1.
Set
a.
b.
2.
3.
4.
Definitions
a. Random Experiment: An experiment that can result in different outcomes, even though it is performed under the
same conditions and in the same manner.
b. Sample Space: This is the set of all possible outcomes.
i. Discrete: Consists of finite or countable infinite set of outcomes
ii. Continuous: Contains an interval (either finite or infinite) of real numbers
c. Event: It is a subset of the sample space of a random experiment.
i. Mutually Exclusive: 𝐸1 ∩ 𝐸2 = ∅
Probability
a. 𝑃(𝐴′) = 1 − 𝑃(𝐴)
b. 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
c. 𝑃(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑃(𝐴) + 𝑃(𝐵) + 𝑃(𝐶) − 𝑃(𝐴 ∩ 𝐵) − 𝑃(𝐴 ∩ 𝐶) − 𝑃(𝐵 ∩ 𝐶) + 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶)
d. 𝑃(𝐵) = 𝑃(𝐵 ∩ 𝐴) + 𝑃(𝐵 ∩ 𝐴′ )
Conditional Probability: 𝑃(𝐴|𝐵) → 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝐴 𝑔𝑖𝑣𝑒𝑛 𝐵.
a.
5.
6.
∪ → 𝑈𝑛𝑖𝑜𝑢𝑛: 𝐼𝑡 𝑖𝑠 𝑙𝑖𝑘𝑒𝑛𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 "𝑂𝑅"
∩ → 𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛: 𝐼𝑡 𝑖𝑠 𝑙𝑖𝑘𝑒𝑛𝑒𝑑 𝑡𝑜 "𝐴𝑁𝐷"
𝑃(𝐵 ∩ 𝐴) = 𝑃(𝐴|𝐵)𝑃(𝐵) ⋯ ∴ ⋯ 𝑃(𝐴|𝐵) =
𝑃(𝐵∩𝐴)
𝑃(𝐵)
b. 𝑃(𝐵) = 𝑃(𝐵|𝐴)𝑃(𝐴) + 𝑃(𝐵|𝐴′)𝑃(𝐴′) → 𝐓𝐨𝐭𝐚𝐥 𝐏𝐫𝐨𝐛𝐚𝐛𝐢𝐥𝐢𝐭𝐲 𝐫𝐮𝐥𝐞: 𝐀 & 𝐀′ 𝐚𝐫𝐞 𝐦𝐮𝐭𝐮𝐚𝐥𝐥𝐲 𝐞𝐱𝐜𝐥𝐮𝐬𝐢𝐯𝐞
Independence: Two events (A & B) are independent if one of the conditions listed below is satisfied;
a. 𝑃(𝐴|𝐵) = 𝑃(𝐴)
b. 𝑃(𝐵|𝐴) = 𝑃(𝐵)
c. 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵)
CIRCUIT PROBLEM:𝐹𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑖𝑡𝑠 𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟.
𝑃(𝐷) = 𝑃(𝐴)𝑃(𝐵) = 0.992 = 0.98
A
B
𝑇ℎ𝑖𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑠 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑐𝑒 “𝑎𝑛𝑑”
0.99
0.99
C
0.8
Fig 2:- Shows the probability of Success
𝐹𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑖𝑡𝑠 𝑖𝑛 𝑃𝑎𝑟𝑎𝑙𝑙𝑒𝑙, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟.
𝑡ℎ𝑒𝑛 𝑓𝑖𝑛𝑑 𝑠𝑢𝑐𝑐𝑒𝑠𝑠 𝑏𝑦 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑛𝑠𝑤𝑒𝑟 𝑓𝑟𝑜𝑚 1.
𝑃(𝐸) = 1 − (1 − 𝑃(𝐶))(1 − 𝑃(𝐷)) = 1 − (0.2)(0.02) = 0.996
𝑇ℎ𝑖𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑠 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑐𝑒 “𝑜𝑟”
𝑃(𝐸) = 𝑃(𝐶) + 𝑃(𝐷) − 𝑃(𝐶)𝑃(𝐷)
7.
a.
b.
c.
d.
8.
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐵 ∩ 𝐴) = 𝑃(𝐵|𝐴)𝑃(𝐴) = 𝑃(𝐴|𝐵)𝑃(𝐵)
𝑃(𝐵|𝐴)𝑃(𝐴) = 𝑃(𝐴|𝐵)𝑃(𝐵)
𝑃(𝐴|𝐵 )𝑃(𝐵)
𝑃(𝐵|𝐴) =
𝑆𝐼𝑁𝐶𝐸 𝑃(𝐴) = 𝑃(𝐴|𝐵)𝑃(𝐵) + 𝑃(𝐴|𝐵′ )𝑃(𝐵′ )
𝑃(𝐴)
𝑃(𝐴|𝐵 )𝑃(𝐵)
𝑃(𝐵|𝐴) =
𝐁 & 𝐁 ′ 𝐚𝐫𝐞 𝐦𝐮𝐭𝐮𝐚𝐥𝐥𝐲 𝐞𝐱𝐜𝐥𝐮𝐬𝐢𝐯𝐞
′
𝑃(𝐴|𝐵 )𝑃(𝐵)+𝑃(𝐴|𝐵 )𝑃(𝐵′ )
Mean (Expected Value):
𝜇 = 𝐸(𝑋) = ∑ 𝑥𝑓(𝑥)
𝑥
9.
