probability distributions for discrete rv

Probability Distributions for Discrete RV

Example:Assume we are drawing cards from a 100 well-shuffled cards withreplacement. We keep drawing until we get a ♠. Let p = P({♠}),i.e. there are 100 · p ♠’s. Assume the successive drawings areindependent and define X = the number of drawings.If we know that there are 20 ♠’s, i.e. p = 0.2, then what is theprobability for us to draw at most 3 times? More than 2 times?

P(X ≤ 3) = p(1)+p(2)+p(3) = 0.2+0.2·0.8+0.2·(0.8)2 = 0.488

P(X > 2) = p(3)+p(4)+p(5)+· · · = 1−p(1)−p(2) = 1−0.2−0.2·0.8 = 0.64

Example:Assume we are drawing cards from a 100 well-shuffled cards withreplacement. We keep drawing until we get a ♠. Let p = P({♠}),i.e. there are 100 · p ♠’s. Assume the successive drawings areindependent and define X = the number of drawings.

If we know that there are 20 ♠’s, i.e. p = 0.2, then what is theprobability for us to draw at most 3 times? More than 2 times?

P(X ≤ 3) = p(1)+p(2)+p(3) = 0.2+0.2·0.8+0.2·(0.8)2 = 0.488

P(X > 2) = p(3)+p(4)+p(5)+· · · = 1−p(1)−p(2) = 1−0.2−0.2·0.8 = 0.64

P(X ≤ 3) = p(1)+p(2)+p(3) = 0.2+0.2·0.8+0.2·(0.8)2 = 0.488

P(X > 2) = p(3)+p(4)+p(5)+· · · = 1−p(1)−p(2) = 1−0.2−0.2·0.8 = 0.64

P(X ≤ 3) = p(1)+p(2)+p(3) = 0.2+0.2·0.8+0.2·(0.8)2 = 0.488

P(X > 2) = p(3)+p(4)+p(5)+· · · = 1−p(1)−p(2) = 1−0.2−0.2·0.8 = 0.64

P(X ≤ 3) = p(1)+p(2)+p(3) = 0.2+0.2·0.8+0.2·(0.8)2 = 0.488

P(X > 2) = p(3)+p(4)+p(5)+· · · = 1−p(1)−p(2) = 1−0.2−0.2·0.8 = 0.64

DefinitionThe cumulative distribution function (cdf) F (x) of a discrete rvX with pmf p(x) is defined for every number x by

F (x) = P(X ≤ x) =∑

y :y≤x

For any number x , F(x) is the probability that the observed valueof X will be at most x .

F (x) = P(X ≤ x) = P(X is less than or equal to x)

p(x) = P(X = x) = P(X is exactly equal to x)

F (x) = P(X ≤ x) =∑

y :y≤x

F (x) = P(X ≤ x) =∑

y :y≤x

Example 3.10 (continued):

F (y) =

0 if y < 1

0.4 if 1 ≤ y < 2

0.7 if 2 ≤ y < 3

0.9 if 3 ≤ y < 4

1 if y ≥ 2

F (y) =

0 if y < 1

0.4 if 1 ≤ y < 2

0.7 if 2 ≤ y < 3

0.9 if 3 ≤ y < 4

1 if y ≥ 2

F (y) =

0 if y < 1

0.4 if 1 ≤ y < 2

0.7 if 2 ≤ y < 3

0.9 if 3 ≤ y < 4

1 if y ≥ 2

Example:Assume we are drawing cards from a 100 well-shuffled cards withreplacement. We keep drawing until we get a ♠. Let α = P({♠}),i.e. there are 100 · α ♠’s. Assume the successive drawings areindependent and define X = the number of drawings. The pmfwould be

p(x) =

{(1− α)x−1 · α x = 1, 2, 3, . . .

