Lecture 8
Joint Distributions
Joint Probability Distributions
• In many case, we are not only interested in one outcome/random
variable but in multiple ones.
• For instance, we may be interested in the Hardness 𝐻, and
Toughness 𝑇 of material released from a certain experiment.
• The sample space is two-dimensional, with elements being pairs
(ℎ, 𝑡).
• In the discrete case, the joint probability distribution is defined
as:
𝑓 ℎ, 𝑡 = 𝑃(𝐻 = ℎ, 𝑇 = 𝑡)
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Discrete Joint Distribution
• The function 𝑓(𝑥, 𝑦) is a joint probability distribution or
probability mass function of the discrete random variables 𝑋and
𝑌if
• 𝑓(𝑥, 𝑦) ≥ 0, for all (𝑥, 𝑦).
• σ𝑥 σ𝑦 𝑓(𝑥, 𝑦) = 1.
• 𝑃 𝑋 = 𝑥, 𝑌 = 𝑦 = 𝑓(𝑥, 𝑦).
• For a region 𝐴 in the 𝑥𝑦 plane, 𝑃 𝑥, 𝑦 ∈ 𝐴 = σ σ𝐴 𝑓(𝑥, 𝑦) .
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Example
•
•
•
•
•
A box contains 3 blue pens, 2 red pens, and 3 green pens.
2 pens are selected at random.
𝑋 is the number of blue pens selected.
𝑌 is the number of red pens selected.
Find:
1. The joint probability function 𝑓(𝑥, 𝑦).
2. 𝑃[(𝑥, 𝑦) ∈ 𝐴], where A is the region (𝑥, 𝑦) (𝑥 + 𝑦) ≤ 1 .
• The possible values for the pairs are:
(0,0), (0,1), (1,0), (1,1), (0,2), and (2,0)
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Solution
1. The number of possible ways of choosing any 2 pins is:
8
= 28
2
The number of ways of selecting 1 red from 2 red pens and 1
2 3
green from 3 green pens is
= 6. Hence, f(0,1) = 6/28=3 /14.
1 1
Similar calculations yield the probabilities for the other cases, which
are presented in Table. Note that the probabilities sum to 1.
The number of cases of choosing x blue and y red is:
3
𝑥
2
𝑦
3
2−𝑥−𝑦
Hence, we can write:
Note that: 𝑥 = 0,1,2; 𝑦 = 0,1,2; and 0 ≤ (𝑥 + 𝑦) ≤ 2.
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None
2. The probability that (𝑥, 𝑦) falls in region 𝐴 is:
𝑃 𝑥, 𝑦 ∈ 𝐴 = 𝑃 𝑥 + 𝑦 ≤ 1
= 𝑓 0,0 + 𝑓 0,1 + 𝑓(1,0)
3
3
9
9
=
+
+
=
28 14 28 14
• The probability distribution 𝑓(𝑥, 𝑦) can be tabulated:
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Continuous Joint Distribution
• The function 𝑓(𝑥, 𝑦) is a joint probability density function of the
continuous random variables 𝑋and 𝑌if
• 𝑓(𝑥, 𝑦) ≥ 0, for all (𝑥, 𝑦).
∞
∞
• ‫׬‬−∞ ‫׬‬−∞ 𝑓 𝑥, 𝑦 𝑑𝑥 𝑑𝑦 = 1.
• 𝑃 𝑥, 𝑦 ∈ 𝐴
= ‫𝑥 𝑓 𝐴׬ ׬‬, 𝑦 𝑑𝑥 𝑑𝑦.
• for any region 𝐴 in the 𝑥𝑦 plane.
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Example
• A privately owned business operates both a drive-in facility and a
walk-in facility.