Variance:
𝜎 2 = 𝑉(𝑋) = 𝐸(𝑋 − 𝜇)2 = ∑(𝑥 − 𝜇)2 𝑓(𝑥) = ∑ 𝑥 2 𝑓(𝑥) − 𝜇 2
𝑥
𝑥
10. Standard Deviation: 𝜎 = √𝜎 2
11. Discrete Uniform Distribution (DUD): This is when every member of the sample space has the same probability.
𝐼𝑓 𝑥1 , 𝑥2 , … , 𝑥𝑛
𝐍𝐨𝐭𝐞: if the values are not whole numbers, multiply by 𝟏𝟎𝒎 to make it whole.
1
𝑓(𝑥𝑖 ) =
𝑛
a. For DUD, the
𝑥𝑛 + 𝑥1
𝒎𝒆𝒂𝒏(𝝁) =
∗ 10−𝑚
2
b. For DUD, the
(𝑥𝑛 − 𝑥1 + 1)2 − 1
𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆(𝝈𝟐 ) =
∗ 10−2𝑚
12
12. Bernoulli Trials:- A random experiment consists of Bernoulli trials if;
The trials are independent.
Each trial results in only two possible outcomes (Success and Failure).
The probability of success in each trial remains constant
a. Binomial Random Variable:- 𝑅𝑒𝑓𝑒𝑟𝑠 𝑡𝑜 𝑥 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠 𝑖𝑛 𝑛 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠
A particularly long traffic light on your morning commute is
𝑓(𝑥) = 𝑃(𝑋 = 𝑥)
𝑥 → 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠
𝑛 𝑥
𝑛−𝑥
green 20% of the time that you approach it. Assume that each
𝑛
→
𝑁𝑢𝑚𝑏𝑒𝑟
𝑜𝑓
𝑡𝑟𝑖𝑎𝑙𝑠
= ( ) 𝑝 (1 − 𝑝)
𝑥
morning represents an independent trial.
𝑝 → 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠
a) Over five mornings, what is the probability that the
𝑀𝑒𝑎𝑛(𝜇) = 𝐸(𝑋) = 𝑛𝑝
light is green on exactly one day?
𝑁𝑜𝑡𝑒: 𝑆𝑢𝑚 𝑎𝑙𝑙 𝑖𝑓 𝑤𝑖𝑡ℎ𝑖𝑛 𝑎 𝑟𝑎𝑛𝑔𝑒
𝑥 = 1 → 𝑂𝑛𝑒 𝑑𝑎𝑦
𝑛 = 5 → 𝑓𝑖𝑣𝑒 𝑚𝑜𝑟𝑛𝑖𝑛𝑔𝑠
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒(𝜎 2 )
𝑷𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝑺𝒖𝒄𝒄𝒆𝒔𝒔
𝑝 = 0.2 → 20%
= 𝑉(𝑋) = 𝑛𝑝(1 − 𝑝)
Page 1 of 6
b. Geometric Random Variable:- 𝑅𝑒𝑓𝑒𝑟𝑠 𝑡𝑜 𝑥 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠 𝑢𝑛𝑡𝑖𝑙 𝑡ℎ𝑒 𝒇𝒊𝒓𝒔𝒕 𝑠𝑢𝑐𝑐𝑒𝑠𝑠
The probability of a successful optical alignment in the
𝑓(𝑥) = 𝑃(𝑋 = 𝑥)
𝑥 → 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠
assembly of an optical data storage product is 0.8. Assume the
= (1 − 𝑝) 𝑥−1 ∗ 𝑝
𝑢𝑛𝑡𝑖𝑙 𝑓𝑖𝑟𝑠𝑡 𝑠𝑢𝑐𝑐𝑒𝑠𝑠
1
trials are independent.
𝑝 → 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠
𝑀𝑒𝑎𝑛(𝜇) = 𝐸(𝑋) =
𝑝
a) What is the probability that the first successful
alignment requires exactly four trials?
𝑁𝑜𝑡𝑒: 𝑆𝑢𝑚 𝑎𝑙𝑙 𝑖𝑓 𝑤𝑖𝑡ℎ𝑖𝑛 𝑎 𝑟𝑎𝑛𝑔𝑒
𝑥 = 4 → 𝑓𝑜𝑢𝑟 𝑡𝑟𝑖𝑎𝑙𝑠
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒(𝜎 2 ) = 𝑉(𝑋)
1−𝑝
𝑝 = 0.8
= 2
𝑝
c. Negative Binomial:- 𝑅𝑒𝑓𝑒𝑟𝑠 𝑡𝑜 𝑥 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠 𝑢𝑛𝑡𝑖𝑙 𝑟 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑖𝑙𝑢𝑟𝑒
𝑓(𝑥) = 𝑃(𝑋 = 𝑥)
𝑟 → 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 ≥ 1
𝑥−1 𝑟
𝑥 → 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠 ≥ 1
𝑥−𝑟
=(
) 𝑝 (1 − 𝑝)
𝑟−1
𝑝 → 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑓𝑎𝑖𝑙𝑢𝑟𝑒
𝑟
𝑀𝑒𝑎𝑛(𝜇) = 𝐸(𝑋) =
𝑝
𝑁𝑜𝑡𝑒: 𝑆𝑢𝑚 𝑎𝑙𝑙 𝑖𝑓 𝑤𝑖𝑡ℎ𝑖𝑛 𝑎 𝑟𝑎𝑛𝑔𝑒
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒(𝜎 2 ) = 𝑉(𝑋)
(1−𝑝)𝑟
𝑷𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒇𝒂𝒊𝒍𝒖𝒓𝒆
= 2
𝑝
13. Hypergeometric: - 𝑅𝑒𝑓𝑒𝑟𝑠 𝑡𝑜 𝑎 𝑠𝑎𝑚𝑝𝑙𝑒 𝑜𝑓 𝑛 − 𝑜𝑏𝑗 𝑡𝑎𝑘𝑖𝑛𝑔 𝑜𝑢𝑡 𝑜𝑓 𝑎 𝑠𝑎𝑚𝑝𝑙𝑒 𝑜𝑓 𝑁 − 𝑜𝑏𝑗
(𝒘𝒊𝒕𝒉𝒐𝒖𝒕 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕) 𝑤𝑖𝑡ℎ 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑖𝑛 𝑥 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡𝑠 𝑓𝑟𝑜𝑚 𝑛.
