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Outline
 Continuous random variables and density
 Common continuous random variables
 Moment generating function
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Continuous Random Variables
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Absolutely continuous random variable has absolutely continuous cdf 𝐹 that satifies
– for finitely many disjoint intervals (𝑎𝑖 , 𝑏𝑖 ), there exists 𝜖 for every given 𝛿 such that
𝑖 𝑏𝑖 − 𝑎𝑖 ≤ 𝛿 implies 𝑖 𝐹(𝑏𝑖 ) − 𝐹(𝑎𝑖 ) ≤ 𝜖
– Absolutely continuous functions
• are continuous. Continuity needs the condition above for a single interval.
• are differentiable almost everywhere.
• map length-0 intervals  length-0 intervals
• has a corresponding 𝑓 called density such that
𝒂
𝑭 𝒂 =
𝒇 𝒖 𝒅𝒖.
−∞

Singularly continuous function or random variable:
– Function: Continuous functions that increase only over sets whose total length is zero.
– Random variable: Sample space is uncountable but the range of the rv is a set with zero length.
E.g., making a probability mass of 1 from an uncountable range of 0 length is singularity
» Cantor function: Cantor set of length 0 → [0,1] of length 1.
2
A continuous random variable has an uncountable sample space
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Seeking a Density
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Exclude from Cantor Set
Exclude from
Cantor Set
Exclude from Cantor Set
Exclude from Cantor Set
Exclude from
Cantor Set
Exclude from Cantor Set
1.Cantor Set
Exclude from
Cantor Set
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Cantor Set  Cantor Function  Cantor Random Variable
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Continuous but Not Absolutely Continuous
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3. Cantor function as a cdf yields
Cantor random variable that is singularly continuous
2.Cantor Function
Cantor function is
constant outside Cantor Set
increasing inside Cantor Set
continuous over all
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Chalice (1991): Any increasing real-valued function on [0,1] that has
a) 𝐹 0 = 0, b) 𝐹 𝑥/3 = 𝐹(𝑥)/2, c) 𝐹 1 − 𝑥 = 1 − 𝐹(𝑥)
is Cantor function.
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Finding Many Densities
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In this course, absolutely continuous rv = absolutely continuous rv
Pdf 𝑓 is derived from cdf 𝐹
– Recall the sequence of probability construction: (Ω, ℑ, P) and then 𝐹 𝑎 : = P(𝑋 ≤ 𝑎)
Does each 𝐹 lead to a unique 𝑓? No!
– Ex: Consider 𝑓0 𝑥 = 1 for 0 ≤ 𝑥 ≤ 1 and over the same range of 𝑥 define a family of
functions parametrized as 𝑓𝑎 𝑥 = 𝑓0 𝑥 + 𝑎 𝐈𝑥=𝑎 for 0 ≤ 𝑎 ≤ 1. 𝐈𝐴 is a binary variable
taking the value of 1 only when 𝐴 is correct.
𝑥
» Note that −∞ 𝑓𝑎 𝑢 𝑑𝑢 = 𝑥 = 𝐹(𝑥) for 0 ≤ 𝑥 ≤ 1, because integral is not affected
by the alterations at countably many points or at the single point {𝑎}.
» Each function in the family {𝑓𝑎 : 0 ≤ 𝑎 ≤ 1} is a density for 𝐹 𝑥 = 𝐈0≤𝑥≤1 𝑥.
– Versions of the last example are the same almost everywhere and differ over a point.
– Can we generate examples that have versions differing over an interval rather than a
single point?
» No, because
𝑓 = 𝑓 almost everywhere
1
2
if and only if
𝑓 𝑥 𝑑𝑥 = 𝐴 𝑓2 𝑥 𝑑𝑥 for every set 𝐴.
𝐴 1
– Can we generate examples that have continuous versions?
» No, because continuous versions must differ over an interval.
utdallas
– 𝑓𝑋 𝑥 = lim (𝐹𝑋 𝑥 − 𝐹𝑋 𝑢 )/(𝑥 − 𝑢) define almost everywhere
𝑢→𝑥
– 0 = P 𝑋 𝜔 = 𝑥 ≠ 𝑓𝑋 𝑥 = arbitrary at countably many points