0 otherwise

Then for any positive interger x, we have

F (x) =∑y≤x

p(y) =x∑

(1− α)(y−1) · α = αx−1∑y=0

(1− α)y

{1− (1− α)x x ≥ 1

0 x < 1

Probability Distributions for Discrete RVExample:Assume we are drawing cards from a 100 well-shuffled cards withreplacement. We keep drawing until we get a ♠. Let α = P({♠}),i.e. there are 100 · α ♠’s. Assume the successive drawings areindependent and define X = the number of drawings. The pmfwould be

p(x) =

{(1− α)x−1 · α x = 1, 2, 3, . . .

0 otherwise

F (x) =∑y≤x

p(y) =x∑

(1− α)(y−1) · α = αx−1∑y=0

(1− α)y

{1− (1− α)x x ≥ 1

0 x < 1

Probability Distributions for Discrete RVExample:Assume we are drawing cards from a 100 well-shuffled cards withreplacement. We keep drawing until we get a ♠. Let α = P({♠}),i.e. there are 100 · α ♠’s. Assume the successive drawings areindependent and define X = the number of drawings. The pmfwould be

p(x) =

{(1− α)x−1 · α x = 1, 2, 3, . . .

0 otherwise

F (x) =∑y≤x

p(y) =x∑

(1− α)(y−1) · α = αx−1∑y=0

(1− α)y

{1− (1− α)x x ≥ 1

0 x < 1

pmf =⇒ cdf:

F (x) = P(X ≤ x) =∑

y :y≤x

It is also possible cdf =⇒ pmf:

p(x) = F (x)− F (x−)

where “x−” represents the largest possible X value that is strictlyless than x .

pmf =⇒ cdf:

F (x) = P(X ≤ x) =∑

y :y≤x

p(x) = F (x)− F (x−)

pmf =⇒ cdf:

F (x) = P(X ≤ x) =∑

y :y≤x

p(x) = F (x)− F (x−)

pmf =⇒ cdf:

F (x) = P(X ≤ x) =∑

y :y≤x

p(x) = F (x)− F (x−)

Proposition

For any two numbers a and b with a ≤ b,

P(a ≤ X ≤ b) = F (b)− F (a−)

where “a−” represents the largest possible X value that is strictlyless than a. In particular, if the only possible values are integersand if a and b are integers, then

P(a ≤ X ≤ b) = P(X = a or a + 1 or . . . or b)

= F (b)− F (a− 1)

Taking a = b yields P(X = a) = F (a)− F (a− 1) in this case.

Proposition

For any two numbers a and b with a ≤ b,

P(a ≤ X ≤ b) = F (b)− F (a−)

where “a−” represents the largest possible X value that is strictlyless than a. In particular, if the only possible values are integersand if a and b are integers, then

P(a ≤ X ≤ b) = P(X = a or a + 1 or . . . or b)

= F (b)− F (a− 1)

Taking a = b yields P(X = a) = F (a)− F (a− 1) in this case.

Example (Problem 23):A consumer organization that evaluates new automobiles customarilyreports the number of major defects in each car examined. Let X denotethe number of major defects in a randomly selected car of a certain type.The cdf of X is as follows:

F (x) =

0 x < 0

0.06 0 ≤ x < 1

0.19 1 ≤ x < 2

0.39 2 ≤ x < 3

0.67 3 ≤ x < 4

0.92 4 ≤ x < 5

0.97 5 ≤ x < 6

1 x ≤ 6

Calculate the following probabilities directly from the cdf: (a)p(2),

(b)P(X > 3) and (c)P(2 ≤ X < 5).

Example (Problem 23):A consumer organization that evaluates new automobiles customarilyreports the number of major defects in each car examined. Let X denotethe number of major defects in a randomly selected car of a certain type.The cdf of X is as follows:

F (x) =

0 x < 0

0.06 0 ≤ x < 1

0.19 1 ≤ x < 2

0.39 2 ≤ x < 3

0.67 3 ≤ x < 4

0.92 4 ≤ x < 5

0.97 5 ≤ x < 6

1 x ≤ 6

Calculate the following probabilities directly from the cdf: (a)p(2),

(b)P(X > 3) and (c)P(2 ≤ X < 5).

probability distributions for discrete rv

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