A state runs a lottery in which six numbers are randomly
𝑓(𝑥) = 𝑃(𝑋 = 𝑥)
𝑁 → 𝑇𝑜𝑡𝑎𝑙 # 𝑜𝑓 𝑜𝑏𝑗.
𝑁−𝐾 𝐾
selected from 40, without replacement. A player chooses six
𝑛
→
𝑇𝑜𝑡𝑎𝑙
#
𝑡𝑎𝑘𝑖𝑛𝑔
𝑜𝑢𝑡
𝑜𝑓
𝑁
(
)( )
𝑛
−
𝑥
𝑥
numbers before the state’s sample is selected.
𝐾
→
𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒
#
𝑜𝑓
𝑠𝑢𝑐𝑐𝑒𝑠𝑠
𝑖𝑛
𝑁
=
𝑁
a) What is the probability that the six numbers chosen
𝑥
→
#
𝑜𝑓
𝑛
𝑜𝑏𝑗
𝑤𝑒
𝑤𝑎𝑛𝑡
(𝐾
𝑜𝑏𝑗
𝑡𝑦𝑝𝑒)
( )
𝐾
𝑛
by a player match all six numbers in the state’s
𝑝 = → 𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑢𝑐. 𝑖𝑛 𝑁
𝑁
sample?
𝑁−𝑛
𝑀𝑒𝑎𝑛(𝜇) = 𝐸(𝑋) = 𝑛𝑝
(
) → 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟
2)
6
𝑁−1
𝑁 = 40
𝐾 = 6 → 𝑠𝑎𝑚𝑝𝑙𝑒
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒(𝜎 = 𝑉(𝑋)
𝑝 = = 0.15
40
𝑁−𝑛
𝑛 = 6 → 𝑐ℎ𝑜𝑠𝑒𝑛
𝑥 = 6 → 𝑚𝑎𝑡𝑐ℎ𝑒𝑠
= 𝑛𝑝(1 − 𝑝) (
)
𝑁−1
14. Poisson Distribution:- This refers to any random experiment with the properties below;
Counting the number of occurrence of an event over a period of time/space.
The # of occurrence of an event in an interval is proportional to the length of the interval.
Events cannot occur simultaneously.
Occurrences of the events are independent for non-overlapping intervals.
𝑓(𝑥) = 𝑃(𝑋 = 𝑥)
𝑒 −𝜆𝑡 (𝜆𝑡) 𝑥
=
𝑥!
𝜆 → 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑜𝑐𝑐𝑢𝑟𝑟𝑒𝑛𝑐𝑒
𝑡 → 𝑇ℎ𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑡𝑖𝑚𝑒 𝑢𝑛𝑖𝑡
𝑥 → # 𝑜𝑓 𝑒𝑣𝑒𝑛𝑡𝑠
𝝀 = 𝒏𝒑
𝑀𝑒𝑎𝑛(𝜇) = 𝐸(𝑋) = 𝜆
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒(𝜎 2 ) = 𝑉(𝑋) = 𝜆
15. Continuous Random Variable (C.R.V)
𝑥
𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥) = ∫ 𝑓(𝑥)𝑑𝑥
−∞
𝑥
𝑀𝑒𝑎𝑛(𝜇) = ∫ 𝑥𝑓(𝑥)𝑑𝑥
−∞
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒(𝜎
2)
𝑥
= ∫ 𝑥 2 𝑓(𝑥)𝑑𝑥 − 𝜇 2
−∞
16. Continuous Uniform Distribution:
𝐼𝑓 𝑥1 , 𝑥2 , … , 𝑥𝑛
1
𝑓(𝑥𝑖 ) =
𝑥𝑛 − 𝑥1
𝐍𝐨𝐭𝐞: if the values are not whole numbers, multiply by 𝟏𝟎𝒎 to make it whole.
a. For CUD, the
b.
𝑚𝑒𝑎𝑛(𝜇) = (
For CUD, the
𝑥𝑛 +𝑥1
2
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒(𝜎 2 ) = (
) ∗ 10−𝑚
(𝑥𝑛 −𝑥1 )2
12
) ∗ 10−2𝑚
17. Normal Distribution:
(𝑥−𝜇)2
1
−
𝑓(𝑥) =
𝑒 2𝜎2
√2𝜋𝜎
𝝁 𝒊𝒔 𝒕𝒉𝒆 𝒎𝒆𝒂𝒏 𝒘𝒉𝒊𝒍𝒆 𝝈𝟐 𝒊𝒔 𝒕𝒉𝒆 𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆.
a.
Note: you can only read from the table when 𝝁 = 𝟎 𝒂𝒏𝒅 𝝈 = 𝟏. standardized…
𝐼𝑓 𝑿 𝑖𝑠 𝑎 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 (𝑛𝑜𝑡 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑)𝑤𝑖𝑡ℎ 𝑬(𝒙) = 𝝁 𝑎𝑛𝑑 𝑽(𝒙) = 𝝈𝟐 , 𝑡ℎ𝑒 𝑅𝑉
𝒙−𝝁
𝒁=
𝑖𝑠 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑛𝑜𝑟𝑚𝑎𝑙 𝑅𝑉 𝑤𝑖𝑡ℎ 𝑬(𝒛) = 𝟎, 𝑽(𝒛) = 𝟏.