Expected value E(𝑋)
– E 𝑋 =

k th moment
– E 𝑋𝑘 =

𝑥𝑓𝑋 𝑥 𝑑𝑥
𝑥
𝑥
𝑥 𝑘 𝑓𝑋 𝑥 𝑑𝑥
Variance V(𝑋)
– V 𝑋 = E 𝑋2 - E 𝑋

2
Ex: For a continuous random variable 𝑋, let f u = 𝐈0≤u≤1 (𝑐𝑢2 + 𝑢).What value of 𝑐
makes 𝑓 a legitimate density so that it integrates to one? Find E 𝑋 .
– 1=𝐹 1 =
– E 𝑋 =
1 3
0 2
1
2
𝑐𝑢
0
+ 𝑢 𝑑𝑢 =
𝑢3 + 𝑢2 𝑑𝑢 =
𝑐
3
3
8
𝑢3 +
𝑢4 +
1
2
1
3
𝑢2
𝑢3
𝑢=1
𝑢=0
𝑢=1
𝑢=0
𝑐
3
3
+
8
1
so 𝑐
2
1
17
= .
3
24
= +
=
3
2
= .
5
Continuous random variable
 Probability density function pdf
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Moments
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Independence
Random variables 𝑋 and 𝑌 are independent if
P 𝑋 ≤ 𝑎, 𝑌 ≤ 𝑏 = P 𝑋 ≤ 𝑎 P 𝑌 ≤ 𝑏

This definition is inherited from the independence of events
P 𝑋 ∈ A, 𝑌 ∈ 𝐵 = P 𝑋 ∈ 𝐴 P 𝑌 ∈ 𝐵

where 𝐴 = (−∞, 𝑎] and 𝐵 = −∞, 𝑏 .
Ex: For two independent variables 𝑋1 , 𝑋2 , we have E 𝑋1 𝑋2 = E 𝑋1 E 𝑋2
– E 𝑋1 𝑋2 =
𝑥 𝑥 𝑓(𝑥1 )𝑓
𝑥1 ,𝑥2 1 2
𝑥2 𝑑𝑥1 𝑑𝑥2 =
𝑥1
𝑥1 𝑓(𝑥1 )𝑑𝑥1
𝑥2
𝑥2 𝑓(𝑥2 )𝑑𝑥2
= E 𝑋1 E 𝑋2


Ex: A random variable is not independent of itself so E X 2 = E 𝑋𝑋 ≠ E 𝑋 E 𝑋
Ex: For two independent variables 𝑋1 , 𝑋2 , we have V 𝑋1 + 𝑋2 = V 𝑋1 + V 𝑋2
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Common Continuous Random Variables
Uniform Random Variable
A uniform random variable 𝑋 takes values between 𝑥𝑢 and 𝑥𝑜 , and the probability of 𝑋 being
in any subinterval of [𝑥𝑢 , 𝑥𝑜 ] of a given length 𝛿 is the same:
P 𝑎 ≤ 𝑋 ≤ 𝑎 + 𝛿 = P 𝑏 ≤ 𝑋 ≤ 𝑏 + 𝛿 for 𝑥𝑢 ≤ 𝑎, 𝑏 ≤ 𝑎 + 𝛿, 𝑏 + 𝛿 ≤ 𝑥𝑜 .
Uniform random variable over [𝑥𝑢 , 𝑥𝑜 ] is denoted by 𝑈(𝑥𝑢 , 𝑥𝑜 ).
Ex: What is the pdf of a uniform random variable over [𝑥𝑢 , 𝑥𝑜 ]?
– Since P 𝑎 ≤ 𝑋 ≤ 𝑎 + 𝛿 = P 𝑏 ≤ 𝑋 ≤ 𝑏 + 𝛿 , the density 𝑓 𝑢 = 𝑐 must be constant over
[𝑥𝑢 , 𝑥𝑜 ].
– We need 1 = 𝐹(𝑥𝑜 ), that is 1 =
𝑓 𝑢 = 𝐈𝑥𝑢 ≤𝑢≤𝑥𝑜

