Transcript Probability theory

Random Variable
Martina Litschmannová
[email protected]
K210
Random Variable
 A random variable is a function or rule that assigns a number to
each outcome of an experiment.
 Basically it is just a symbol that represents the outcome of an
experiment.
Ω
-3
-2
-1
0
1
2
3
ℝ
Random Variable
 A random variable is a function or rule that assigns a number to
each outcome of an experiment.
 Basically it is just a symbol that represents the outcome of an
experiment.
For example:
 X = number of heads when the experiment is flipping a coin 20
times.
 R = the number of miles per gallon you get on your auto during a
family vacation.
 Y = the amount of medication in a blood pressure pill.
 V = the speed of an auto registered on a radar detector.
Types of Random Variable
Discrete Random Variable
 usually count data [Number of]
 one that takes on a countable number of values – this means you
can sit down and list all possible outcomes without missing any,
although it might take you an infinite amount of time.
For example:
 X = values on the roll of two dice: X has to be either 2, 3, 4, …, or 12.
 Y = number of accidents in Ostrava during a week: Y has to be 0, 1,
2, 3, 4, 5, 6, 7, 8, ……………”real big number”
Continuous Random Variable
 usually measurement data [time, weight, distance, etc]
 one that takes on an uncountable number of values – this means
you can never list all possible outcomes even if you had an infinite
amount of time.
For example:
 time it takes you to drive home from class: X > 0, might be 30.1
minutes measured to the nearest tenth but in reality the actual
time is 30.10000001…………………. minutes.
Exercise: try to list all possible numbers between 0 and 1.
Probability distribution
 A probability distribution (probability mass function, cumulative
distribution function or density function) is a table, formula, or
graph that describes the values of a random variable and the
probability associated with these values.
 An upper-case letter will represent the name of the random
variable, for example X.
 Its lower-case counterpart, x, will represent the value of the
random variable.
Probability Distribution
of Discrete Random Variable
Probability mass (density) function

A set of probability value 𝑝𝑖 assigned to each of the
values taken by the discrete random variable 𝑥𝑖 .

0 ≤ 𝑝𝑖 ≤ 1 and
𝑎𝑙𝑙 𝑖 𝑝𝑖
= 1.
 Probability : 𝑝𝑖 = 𝑃 𝑋 = 𝑥𝑖 = 𝑃 𝑥𝑖
Probability mass function
For example:
X = number of heads in 3 flips of unfair coin (P(H)=0,1)
 The probability that the random variable X will equal x is:
∀𝑥 ∈ 0; 1; 2; 3 : 𝑃 𝑋 = 𝑥 =
x
𝑃 𝑥
3
∙ 0,1𝑥
𝑥
0
1
2
3
0,729 0,243 0,027 0,001
Developing Probability Mass Function
Probability Mass Function can be estimated from relative frequencies.
Consider the discrete (countable) number of televisions per household
(X) from US survey data.
1218 /101501 = 0,012
# of Televisions # of Households
0
1 218
1
32 379
2
37 961
3
19 387
4
7 714
5
2 842
Total
101 501
x
0
1
2
3
4
5
Total
P(x)
0,012
0,319
0,374
0,191
0,076
0,028
1,000
e.g. 𝑃 𝑋 = 0 = 𝑃 0 = 0,012 = 1,2%
1. What is the probability there is at least one television but no more
than three in any given household?
# of Televisions # of Households
0
1 218
1
32 379
2
37 961
3
19 387
4
7 714
5
2 842
Total
101 501
x
0
1
2
3
4
5
Total
P(x)
0,012
0,319
0,374
0,191
0,076
0,028
1,000
1. What is the probability there is at least one television but no more
than three in any given household?
# of Televisions # of Households
0
1 218
1
32 379
2
37 961
3
19 387
4
7 714
5
2 842
Total
101 501
x
0
1
2
3
4
5
Total
P(x)
0,012
0,319
0,374
0,191
0,076
0,028
1,000
𝑃 1 ≤ 𝑋 ≤ 3 = 𝑃 1 + 𝑃 2 + 𝑃 3 = 0,319 + 0,374 + 0,191 = 𝟎, 𝟖𝟖𝟒
2. A mutual fund sales person knows that there is 20% chance of
closing a sale on each call she makes. What is the probability
distribution of the number of sales if she plans to call three
customers?