𝝈
18. Normal Approximation to Binomial Distribution: Use iff both 𝒏𝒑 > 𝟓 𝒂𝒏𝒅 𝒏𝒑(𝟏 − 𝒑) > 𝟓
𝒙−𝝁
𝒙−𝒏𝒑
a. 𝐼𝑓 𝑋 𝑖𝑠 𝑎 𝑏𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑑𝑖𝑠𝑡. 𝒁 =
=
𝝈
b.
c.
√𝒏𝒑(𝟏−𝒑)
𝑇𝑜 𝑢𝑠𝑒 𝑡ℎ𝑖𝑠, 𝑦𝑜𝑢 𝑚𝑢𝑠𝑡 𝑢𝑠𝑒 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 0.5. (𝑖. 𝑒. 𝑓𝑜𝑟 ≤ 𝒙, 𝑢𝑠𝑒 + 𝟎. 𝟓 𝑎𝑛𝑑 𝑓𝑜𝑟 ≥ 𝒙 𝑢𝑠𝑒 − 𝟎. 𝟓
𝒙+𝟎.𝟓−𝒏𝒑
𝑓𝑜𝑟 (𝑋 ≤ 𝑥) 𝑢𝑠𝑒 𝑃(𝑋 ≤ (𝑥 + 0.5)) … ∴ 𝑷(𝒛 ≤ 𝒙) =
√𝒏𝒑(𝟏−𝒑)
Page 2 of 6
19. Normal Approximation to Poison RV: Use iff both 𝝀 > 𝟓
𝒙±𝟎.𝟓−𝝁
𝒙±𝟎.𝟓−𝝀
a. 𝒁 =
=
𝑠𝑒𝑒 𝑟𝑒𝑎𝑠𝑜𝑛 𝑓𝑜𝑟 ± 0.5 𝑖𝑛 𝑏𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑑𝑖𝑠𝑡.
𝝈
√𝝀
20. Exponential Distribution: Amount of wait time until first count is obtained use integral.
𝑖𝑓 𝒇(𝒙) = 𝝀𝒆−𝝀𝒙 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ ∞
𝑎
𝑭(𝒂) = 𝑓(𝑥 ≤ 𝑎) = ∫0 𝜆𝑒 −𝜆𝑥 𝑑𝑥 = 𝟏 − 𝒆−𝝀𝒂 𝒄𝒐𝒖𝒍𝒅 𝒃𝒆 𝒑𝒐𝒊𝒔𝒐𝒏 𝒄𝒉𝒆𝒄𝒌.
∞
𝑚𝑒𝑎𝑛(𝝁) =
𝑭(𝒂) = 𝑓(𝑥 ≥ 𝑎) = ∫ 𝜆𝑒 −𝜆𝑥 𝑑𝑥 = 𝒆−𝝀𝒂
𝟏
1
= 𝝈 𝑎𝑛𝑑 𝜎 2 = 2
𝝀
𝜆
𝑚𝑒𝑎𝑛 𝑢𝑛𝑡𝑖𝑙 𝑏 𝑢𝑛𝑖𝑡𝑠 =
𝑓(𝑥 < 𝑎) = ∑𝑎−1
0
𝑎
Lack of Memory Property:
𝑊ℎ𝑒𝑟𝑒;
′𝑎′ → 𝑡𝑖𝑚𝑒 𝑎𝑙𝑟𝑒𝑎𝑑𝑦 𝑝𝑎𝑠𝑡
𝑃(𝑋 < (𝑎 + 𝑏)|𝑋 > 𝑎)
′𝑏 ′ → 𝑒𝑥𝑡𝑟𝑎 𝑡𝑖𝑚𝑒
= 𝑃(𝑋 < 𝑏)
−𝝀𝒃
𝜆 → 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑜𝑐𝑐𝑢𝑟𝑟𝑒𝑛𝑐𝑒
=𝟏−𝒆
𝑏
𝜆
𝑒−𝜆𝑡 (𝜆𝑡)
𝑥!
𝑥
a.
What is the probability that EQ is detected before 2 years
if it doesn’t occur the first year? EQ occurs at a rate of 1.5
yearly.
𝑎 = 1, 𝑏 = 2 − 1 = 1, 𝜆 = 1.5
21. 𝑮𝒂𝒎𝒎𝒂 𝑫𝒊𝒔𝒕. : 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝒘𝒂𝒊𝒕 𝒕𝒊𝒎𝒆 𝒖𝒏𝒕𝒊𝒍 ‘𝒓’ 𝒄𝒐𝒖𝒏𝒕𝒔 𝑖𝑠 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 (𝑟 > 1).
3
1
1
a. 𝐹𝑜𝑟 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠; 𝚪(𝒓) = (𝒓 − 𝟏)𝚪(𝒓−𝟏) … → Γ ( ) = ( ) Γ ( )
2
2
2
b. 𝑓𝑜𝑟 𝐼𝑛𝑡𝑒𝑔𝑒𝑟𝑠; 𝚪(𝒓) = (𝒓 − 𝟏)! … → Γ(5) = 4! = 24
𝟏
𝟏
𝑵𝒐𝒕𝒆: 𝚪(𝟎) = 𝟏 𝒂𝒏𝒅 𝚪 ( ) = √𝝅 … → 𝒕𝒉𝒆 𝒐𝒏𝒍𝒚 𝒂𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏 𝒊𝒅𝒆𝒏𝒕𝒊𝒕𝒚 𝒊𝒔 .