• On a randomly selected day, let X and Y , respectively, be the
times that the drive-in and the walk-in facilities are in use, and
suppose that the joint density function is:
A. Verify the conditions of the joint density function.
1 1
1
2 4
2
B. Find 𝑃 𝑥, 𝑦 ∈ 𝐴 , where 𝐴 = (𝑥, 𝑦) 0 < 𝑥 < , < 𝑦 <
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Solution
A. It is easy to show that 𝑓(𝑥, 𝑦) ≥ 0 is satisfied.
The second condition is:
∞
න
∞
1
1
න 𝑓 𝑥, 𝑦 𝑑𝑥 𝑑𝑦 = න න
−∞ −∞
0
1
=න
0
0
2
(2𝑥 + 3𝑦) 𝑑𝑥 𝑑𝑦
5
2 6𝑦
+
𝑑𝑦
5 5
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The Mean/Expected Value
• Let 𝑋be a random variable with probability distribution 𝑓(𝑥). The
mean, or expected value, of 𝑋is:
𝜇𝑥 = 𝐸 𝑋 = ෍ 𝑥𝑓(𝑥)
𝑥
• if 𝑋 is discrete. And:
+∞
𝜇𝑥 = 𝐸 𝑋 = න
𝑥𝑓 𝑥 𝑑𝑥
−∞
• if 𝑋 is continuous.
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The Mean of a Random Variable
• If a fair coin is tossed twice, the sample space is:
𝑆 = {𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇}
• These probabilities are the relative frequencies in the long
run. Hence,
This result means that a person who tosses 2 coins over and
over again will, on the average, get 1 head per toss.
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Example
• Introduction
• Consider tossing two coins 16 times and X is the number of
heads that occur per toss
• The possible outcomes for every toss are: 0, 1, and 2.
• If they occur exactly: 4, 7, and 5 times, respectively.
• The mean (or mathematical expectation) is:
• Or:
• Hence, by knowing the outcomes and their relative
frequencies, we can calculate the average.
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Example
• A certain box contains 7 components; of which 3 are defective.
• A sample of 3 is chosen by the inspector.
• What is the expected value for the number of good items in the
sample (𝑋)?
Solution:
• The probability distribution for X is:
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• Hence:
𝑓(0) = 1/35, 𝑓(1) = 12/35, 𝑓(2) = 18/35, 𝑓(3) = 4/35
• The expected value in the long run is:
1
12
18
4
𝜇𝑥 = 𝐸 𝑋 = 0 ×
+1×
+2×
+3×
= 1.7
35
35
35
35
• Hence, the average over the long run is 1.7.
Thus, if a sample of size 3 is selected at random over and over again
from a lot of 4 good components and 3 defective components, it will
contain, on average, 1.7 good components.
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Example
• 𝑋 denotes the life in hours of a certain electronic device.
• The PDF is:
• Find the expected life of this type of device.
Solution:
• The expected life is:
+∞
𝐸 𝑋 =න
100
+∞ 20,000
20,000
−20,000
𝑥
𝑑𝑥 = න
𝑑𝑥 =
ቤ
3
2
𝑥
𝑥
𝑥
100
20000
100
Note:
+∞
=
100
= 200
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Functions of Random Variables
• Let 𝑋be a random variable with probability distribution 𝑓(𝑥). The
expected value of the random variable 𝑔(𝑋)is
𝜇𝑔 𝑋 = 𝐸 𝑔(𝑋) = ෍ 𝑔 𝑥 𝑓(𝑥)
𝑥
• if 𝑋 is discrete, and
+∞
𝜇𝑔 𝑋 = 𝐸 𝑔(𝑋) = න
𝑔 𝑥 𝑓 𝑥 𝑑𝑥
−∞
• if 𝑋 is continuous.
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Example
• Suppose that the number of cars 𝑋that pass through a car wash
between 4: 00𝑝𝑚and 5: 00 𝑝𝑚 on any sunny Friday has the
following probability distribution.
• Let 𝑔(𝑥) = (2𝑥 − 1) represent the amount of money in pounds
paid to the attendant by the manager.
• Find the attendant’s expected earnings during this 1 hour
period.
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Solution
• Let the probability distribution of 𝑋 be denoted
by 𝑓(𝑥).
• The attendant can expect:
= 12.67 Pounds
Example
• Let 𝑋 be a random variable with density function
• Find the expected value of 𝑔(𝑋) = (4𝑋 + 3).
• Solution:
=8
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