1

1-p

Page 9: Discrete Random Variables 2

Binomial Random Variable

Experiment: number of tosses: 4 probability of heads: ¾ X = number of heads

TH

TH

TH

TH

TH

TH

TH

TH

H

T

H

T

H

T

H

T

T

H

T

H

H

T

HHHHHHHT

HHTHHHTT

HTHHHTHT

HTTHHTTT

THHHTHHT

THTHTHTT

TTHHTTHT

TTTHTTTT

Generalized Experiment: number of tosses: n probability of heads: p

P(X = k) = ?

P(X=2) ?P(X=3) ?

¾

¼

¾

¼

¾

¼(n C k) pk (1-p)n-k

6 x (¾)2 x (¼)2 = 27/128 = 0.21 4 x (¾)3 x (¼) = 108/256 = 0.42

¼ ¼

¾¾

Page 10: Discrete Random Variables 2

Binomial Random Variable (cont)

k

k

PMF:

CDF:

E(X) = np

Variance(X) = np(1-p)

Page 11: Discrete Random Variables 2

Geometric Random Variable

Experiment: number of tosses: 3 probability of heads: ¾ X = number of tosses until you get heads

H1

T1

P(H1) = 3/4

P(T1) = 1/4

H2

T2

P(H2|T1) = 3/4

P(T2 | T1) = 1/4

P(T1 H2) = 12/64P(H1) = 48/64

H3

T3

P(H2|T1 T2) = 3/4

P(T3 | T1 T2) = 1/4

P(T1 T2 H2) = 3/64P(X=3) = ?

Generalized Experiment: number of tosses: n probability of heads: pP(X = k) = ?

Page 12: Discrete Random Variables 2

Geometric Random Variable (cont)

PMF:

CDF:

E(X) =

Variance(X) =

Page 13: Discrete Random Variables 2
Page 14: Discrete Random Variables 2
Page 15: Discrete Random Variables 2

Independence of Random Variables

Page 16: Discrete Random Variables 2

Some thoughts

What does it mean for 2 experiments to be independent?

How do you derive the properties of binomial random variables from Bernoulli random variables?

Page 17: Discrete Random Variables 2

Other topics: - PMFs of more than one random variable - conditional PMF