𝟐
𝟐
𝑢𝑛𝑖𝑡𝑠
a) What is the mean time until a packet is
𝜆𝑟 𝑥 𝑟−1 𝑒 −𝜆𝑥
𝜆
→
𝑟𝑎𝑡𝑒
𝑜𝑓
𝑜𝑐𝑐𝑢𝑟𝑟𝑒𝑛𝑐𝑒
(
)
𝑴𝒂𝒔𝒔 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏; 𝑓(𝑥) =
formed, that is, until five messages have
𝑡𝑖𝑚𝑒
Γ(𝑟)
𝑥 → 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝜆.
arrived at the node?
𝑪𝒖𝒎𝒎𝒖𝒍𝒂𝒕𝒊𝒗𝒆;
b) What is the probability that a packet is
𝑘 → # 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 𝑎𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝜆
𝑘−1 −𝜆𝑥
formed in less than 10 seconds?
𝑟 → # 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 𝑢𝑠𝑒𝑑 𝑖𝑛 𝜆 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛
𝑒 (𝜆𝑥)𝑘
𝐹(𝑘) = 𝑃(𝑋 > 𝑘) = ∑
4-116. In a data communication system, several
messages that arrive at a node are bundled into
a packet before they are transmitted over the
network. Assume the messages arrive at the
node according to a Poisson process with 𝝉 =
𝟑𝟎 messages per minute. Five messages are
used to form a packet.
𝑘!
𝑘=0
𝑟
𝜆
𝑟
𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆 (𝝈𝟐 ) = 2
𝜆
𝑴𝒆𝒂𝒏(𝝁) =
𝜆 = 30
𝑚𝑒𝑠𝑠𝑎𝑔𝑒𝑠
𝑚𝑖𝑛𝑢𝑡𝑒𝑠
10 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝑥=
60 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
𝑘 = (≥ 5)
𝑟=5
→ 𝑓𝑟𝑜𝑚 𝑝𝑎𝑟𝑡 𝐵
22. Weibull Dist. : 𝑇𝑖𝑚𝑒 𝑢𝑛𝑡𝑖𝑙 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 𝑜𝑓 𝑚𝑎𝑛𝑦 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑝ℎ𝑦𝑠𝑖𝑐𝑎𝑙 𝑠𝑦𝑠𝑡𝑒𝑚𝑠. 𝐿𝑜𝑜𝑘 𝑓𝑜𝑟 𝛃 𝑎𝑛𝑑/𝑜𝑟 𝜹.
𝑴𝒂𝒔𝒔 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏;
(𝛽−1)
𝑥 𝛽
𝛽 𝑥
𝑓(𝑥) = [( ) ( )
] [𝑒 −(𝛿) ]
𝛿 𝛿
𝑪𝒖𝒎𝒎𝒖𝒍𝒂𝒕𝒊𝒗𝒆;
𝐹(𝑥) = 𝑃(𝑋 < 𝑥) = 1 − 𝑒
𝑃(𝑋 > 𝑥) =
𝑥 𝛽
−( )
𝛿
𝑥 𝛽
−( )
𝑒 𝛿
𝑴𝒆𝒂𝒏(𝝁) = 𝛿Γ
1
(1+ )
𝛽
𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆 (𝝈𝟐 )
2
= 𝛿 2Γ
2
(1+ )
𝛽
− 𝛿 2 [Γ
1 ]
(1+ )
𝛽
𝛿 → 𝐴𝑙𝑤𝑎𝑦𝑠 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛
𝛽 → 𝐴𝑙𝑤𝑎𝑦𝑠 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛
𝑥 → 𝑇ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑎𝑠𝑘𝑒𝑑 𝑓𝑜𝑟
𝚪 → 𝑻𝒓𝒆𝒂𝒕 𝒂𝒔 𝒈𝒂𝒎𝒎𝒂 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏.
4-135. Suppose the lifetime of a component (in
hours) is modeled with a Weibull distribution with
𝜷 = 𝟐 and𝜹 = 𝟒𝟎𝟎𝟎. Determine the following:
(a) 𝑃(𝑋 > 3000)
(b) 𝑃(𝑋 > 6000 | 𝑋 > 3000)
(c) value for x such that 𝑃(𝑋 > 𝑥) = 0.9
−(
3000 2
−(
)
4000
= 0.5698
𝑃(𝐴 ∩ 𝐵)
𝐹𝑜𝑟 𝑝𝑎𝑟𝑡 𝒃); 𝑈𝑠𝑖𝑛𝑔 𝑃(𝐴|𝐵) =
𝑃(𝐵)
𝑃((𝑋 > 6000) ∩ (𝑋 > 3000)) = 𝑃(𝑋 > 6000)
𝑃((𝑋 > 6000) | (𝑋 > 3000)) =
=
6000 2
)
4000
3000 2
)
−(
𝑒 4000
𝑒
−(
=
0.1054
0.5698
𝑃(𝑋>6000)
𝑃(𝑋>3000)
= 0.1850
𝐹𝑜𝑟 𝑃𝑎𝑟𝑡 𝒄); 𝑃(𝑋 > 𝑥) = 𝑒
𝛽
−(𝑥)
𝛿
𝛽
𝑥 = (− ln(𝑃(𝑋 > 𝑥))) (𝛿)
𝑥 = (− ln(0.9))2 (4000)
= (0.0111)(4000) = 44.4 ℎ𝑜𝑢𝑟𝑠
𝐹𝑜𝑟 𝑝𝑎𝑟𝑡 𝒂); 𝑥 = 3000
𝑃(𝑋 > 3000) = 1 − (1 − 𝑒
𝑃(𝑋 > 3000) = 𝑒
3000 2
)
4000 )