= c xo − xu . Hence,
1
𝑥𝑜 −𝑥𝑢
Find E(𝑈(𝑥𝑢 , 𝑥𝑜 )).
E 𝑈 𝑥𝑢 , 𝑥𝑜

𝑥𝑜
𝑐𝑑𝑢
𝑥𝑢
=
𝑥𝑜
1
𝑢
𝑑𝑢
𝑥𝑢
𝑥𝑜 −𝑥𝑢
=
1 𝑢2 𝑢=𝑥𝑜
𝑥𝑜 −𝑥𝑢 2 𝑢=𝑥𝑢
=
𝑥𝑢 +𝑥𝑜
2
.
Find P(𝑋1 = 𝑈 0,1 ≥ 𝑋2 = 𝑈(0,1)). Note each Xi = 𝑈 0,1 is independent and identical.
𝑥2
1
𝑥1 ≥ 𝑥2
P 𝑋1 ≥ 𝑋2 =
𝑥1
𝑥1 ≥𝑥2
1 1 𝑑𝑥1 𝑑𝑥2 =
0
𝑥1
0
1
𝑑𝑥2 𝑑𝑥1 =
0
𝑥12
𝑥1 𝑑𝑥1 =
2
𝑥1 =1
𝑥1 =0
=
1
2
8
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A uniform random variable 𝑋 takes values between 0 and ∞, and the probability of 𝑋 being in any
subinterval of [𝑎, 𝑎 + 𝛿] of a given length 𝛿 exponentially decreases as 𝑎 increases:
P 𝑎 ≤ 𝑋 ≤ 𝑎 + 𝛿 = exp −𝜆 𝑎 − 𝑏
P 𝑏 ≤ 𝑋 ≤ 𝑏 + 𝛿 for 0 ≤ 𝑎 ≤ 𝑏 and 𝛿 ≥ 0.
Here 𝜆 is the parameter of the distribution.
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Start with the equality above, and first 𝛿 → ∞ and then 𝑎 = 0: P 𝑋 ≥ 𝑏 = exp(−𝜆𝑏)
The cdf and pdf are respectively 𝐹𝑋 𝑥 = 𝐈𝑥≥0 [1 − exp −𝜆𝑥 ] and 𝑓𝑋 𝑥 = 𝐈𝑥≥0 𝜆exp(−𝜆𝑥)
Ex: Find moments of Exponential distribution
–
∞
𝑥 𝜆exp −𝜆𝑥 =
0
∞
𝑋 2 = 0 𝑥 2 𝜆exp −𝜆𝑥
E 𝑋 =
= −𝑥 2 exp
∞
0
+
0−
exp −𝜆𝑥
𝜆
=
1
𝜆
∞
𝑥𝜆exp
𝜆 0
2
E
–
As an induction hypothesis suppose that E 𝑋
–
–
Hypothesis holds for 𝑘 = 1.
It also holds for k = 2, which we consider for getting intuition. The step of k = 2 can be dropped.
–
E 𝑋 𝑘+1 =
∞
0
–
𝜆
∞ 𝑘
𝑥 𝜆exp
0
−𝜆𝑥 𝑑𝑥 =
−𝜆𝑥 𝑑𝑥 =
2
𝜆2
= 𝑘!/𝜆 .
𝑥 𝑘+1 𝜆exp −𝜆𝑥 = −𝑥 𝑘+1 exp −𝜆𝑥
𝑘+1
𝑑𝑥 = 0 +
∞
0
–
=0+

