Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
Example:
Assume we are drawing cards from a 100 well-shuffled cards with
replacement. We keep drawing until we get a ♠. Let p = P({♠}),
i.e. there are 100 · p ♠’s. Assume the successive drawings are
independent and define X = the number of drawings.
Probability Distributions for Discrete RV
Example:
Assume we are drawing cards from a 100 well-shuffled cards with
replacement. We keep drawing until we get a ♠. Let p = P({♠}),
i.e. there are 100 · p ♠’s. Assume the successive drawings are
independent and define X = the number of drawings.
If we know that there are 20 ♠’s, i.e. p = 0.2, then what is the
probability for us to draw at most 3 times? More than 2 times?
Probability Distributions for Discrete RV
Example:
Assume we are drawing cards from a 100 well-shuffled cards with
replacement. We keep drawing until we get a ♠. Let p = P({♠}),
i.e. there are 100 · p ♠’s. Assume the successive drawings are
independent and define X = the number of drawings.
If we know that there are 20 ♠’s, i.e. p = 0.2, then what is the
probability 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
Probability Distributions for Discrete RV
Example:
Assume we are drawing cards from a 100 well-shuffled cards with
replacement. We keep drawing until we get a ♠. Let p = P({♠}),
i.e. there are 100 · p ♠’s. Assume the successive drawings are
independent and define X = the number of drawings.
If we know that there are 20 ♠’s, i.e. p = 0.2, then what is the
probability 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
Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
Definition
The cumulative distribution function (cdf) F (x) of a discrete rv
X with pmf p(x) is defined for every number x by
X
F (x) = P(X ≤ x) =
p(y )
y :y ≤x
For any number x, F(x) is the probability that the observed value
of X will be at most x.
Probability Distributions for Discrete RV
Definition
The cumulative distribution function (cdf) F (x) of a discrete rv
X with pmf p(x) is defined for every number x by
X
F (x) = P(X ≤ x) =
p(y )
y :y ≤x
For any number x, F(x) is the probability that the observed value
of 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)
Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
Example 3.10 (continued):


0





0.4
F (y ) = 0.7



0.9




1
if
if
if
if
if
y <1
1≤y <2
2≤y <3
3≤y <4
y ≥2
Probability Distributions for Discrete RV
Example 3.10 (continued):


0





0.4
F (y ) = 0.7



0.9




1
if
if
if
if
if
y <1
1≤y <2
2≤y <3
3≤y <4
y ≥2
Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
Example:
Assume we are drawing cards from a 100 well-shuffled cards with
replacement. We keep drawing until we get a ♠. Let α = P({♠}),
i.e. there are 100 · α ♠’s. Assume the successive drawings are
independent and define X = the number of drawings. The pmf
would be
(
(1 − α)x−1 · α x = 1, 2, 3, . . .
p(x) =
0
otherwise
Probability Distributions for Discrete RV
Example:
Assume we are drawing cards from a 100 well-shuffled cards with
replacement. We keep drawing until we get a ♠. Let α = P({♠}),
i.e. there are 100 · α ♠’s. Assume the successive drawings are
independent and define X = the number of drawings. The pmf
would be
(
(1 − α)x−1 · α x = 1, 2, 3, . . .
p(x) =
0
otherwise
Then for any positive interger x, we have
F (x) =
X
y ≤x
p(y ) =
x
x−1
X
X
(1 − α)(y −1) · α = α
(1 − α)y
y =1
(
1 − (1 − α)x
=
0
y =0
x ≥1
x <1
Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
Example:
Assume we are drawing cards from a 100 well-shuffled cards with
replacement. We keep drawing until we get a ♠. Let α = P({♠}),
i.e. there are 100 · α ♠’s. Assume the successive drawings are
independent and define X = the number of drawings. The pmf
would be
Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
pmf =⇒ cdf:
F (x) = P(X ≤ x) =
X
y :y ≤x
p(y )
Probability Distributions for Discrete RV
pmf =⇒ cdf:
F (x) = P(X ≤ x) =
X
y :y ≤x
It is also possible cdf =⇒ pmf:
p(y )
Probability Distributions for Discrete RV
pmf =⇒ cdf:
F (x) = P(X ≤ x) =
X
p(y )
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 strictly
less than x.
Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
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 strictly
less than a. In particular, if the only possible values are integers
and 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.
Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
Example (Problem 23):
A consumer organization that evaluates new automobiles customarily
reports the number of major defects in each car examined. Let X denote
the number of major defects in a randomly selected car of a certain type.
The cdf of X is as follows:


0
x <0




0.06 0 ≤ x < 1



0.19 1 ≤ x < 2



0.39 2 ≤ x < 3
F (x) =

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).