23. Lognormal Dist. : 𝐿𝑜𝑜𝑘 𝑓𝑜𝑟 θ 𝑎𝑛𝑑/𝑜𝑟 𝝎.
𝑴𝒂𝒔𝒔 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏;
(ln(𝑥)−𝜃)2
1
−(
)
2𝜔2
𝑓(𝑥) = [
] [𝑒
]
𝑥𝜔√2𝜋
𝑪𝒖𝒎𝒎𝒖𝒍𝒂𝒕𝒊𝒗𝒆;
ln(𝑥) − 𝜃
𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥) = Φ [
]
𝜔
𝜔2
(𝜃+ )
2
𝑒
𝑴𝒆𝒂𝒏(𝝁) =
2
2
𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆 (𝝈𝟐 ) = 𝑒 (2𝜃+𝜔 ) (𝑒 𝜔 − 1)
𝑥 → 𝑇ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑎𝑠𝑘𝑒𝑑 𝑓𝑜𝑟
𝜔 → 𝐴𝑙𝑤𝑎𝑦𝑠 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛
𝜃 → 𝐴𝑙𝑤𝑎𝑦𝑠 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛
𝚽 → 𝒁 𝒔𝒄𝒐𝒓𝒆 (𝒄𝒉𝒆𝒄𝒌 𝑵𝒐𝒓𝒎𝒂𝒍 𝒕𝒂𝒃𝒍𝒆)
𝒂); 𝑥 = 13,300
ln(13300)−5
𝑃(𝑋 < 13,300) = Φ [
]
3
Φ[1.50] = 0.9332
𝒃);
4-138. Suppose that X has a lognormal
distribution with parameters 𝜃 = 5 and 𝜔2 =
9. Determine the following:
(a) 𝑃(𝑋 < 13,300)
(b) The value for 𝑥 such that P(X ≤ x) = 0.95
(c) The mean and variance of 𝑋
Φ [ln 𝑥𝜔−𝜃] = 0.95
( )
Φ[1.64] = 0.95 …
ln(𝑥)−5
3
= 1.64
𝑥 = 𝑒 3(1.65)+5 = 20,952
9
2
(5+ )
𝑎𝑛𝑑 𝝈𝟐 = 𝑒 (2(5)+9) (𝑒 9 − 1)
𝝁 = 13,360 𝑎𝑛𝑑 𝝈𝟐 = 1.45 × 1012
𝒄); 𝝁 = 𝑒
24. Beta Dist. : 𝐿𝑜𝑜𝑘 𝑓𝑜𝑟 𝛂 𝑎𝑛𝑑 𝜷.
𝑴𝒂𝒔𝒔 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏;
Γ(𝛼+𝛽) 𝛼−1
𝑓(𝑥) =
𝑥
(1 − 𝑥)𝛽−1
Γ(𝛼) Γ(𝛽)
𝑪𝒖𝒎𝒎𝒖𝒍𝒂𝒕𝒊𝒗𝒆;
𝑥
Γ(𝛼+𝛽) 𝛼−1
𝑃(𝑋 < 𝑥) = ∫ (
𝑥
(1 − 𝑥)𝛽−1 ) 𝑑𝑥
Γ(𝛼) Γ(𝛽)
0
1
Γ(𝛼+𝛽) 𝛼−1
𝑃(𝑋 > 𝑥) = ∫ (
𝑥
(1 − 𝑥)𝛽−1 ) 𝑑𝑥
Γ(𝛼) Γ(𝛽)
𝑥
𝑴𝒆𝒂𝒏(𝝁) =
𝛼
𝛼+𝛽
𝛼𝛽
𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆
=
(𝛼 + 𝛽) 2 (𝛼 + 𝛽 + 1)
𝛼−1
𝑴𝒐𝒅𝒆 =
𝛼+𝛽−2
(𝝈𝟐 )
𝛼 → 𝐴𝑙𝑤𝑎𝑦𝑠 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛
𝛽 → 𝐴𝑙𝑤𝑎𝑦𝑠 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛
𝚪 → 𝑻𝒓𝒆𝒂𝒕 𝒂𝒔 𝒈𝒂𝒎𝒎𝒂 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏.
𝑥 → 𝑇ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑎𝑠𝑘𝑒𝑑 𝑓𝑜𝑟
𝑵𝒐𝒕𝒆: 𝟎 ≤ 𝒙 ≤ 𝟏 𝒂𝒍𝒘𝒂𝒚𝒔.
4-150. Suppose X has a beta distribution
with parameters 𝜶 = 𝟏 and 𝜷 = 𝟒. 𝟐.
Determine the following:
(a) 𝑃(𝑋 < 0.25)
(b) 𝑃(0.5 < 𝑋 )
(c) mean and variance
𝒂)
𝑥 < 0.25 & 𝑐𝑢𝑚𝑚.
0.25
𝑃(𝑋 < 0.25) = ∫0
(
Γ(5.2)
Γ(1) Γ(4.2)
𝑥 0 (1 − 𝑥)3.2 ) 𝑑𝑥
.25
𝑃(𝑋 < 0.25) =
(4.2)Γ(4.2)
3.2
∫ (1 − 𝑥) 𝑑𝑥
Γ(4.2)
0
𝑃(𝑋 < 0.25) = 4.2 [−
(1−𝑥)
4.2
0.25
4.2
]
0
−4.2
4.2
(0.75 − 1)
4.2
= −(0.754.2 − 1) = 0.7013
𝑃(𝑋 < 0.25) =
𝒃)
𝑥 > 0.5 & 𝑐𝑢𝑚𝑚.