∞
exp −𝜆𝑥 𝑑𝑥 =
0
∞
−𝜆𝑥 ∞
0 + 0 2𝑥 exp −𝜆𝑥
𝑘
𝑘
−𝑥exp −𝜆𝑥
𝑘+1
𝜆
∞
0
∞
𝑘
0
𝑘+1 !
+
E(𝑋 𝑘 ) =
+ 1 𝑥 𝑘 exp −𝜆𝑥 𝑑𝑥
𝜆𝑘+1
The last line establishes the induction hypothesis for 𝑘 + 1.
Exponential distribution has memoryless property: P 𝑋 > 𝑎 + 𝑏 = P(𝑋 > 𝑎)P(𝑋 > 𝑏).
–
If a machine is functioning at time 𝑎 (has not failed over [0, 𝑎]), the probability that it will be functioning at
time 𝑎 + 𝑏 (will not fail over [𝑎, 𝑎 + 𝑏]) is the same probability that it functioned for the first 𝑏 time units
after its installation.
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Common Continuous Random Variables
Exponential Random Variable
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Slower decay
Exponential decay
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For slower decay in the tail P(𝑋 ≥ 𝑥), note exp 𝜆𝑥 ≥ 𝑥 𝑎 or
exp −𝜆𝑥 ≥ 𝑥 −𝑎 for sufficiently large 𝑥.
Slower decay yields a heavy tail distribution
Seek a random variable with P(𝑋 ≥ 𝑥) proportional to 𝑥 −𝛼 for 𝑥 ≥
𝑥𝑢 and 𝛼 ≥ 0.
Since P 𝑋 ≥ 𝑥𝑢 = 1, set P 𝑋 ≥ 𝑥 = (𝑥𝑢 /𝑥)𝛼 for 𝑥 ≥ 𝑥𝑢 .
The cdf and pdf are respectively 𝐹𝑋 𝑥 = 𝐈𝑥≥𝑥𝑢 [1 − (𝑥𝑢 /𝑥)𝛼 ] and
𝑓𝑋 𝑥 = 𝐈𝑥≥𝑥𝑢
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Two pdfs
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𝛼
𝛼𝑥𝑢
𝑥 𝛼+1
Ex: Find E 𝑋 and moments
–
For 𝛼 > 1, E 𝑋 =
𝛼
∞ 𝛼𝑥𝑢
𝑥
𝑑𝑥
𝑥𝑢 𝑥 𝛼+1
=−
–
For 𝛼 = 1, E 𝑋 =
∞
𝑥𝑢
𝑥
𝑑𝑥
𝑥𝑢 𝑥 1+1
= 𝑥𝑢 ln 𝑥
–
Since
–
Heavy tail causes diverging moments.
1
𝑥𝑓
0
𝛼
𝛼
𝑥𝑢
𝛼−1 𝑥 𝛼−1
∞
𝑥𝑢
∞
𝑥𝑢
=
𝛼
𝑥
𝛼−1 𝑢
→ ∞. For 𝛼 < 1, E 𝑋 = 𝛼
𝛼
∞
𝑥𝑢
𝑥
𝑑𝑥
𝑥𝑢 𝑥 𝛼+1
>𝛼
∞
𝑥𝑢
𝑥
𝑑𝑥
𝑥𝑢 𝑥 1+1
→∞
𝑥 𝑑𝑥 is finite and 𝑥 𝑘 ≥ 𝑥 for 𝑥 ≥ 1, diverging mean implies diverging moments
Heavy tail in practice
–
–
–
–
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Common Continuous Random Variables
Pareto Random Variable
Wealth distribution in a population: there are plenty of individuals with very large amount of wealth.
Phone conversation duration: there are plenty of phone conversations with a very long duration.
Number of tweets sent by an individual: there are plenty of individuals sending a large number of tweets.
Number of failures caused by a root cause: there are plenty of root causes leading to a large number of failures.
This is known as Pareto rule in Quality Management.
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utdallas
For nonnegative integer 𝑛, 𝑛! = 𝑛 𝑛 − 1 𝑛 − 2 … 2 1 and 0! = 1.
This factorial definition does not help for real numbers or negative integers. Let us extend the definition.
Let us define Gamma function for any real number 𝑛,
∞
Γ 𝑛 =
0