 Random Variable: X … # Sales made in 3 Attemts
 Let S denote closing a sale. Probability of closing a sale is 𝑃 𝑆 =
0,20.
 Thus F = 𝑆 is not closing a sale, and 𝑃 𝐹 = 0,80.
Seems reasonable to assume that sales are independent.
2. A mutual fund sales person knows that there is 20% chance of
closing a sale on each call she makes. What is the probability
distribution of the number of sales if she plans to call three
customers?
 Sample space Ω … list of possible outcomes
Ω = 𝑆𝑆𝑆, 𝑆𝑆𝐹, 𝑆𝐹𝑆, 𝑆𝐹𝐹, 𝐹𝑆𝑆, 𝐹𝑆𝐹, 𝐹𝐹𝑆, 𝐹𝐹𝐹
 𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑆𝑆𝑆
𝑆𝑆𝐹
𝑆𝐹𝑆
𝑆𝐹𝐹
𝐹𝑆𝑆
𝐹𝑆𝐹
𝐹𝐹𝑆
𝐹𝐹𝐹
= 0,20 ∙ 0,20 ∙ 0,20 = 0,008
= 0,20 ∙ 0,20 ∙ 0,80 = 0,032
= 0,20 ∙ 0,80 ∙ 0,20 = 0,032
= 0,20 ∙ 0,80 ∙ 0,80 = 0,128
= 0,80 ∙ 0,20 ∙ 0,20 = 0,032
= 0,80 ∙ 0,20 ∙ 0,80 = 0,128
= 0,80 ∙ 0,80 ∙ 0,20 = 0,128
= 0,80 ∙ 0,80 ∙ 0,80 = 0,512
2. A mutual fund sales person knows that there is 20% chance of
closing a sale on each call she makes. What is the probability
distribution of the number of sales if she plans to call three
customers?
 X … # Sales made in 3 Attemts
 𝑃 𝑆𝑆𝑆 = 0,20 ∙ 0,20 ∙ 0,20 = 0,008
𝑃 𝑆𝑆𝐹 = 0,20 ∙ 0,20 ∙ 0,80 = 0,032
𝑃 𝑆𝐹𝑆 = 0,20 ∙ 0,80 ∙ 0,20 = 0,032
𝑃 𝑆𝐹𝐹 = 0,20 ∙ 0,80 ∙ 0,80 = 0,128
𝑃 𝐹𝑆𝑆 = 0,80 ∙ 0,20 ∙ 0,20 = 0,032
𝑃 𝐹𝑆𝐹 = 0,80 ∙ 0,20 ∙ 0,80 = 0,128
𝑃 𝐹𝐹𝑆 = 0,80 ∙ 0,80 ∙ 0,20 = 0,128
𝑃 𝐹𝐹𝐹 = 0,80 ∙ 0,80 ∙ 0,80 = 0,512
 𝑃 𝑋 = 0 = 𝑃 𝐹𝐹𝐹 = 0,512
2. A mutual fund sales person knows that there is 20% chance of
closing a sale on each call she makes. What is the probability
distribution of the number of sales if she plans to call three
customers?