1
Γ(5.2)
𝑃(𝑋 > 0.5) = ∫ (
𝑥 0 (1 − 𝑥)3.2 ) 𝑑𝑥
Γ(1) Γ(4.2)
0.5
1
(4.2)Γ(4.2)
3.2
𝑃(𝑋 > 0.5) =
∫(1 − 𝑥) 𝑑𝑥
Γ(4.2)
0.5
𝑃(𝑋 > 0.5) = 4.2 [−
4.2
(1 − 𝑥)
1
]
4.2
0.5
−4.2
4.2
𝑃(𝑋 ≤ 0.25) =
(0.5 − 1)
4.2
= −(0 − 0.54.2 ) = 0.0544
1
4.2
;
𝝈𝟐 =
5.2
6.2(5.2)2
𝝁 = 0.1923 ; 𝝈𝟐 = 0.0251
0
𝑚𝑜𝑑𝑒 =
=0
5.2 − 2
𝒄);
𝝁=
Page 3 of 6
25. Joint Probability:
∞
∞
∫−∞ ∫−∞ 𝒇𝑿𝒀 (𝒙, 𝒚)𝒅𝒙 𝒅𝒚 = 𝟏
𝒇𝑿𝒀 (𝒙, 𝒚) ≥ 𝟎 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥, 𝑦
𝑺𝒐𝒍𝒗𝒊𝒏𝒈 𝒇𝒐𝒓 𝒄:
𝒄 ∑∞
−∞ 𝒇𝑿𝒀 (𝒙, 𝒚) = 𝟏
𝑼𝒔𝒆 𝒕𝒉𝒆𝒔𝒆 𝒘𝒉𝒆𝒏 𝒈𝒊𝒗𝒆𝒏 𝒓𝒂𝒏𝒈𝒆𝒔 𝒂𝒓𝒆𝒏′ 𝒕𝒈𝒊𝒗𝒆𝒏. (𝑿 = 𝟏, 𝟐, 𝟑, … 𝒏 𝒂𝒏𝒅 𝒀 = 𝟏, 𝟐, 𝟑, … 𝒎)
𝑥𝑛
𝑴𝒆𝒂𝒏
𝑬(𝑿) = ∑(𝑃𝑋𝑌 × 𝑥𝑖 ) & 𝐸(𝑋) = ∑[(𝑀. 𝑃)𝑋 × 𝑥𝑖 ]
𝑥0
𝑥𝑚 𝑦𝑛
𝑦𝑛
𝑦𝑛
𝑥𝑛
𝑬(𝑿𝒀) = ∑ [(𝑥𝑖 )(𝑦𝑗 )𝑃𝑖𝑗 ]
𝑬(𝒀) = ∑(𝑃𝑋𝑌 × 𝑦𝑖 ) & 𝐸(𝑋) = ∑[(𝑀. 𝑃)𝑦 × 𝑦𝑖 ]
𝑦0
𝑥0
𝑥𝑛
𝑦0
𝑥𝑖 𝑦𝑗
𝑦𝑛
2
2
𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆 𝑽(𝑿) = ∑ [(𝑥𝑖 − 𝐸(𝑋)) × 𝑃𝑋𝑌 ]
𝑽(𝑿𝒀) = [𝐸(𝑋𝑌) − 𝐸(𝑋)𝐸(𝑌)]2
𝑽(𝒀) = ∑ [(𝑦𝑖 − 𝐸(𝑌)) × 𝑃𝑋𝑌 ]
𝑥0
𝑦0
𝑃(𝑋, 𝑌) = ∬ 𝑓𝑋𝑌 (𝑥, 𝑦)𝑑𝑥 𝑑𝑦 → 𝒚𝒐𝒖𝒓 𝒍𝒊𝒎𝒊𝒕𝒔 𝒅𝒆𝒑𝒆𝒏𝒅 𝒕𝒉𝒆 𝒐𝒓𝒅𝒆𝒓 𝒐𝒇 𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒕𝒊𝒐𝒏
𝑅
𝑶𝒏𝒍𝒚 𝑼𝒔𝒆 𝒕𝒉𝒆𝒔𝒆 𝒘𝒉𝒆𝒏 𝒈𝒊𝒗𝒆𝒏 𝒓𝒂𝒏𝒈𝒆𝒔 (𝒂 < 𝑿 < 𝒃 𝒂𝒏𝒅 𝒄 < 𝒀 < 𝒅)
∞
∞
∞
𝑬(𝑿) = ∫ ∫ 𝒙 𝑓𝑋𝑌 (𝑥, 𝑦)𝑑𝑥 𝑑𝑦
𝑴𝒆𝒂𝒏
−∞ −∞
∞ ∞
−∞ −∞
∞
𝑽(𝑿) = ∫ ∫ 𝒙𝟐 𝑓𝑋𝑌 (𝑥, 𝑦)𝑑𝑥 𝑑𝑦
𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆
∞
∞
𝑬(𝒀) = ∫ ∫ 𝒚 𝑓𝑋𝑌 (𝑥, 𝑦)𝑑𝑥 𝑑𝑦
−∞ −∞
∞
𝑽(𝒀) = ∫ ∫ 𝒚𝟐 𝑓𝑋𝑌 (𝑥, 𝑦)𝑑𝑥 𝑑𝑦
−∞ −∞
∞
𝑬(𝑿𝒀) = ∫ ∫ 𝒙𝒚 𝑓𝑋𝑌 (𝑥, 𝑦)𝑑𝑥 𝑑𝑦
𝑽(𝑿𝒀) = [𝐸(𝑋𝑌) − 𝐸(𝑋)𝐸(𝑌)]2
−∞ −∞
1. Marginal Probability:
𝑀. 𝑃 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑋
𝑦𝑛
𝑓(𝑥) (𝑎) = ∑ 𝑓𝑋𝑌 (𝑎, 𝑦) → 𝑅𝑒𝑝𝑙𝑎𝑐𝑒 𝒙 𝑤𝑖𝑡ℎ 𝒂
𝑦0
𝑦
𝑓𝑋 (𝑥) = ∫ 𝑓𝑋𝑌 (𝑥, 𝑦) 𝑑𝑦
𝑎
𝑏
𝒘𝒉𝒆𝒓𝒆 𝒂 < 𝒚 < 𝒃
∞
𝑃(𝑎 < 𝑋 < 𝑏) = ∫ ∫ 𝑓𝑋𝑌 (𝑥, 𝑦)𝑑𝑦 𝑑𝑥
𝑎 −∞
𝑀. 𝑃 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑌
𝑥𝑛
𝑓(𝑦) (𝑏) = ∑ 𝑓𝑋𝑌 (𝑥, 𝑏) → 𝑅𝑒𝑝𝑙𝑎𝑐𝑒 𝒚 𝑤𝑖𝑡ℎ 𝒃
𝑥
𝑥0
𝑓𝑌 (𝑦) = ∫ 𝑓𝑋𝑌 (𝑥, 𝑦) 𝑑𝑥
𝑎
𝑏
𝒘𝒉𝒆𝒓𝒆 𝒂 < 𝒙 < 𝒃
𝑓(𝑥) (3) = ⟨𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 (𝑥 = 3)⟩ = 0.55
𝑓(𝑦) (2) = ⟨𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓( 𝑦 = 2)⟩ = 0.25
∞
𝑃(𝑎 < 𝑌 < 𝑏) = ∫ ∫ 𝑓𝑋𝑌 (𝑥, 𝑦)𝑑𝑥 𝑑𝑦
𝑎 −∞
2. Conditional Probability:
𝑓𝑌|𝑥 (𝑦) =
𝑓𝑋𝑌 (𝑥, 𝑦)
𝑓𝑋 (𝑥)
∞
→ 𝑔𝑖𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓𝑟𝑜𝑚 𝑞𝑥𝑛
→ 𝑵𝒐 𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒕𝒊𝒐𝒏 𝒏𝒆𝒆𝒅𝒆𝒅 𝒉𝒆𝒓𝒆
→ 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑑𝑒𝑟𝑖𝑣𝑒𝑑 𝑎𝑏𝑜𝑣𝑒
𝑃(𝑌 > 𝐵|𝑋 = 𝑎) = ∫ 𝑓𝑌|𝑥 (𝑦) 𝑑𝑦 → 𝑃𝑙𝑢𝑔 𝑖𝑛 𝒂 𝑓𝑜𝑟 𝑎𝑙𝑙 𝒙 𝑡ℎ𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡 𝑡ℎ𝑒 𝑟𝑒𝑠𝑡.
𝐵
𝑬(𝒀|𝒙) = ∫ 𝒚 𝑓𝑌|𝑥 (𝑦) 𝑑𝑦
𝑬(𝑿|𝒚) = ∫ 𝒙 𝑓𝑌|𝑥 (𝑥) 𝑑𝑥
𝑦
𝑷(𝑿 = 𝒂|𝒀 = 𝒃) =
𝑥
𝑉(𝑌|𝑥) = [∫ 𝑦 2 𝑓𝑌|𝑥 (𝑦) 𝑑𝑦] − [𝐸(𝑌|𝑥)]2
𝑦
𝑉(𝑌|𝑥) = [∫ 𝑦 2 𝑓𝑌|𝑥 (𝑦) 𝑑𝑦] − [𝐸(𝑌|𝑥)]2
𝑦
(𝒂 + 𝒃 + 𝒄)!
𝒃
𝑷(𝑿 = 𝒂, 𝒀 = 𝒃, 𝒁 = 𝒄) = (
) (𝒑𝒙 )𝒂 (𝒑𝒚 ) (𝒑𝒛 )𝒄
𝒂! 𝒃! 𝒄!
Covariance (𝝈𝑿𝒀 ):
𝜎𝑋𝑌 = 𝐸(𝑋𝑌) − 𝐸(𝑋)𝐸(𝑌)
𝒃
(𝒂 + 𝒃)!
(𝒑𝒙 )𝒂 (𝒑𝒚 )
𝒂! 𝒃!
𝑷(𝒀 = 𝒃) → 𝑼𝒔𝒆 𝒃𝒊𝒏𝒐𝒎𝒊𝒂𝒍
𝒃
(𝒂 + 𝒃)!
(𝒑𝒙 )𝒂 (𝒑𝒚 ) (𝒑𝒛 )𝒄
𝒂!
𝒃!
𝑷(𝑿 = 𝒂|𝒀 = 𝒃, 𝒁 = 𝒄) =
𝑷((𝒀 + 𝒁) = (𝒃 + 𝒄))
Correlation (𝝆𝑿𝒀 ):
𝜌𝑋𝑌 =
𝜎𝑋𝑌
𝜎𝑋 𝜎𝑌
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