Then

Γ 𝑛+1 =







∞ 𝑛
𝑥 exp −𝑥 𝑑𝑥
0
∞
= 0 − 0 𝑛𝑥 𝑛−1 −exp −𝑥 𝑑𝑥
∞
= 𝑛 0 𝑥 𝑛−1 exp −𝑥 𝑑𝑥 = 𝑛Γ 𝑛
∞
Γ 1 = 0 exp −𝑥 𝑑𝑥 = 1
∞
1
Ex: Γ
= 0 𝑥 −1/2 exp −𝑥 𝑑𝑥
2
∞
2𝑢
= 0 exp −𝑢2
𝑑𝑢
𝑢
∞
= 2 0 exp −𝑢2 𝑑𝑢 = 𝜋
7
5
5
53
3
Ex: Γ
= Γ
= Γ
2
2
2
22
2
=
531
1
Γ
222
2
=
15 𝜋
8
Ex: Γ 0 = ∞, by lower bounding
the limit with a series diverging to ∞.
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Gamma Function
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𝑥 𝑛−1 exp −𝑥 𝑑𝑥
–
𝑓𝑋 𝑥 =
𝑑
𝑑𝑥
𝑥
𝜆exp
0
−𝜆𝑢 𝑑𝑢
𝑢 𝜆exp −𝜆𝑢 𝑑𝑢 =
𝑥
(1 −
0
exp −𝜆(𝑥 − 𝑢) )𝜆exp −𝜆𝑢 𝑑𝑢
𝑑𝑢 = −exp −𝜆𝑥 + 1 − 𝜆𝑥exp(−𝜆𝑥)
P 𝑋 ≤ 𝑥 = 𝜆exp −𝜆𝑥 − 𝜆exp −𝜆𝑥 + 𝜆2 exp −𝜆𝑥 = 𝜆exp −𝜆𝑥 𝜆𝑥 .
Add up 𝛼 many independent 𝐸𝑥𝑝𝑜(𝜆) variables and end up with a distribution parameterized by 𝛼, 𝜆 .
The pdf of Gamma(𝛼, 𝜆) is
𝑓𝑋 𝑥 =

𝑥
P 𝑋2 ≤ 𝑥 −
0
𝑥
− 0 𝜆exp −𝜆𝑥
P 𝑋 ≤ 𝑥 = P 𝑋1 + 𝑋2 ≤ 𝑥 =
=

utdallas
An Erlang rv is the sum of an integer number of exponential random variables.
Ex: Add up two independent 𝑋1 , 𝑋2 ∼ 𝐸𝑥𝑝𝑜(𝜆) , what is the cdf of 𝑋 = 𝑋1 + 𝑋2 ?
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Common Continuous Random Variables
Erlang and Gamma Random Variables
𝜆exp −𝜆𝑥 𝜆𝑥 𝛼−1
Γ(𝛼)
for 𝑥 ≥ 0 and 𝛼 > 0
Ex: UTD's SOM has routers at every floor. Since the service must be provided without an interruption, a failed router is
replaced immediately. We are told that the first floor had 7 failed routers and running the 8th now. On the other hand,
the second floor had 4 routers and running the 5th now.
– Suppose that all these routers are identical with Expo(𝜆) lifetime with parameter 𝜆 measured in 1/year.
– P(Gamma(8,𝜆)≥10) represents the unconditional probability that the routers on the first floor will last for 10 years.
– If we are told that the SOM building is in operation for 8 years, then this probability can be revised as
P(Gamma(8,𝜆)≥10| Gamma(8,𝜆)≥8).
– Similar probabilities for the second floor are P(Gamma(5,𝜆)≥10) and P(Gamma(5,𝜆)≥10| Gamma(5,𝜆)≥8).
For integer 𝛼, find E(Gamma(𝛼,𝜆)) and V(Gamma(𝛼,𝜆)). We can start with the first two moments of Exponential
random variable: E(Expo(𝜆))=1/𝜆 and E(Expo 𝜆 2 )=2/𝜆2 , which imply V(Expo(𝜆))=1/𝜆2 .
– Since 𝛼 is integer, we can think of Gamma as sum of 𝛼 independent exponential random variables so that we can
apply the formulas for the expected value and variance of sum of independent random variables:
E(Gamma(𝛼, 𝜆))=𝛼/𝜆 and V(Gamma(𝛼, 𝜆))=𝛼/ 𝜆2 .
𝑑𝑢 =
1 𝑘−1
𝑢
0
1−𝑢
𝑚−1
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𝑘−1

Beta function 𝐵 𝑚, 𝑘 =

Also 𝐵 𝑘, 𝑚 =

We consider Bernoulli trials where the success probability 𝑃 is constant but unknown. After 𝑛 trials,
suppose that we observe 𝑚 success. Given these 𝑚 successes we can update the distribution of 𝑃, we are
seeking P(𝑃 = 𝑝 𝐵𝑖𝑛(𝑛, 𝑃) = 𝑚):
Γ 𝑘 Γ(𝑚)
.
Γ 𝑘+𝑚
1−𝑢
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1 𝑚−1
𝑢
0
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Common Continuous Random Variables
Beta Random Variable
𝑑𝑢 = 𝐵(𝑘, 𝑚).
This function works as a normalization constant.
– P 𝑃 = 𝑝 𝐵𝑖𝑛 𝑛, 𝑃 = 𝑚 =
P 𝑃=𝑝,𝐵𝑖𝑛 𝑛,𝑃 =𝑚
P Bin n,P =m
=
P 𝑃=𝑝,𝐵𝑖𝑛 𝑛,𝑃 =𝑚
0<𝑢<1 P 𝑃=𝑢,Bin n,P =m
– We approximate P(𝑃 = 𝑝, 𝐵𝑖𝑛(𝑛, 𝑃) = 𝑚) by 𝑓𝑃 (𝑝)P(𝐵𝑖𝑛(𝑛, 𝑃) = 𝑚) which leads to
0<𝑢<1 P 𝑃 = 𝑢, 𝐵𝑖𝑛 𝑛, 𝑃 = 𝑚 ≈
1
𝑓
0 𝑃
𝑢 P 𝐵𝑖𝑛 𝑛, 𝑢 = 𝑚 𝑑𝑢
where 𝑓𝑃 is the pdf of the success probability 𝑃.
–
–
For 𝑋 ≡ 𝑃 𝐵𝑖𝑛 𝑛, 𝑃 = 𝑚
𝑓𝑃 𝑝 P 𝐵𝑖𝑛 𝑛, 𝑝 = 𝑚
𝑓𝑋 𝑝 = 1
=
𝑓
𝑢
P
𝐵𝑖𝑛
𝑛,
𝑢
=
𝑚
𝑑𝑢
0 𝑃
𝑛 𝑚
𝑓𝑃 𝑝 𝐶𝑚
𝑝 1−𝑝
1
𝑓
0 𝑃
𝑛 𝑚
𝑢 𝐶𝑚
𝑢 1−𝑢
𝑛−𝑚
𝑛−𝑚 𝑑𝑢
=
𝑓𝑃 𝑝 𝑝𝑚 1 − 𝑝
1
𝑓
0 𝑃
𝑢 𝑢𝑚 1 − 𝑢
𝑛
𝑛−𝑚 𝑑𝑢
To simplify further suppose 𝑓𝑃 is uniform,
𝑓𝑋 𝑝 =
𝑝𝑚 1−𝑝 𝑛
1 𝑚
0 𝑢
1−𝑢 𝑛 𝑑𝑢
=
𝑝𝑚 1−𝑝 𝑛−𝑚
, posterior probability of success given 𝑚 successes out of 𝑛
𝐵(𝑚+1,𝑛−𝑚+1)

In general, Beta random variable has 𝑓𝑋 𝑥 =

Ex: E 𝑋 =
1 𝑥 𝛼−1 1−𝑥 𝜆−1
𝑥
𝑑𝑥
0
𝐵(𝛼,𝜆)
=
1
𝐵 𝛼,𝜆
𝑥 𝛼−1 1−𝑥 𝜆−1
B(𝛼,𝜆)
𝐵 𝛼 + 1, 𝜆 =
for 0 ≤ 𝑥 ≤ 1 and 𝛼, 𝜆 > 0
Γ 𝛼+𝜆 Γ 𝛼+1 Γ 𝜆
Γ 𝛼 Γ(𝜆) Γ 𝛼+𝜆+1
=
𝛼
𝛼+𝜆
1
𝑥−𝜇
𝑓𝑋 𝑥 =
exp −
2𝜎 2
2𝜋 𝜎


utdallas
The cdf can be computed only numerically.
Ex: The density integrates to 1. First note that
∞
1
exp
−∞ 2𝜋 𝜎

2
On the other hand,
𝑧12
exp −
𝑑𝑧1
2
−∞
−
𝑥−𝜇 2
2𝜎2
𝑑𝑥 =
∞
exp
2𝜋 −∞
1
𝑧22
exp −
𝑑𝑧2 =
2
−∞
∞
∞
∞
−∞
−
𝑧2
2
𝑑𝑧 with 𝑧 = (𝑥 − 𝜇)/𝜎
𝑧12 + 𝑧22
exp −
𝑑𝑧2 𝑑𝑧1 =
2
−∞
∞
2𝜋
0
∞
0
𝑟2
𝑟exp −
𝑑𝑟𝑑𝜃
2

The last equality is from transformation to polar coordinates 𝑟 2 = 𝑧12 + 𝑧22 and 𝜃 = arccos

Finally,

Ex: E 𝑋 =
2𝜋 ∞
𝑟exp
0
0
∞
1
𝑥
exp
−∞
2𝜋 𝜎
−
−
𝑟2
2
𝑥−𝜇 2
2𝜎2
=𝜇
∞
𝑑𝑟𝑑𝜃 =
𝑑𝑥 =
1
−∞ 2𝜋
2𝜋
∞
𝑑𝜃
𝑟exp
0
0
∞
(𝜇
−∞
exp −
=1
𝑧2
2
+ 𝑧𝜎)
𝑑𝑥 + 𝜎
1
−
𝑟2
2
𝑑𝑟 = 2𝜋
exp −
2𝜋
∞
𝑧
−∞ 2𝜋
𝑧2
2
∞
exp
0
𝑑𝑥 = 𝜇
exp −
=0
𝑧2
2
−𝑢 𝑑𝑢 = 2𝜋.
∞ 1
−∞ 2𝜋
𝑑𝑥 = 𝜇
𝑧1
𝑟
exp −
𝑧2
2
𝑑𝑥
13

Normal density is symmetric around its mean (mode) and has light tails.
The range for the normal random variable 𝑁𝑜𝑟𝑚𝑎𝑙(𝜇, 𝜎 2 ) is all real numbers and the density is
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
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Common Continuous Random Variables
Normal Random Variable
P 𝑎𝑋 + 𝑏 ≤ 𝑢 = P 𝑋 ≤
𝑑
P
𝑑𝑢
𝑋≤
𝑢−𝑏
𝑎
𝑢−𝑏
𝑎
=
and differentiate with respect to 𝑢
𝑢−𝑏
𝑑
𝑎
−∞
𝑑𝑢
1
exp
2𝜋 𝜎
We obtain the density for 𝑁𝑜𝑟𝑚𝑎𝑙 𝑎𝜇

𝑥−𝜇 2
− 2𝜎2 𝑑𝑥
+ 𝑏, 𝑎2 𝜎 2 .
=
1
exp
2𝜋 𝑎𝜎
−
𝑢−𝑏−𝑎𝜇 2
2𝑎2 𝜎2
Ex: Partial expected value for 𝑋 ∼ 𝑁𝑜𝑟𝑚𝑎𝑙 𝜇, 𝜎 2 upto 𝑎.
𝑎
𝑎
1
𝑥−𝜇 2
1
𝑧2
E 𝑋I𝑋≤𝑎 =
𝑥
exp −
𝑑𝑥 =
(𝜇 + 𝑧𝜎)
exp −
𝑑𝑥
2
2𝜎
2
2𝜋 𝜎
2𝜋
−∞
−∞
= 𝜇F𝑁𝑜𝑟𝑚𝑎𝑙
0,1
= 𝜇F𝑁𝑜𝑟𝑚𝑎𝑙
𝜇
1
+𝜎
exp −𝑢 𝑑𝑢
𝜎
2𝜋
∞
𝑎−𝜇
𝑎−𝜇
− 𝜎𝑓𝑁𝑜𝑟𝑚𝑎𝑙 0,1
𝜎
𝜎
𝑎−
0,1
𝑎−𝜇 2
/2
𝜎
utdallas
14
Ex: What is the distribution of 𝑎𝑋 + 𝑏 if 𝑋 is 𝑁𝑜𝑟𝑚𝑎𝑙 𝜇, 𝜎 2 . We can consider
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
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Common Continuous Random Variables
Normal Random Variable


Ex: Find the moment E(𝑋 𝑘 ).
∞
𝑦−𝜇 2
E 𝑋 = E exp 𝑘𝑌 =
exp 𝑘𝑦
exp −
𝑑𝑦
2
2𝜎
2𝜋
𝜎
−∞
∞
1
𝑢2
= exp 𝑘𝜇
exp 𝑘𝑢
exp − 2 𝑑𝑢
2𝜎
2𝜋
𝜎
−∞
∞
2 2
2 2
𝑘 𝜎
1
(𝑢 − 𝑘𝜎 )
𝑘2𝜎2
= exp 𝑘𝜇 exp
exp −
𝑑𝑢 = exp 𝑘𝜇 exp
2
2
2𝜎
2
2𝜋
𝜎
−∞
𝑘
1
=1
Finally, E 𝑋 𝑘 = exp 𝑘𝜇 +
1
2
𝑘𝜎
2
utdallas
15

𝑋 is lognormal random variable if 𝑋 = exp(𝑌) for 𝑌 ∼ 𝑁𝑜𝑟𝑚𝑎𝑙(𝜇, 𝜎 2 )
The range for lognormal is nonnegative reals. This is an advantage for modelling price, demand.
Ex: Find the pdf of 𝑋
ln 𝑥
1
𝑦−𝜇 2
P 𝑋 ≤ 𝑥 = P 𝑌 ≤ ln 𝑥 =
exp −
𝑑𝑦
2𝜎 2
2𝜋 𝜎
−∞
𝑑
1 1
ln 𝑥 − 𝜇 2
𝑓𝑋 𝑥 =
P 𝑋≤𝑥 =
exp −
𝑑𝑥
𝑥 2𝜋 𝜎
2𝜎 2
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
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Common Continuous Random Variables
Lognormal Random Variable




utdallas
Given rvs 𝑋, 𝑌 , the mixture 𝑍 can be obtained as
𝑋 with probability 𝑝
𝑍=
𝑌 with probability 1 − 𝑝
Then we have 𝐹𝑍 𝑧 = 𝑝𝐹𝑋 𝑧 + 1 − 𝑝 𝐹𝑌 (𝑧).
Mixing allows us to create new rvs and obtain richer examples.
Ex: Consider the mixture of a mass of 𝑞 at zero with 𝑒𝑥𝑝𝑜 𝜆 random variable whose cdf is
𝑞 if 𝑧 = 0
𝐹𝑍 (𝑧) =
𝑞 + (1 − 𝑞)(1 − 𝑒𝑥𝑝(−𝜆𝑧)) if 𝑧 > 0
Random variables derived from others through max or min operators can have masses at the end of their
ranges.
–
–

16

Page
Mixtures of Random variables
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Product returns due to unmet expectations to a retailer can be treated as negative demand, then the demand 𝐷 ∈ ℜ .
If you want only the new customer demand for products, you should consider max{0, 𝐷} which has a mass at 0.
In general, any random variable can be represented as a combination of cdfs of a discrete rv, a singularly
continuous rv and an absolutely continuous rv.


utdallas
Random variable 𝑋 can be described uniquely by all of its moments: E 𝑋 , E 𝑋 2 , … , E 𝑋 𝑘 , …
A function that yields all of the moments uniquely identifies the associated random variable
A such function is moment generating function
𝑚𝑋 (𝑡) =


∞
−∞ exp
Since exp 𝑡𝑥 = 1 + 𝑡𝑥 +
𝑡𝑥 2
2!
𝑚𝑋 𝑡 =
∞
𝑥=−∞
𝑡𝑥 2
2!
=
∞ 𝑡
𝑖=0 𝑖!
𝑖
1 + 𝑡𝑥 +
𝑡𝑥 𝑝𝑋 (𝑥) and 𝑚𝑋 (𝑡) =
+
𝑡𝑥 3
3!
+
𝑡𝑥 3
3!
E(𝑋 i )
Insert some examples: To be added
+ ⋯+
𝑡𝑥 𝑖
𝑖!
+⋯
+ ⋯+
𝑡𝑥 𝑖
𝑖!
+⋯
∞
exp
−∞
𝑝𝑋 𝑥
𝑡𝑥 𝑓𝑋 𝑥 𝑑𝑥
17

Page
Moment Generating Functions
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

Continuous random variables and density
Common continuous random variables
Moment generating function
utdallas
18

Page
Summary
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