 X … # Sales made in 3 Attemts
 𝑃 𝑆𝑆𝑆 = 0,20 ∙ 0,20 ∙ 0,20 = 0,008
𝑃 𝑆𝑆𝐹 = 0,20 ∙ 0,20 ∙ 0,80 = 0,032
𝑃 𝑆𝐹𝑆 = 0,20 ∙ 0,80 ∙ 0,20 = 0,032
𝑃 𝑆𝐹𝐹 = 0,20 ∙ 0,80 ∙ 0,80 = 0,128
𝑃 𝐹𝑆𝑆 = 0,80 ∙ 0,20 ∙ 0,20 = 0,032
𝑃 𝐹𝑆𝐹 = 0,80 ∙ 0,20 ∙ 0,80 = 0,128
𝑃 𝐹𝐹𝑆 = 0,80 ∙ 0,80 ∙ 0,20 = 0,128
𝑃 𝐹𝐹𝐹 = 0,80 ∙ 0,80 ∙ 0,80 = 0,512
 𝑃 𝑋 = 1 = 𝑃 𝑆𝐹𝐹 ∪ 𝐹𝑆𝐹 ∪ 𝐹𝐹𝑆 = 0,128 + 0,128 + 0,128 = 0,374
2. A mutual fund sales person knows that there is 20% chance of
closing a sale on each call she makes. What is the probability
distribution of the number of sales if she plans to call three
customers?
 X … # Sales made in 3 Attemts
 𝑃 𝑆𝑆𝑆 = 0,20 ∙ 0,20 ∙ 0,20 = 0,008
𝑃 𝑆𝑆𝐹 = 0,20 ∙ 0,20 ∙ 0,80 = 0,032
𝑃 𝑆𝐹𝑆 = 0,20 ∙ 0,80 ∙ 0,20 = 0,032
𝑃 𝑆𝐹𝐹 = 0,20 ∙ 0,80 ∙ 0,80 = 0,128
𝑃 𝐹𝑆𝑆 = 0,80 ∙ 0,20 ∙ 0,20 = 0,032
𝑃 𝐹𝑆𝐹 = 0,80 ∙ 0,20 ∙ 0,80 = 0,128
𝑃 𝐹𝐹𝑆 = 0,80 ∙ 0,80 ∙ 0,20 = 0,128
𝑃 𝐹𝐹𝐹 = 0,80 ∙ 0,80 ∙ 0,80 = 0,512
 𝑃 𝑋 = 2 = 𝑃 𝑆𝑆𝐹 ∪ 𝑆𝐹𝑆 ∪ 𝐹𝑆𝑆 = 0,032 + 0,032 + 0,032 = 0,096
2. A mutual fund sales person knows that there is 20% chance of
closing a sale on each call she makes. What is the probability
distribution of the number of sales if she plans to call three
customers?
 X … # Sales made in 3 Attemts
 𝑃 𝑆𝑆𝑆 = 0,20 ∙ 0,20 ∙ 0,20 = 0,008
𝑃 𝑆𝑆𝐹 = 0,20 ∙ 0,20 ∙ 0,80 = 0,032
𝑃 𝑆𝐹𝑆 = 0,20 ∙ 0,80 ∙ 0,20 = 0,032
𝑃 𝑆𝐹𝐹 = 0,20 ∙ 0,80 ∙ 0,80 = 0,128
𝑃 𝐹𝑆𝑆 = 0,80 ∙ 0,20 ∙ 0,20 = 0,032
𝑃 𝐹𝑆𝐹 = 0,80 ∙ 0,20 ∙ 0,80 = 0,128
𝑃 𝐹𝐹𝑆 = 0,80 ∙ 0,80 ∙ 0,20 = 0,128
𝑃 𝐹𝐹𝐹 = 0,80 ∙ 0,80 ∙ 0,80 = 0,512
 𝑃 𝑋 = 3 = 𝑃 𝑆𝑆𝑆 = 0,008
2. A mutual fund sales person knows that there is 20% chance of
closing a sale on each call she makes. What is the probability
distribution of the number of sales if she plans to call three
customers?
 X … # Sales made in 3 Attemts
 𝑃 𝑆𝑆𝑆 = 0,20 ∙ 0,20 ∙ 0,20 = 0,008
𝑃 𝑆𝑆𝐹 = 0,20 ∙ 0,20 ∙ 0,80 = 0,032
𝑃 𝑆𝐹𝑆 = 0,20 ∙ 0,80 ∙ 0,20 = 0,032
𝑃 𝑆𝐹𝐹 = 0,20 ∙ 0,80 ∙ 0,80 = 0,128
𝑃 𝐹𝑆𝑆 = 0,80 ∙ 0,20 ∙ 0,20 = 0,032
𝑃 𝐹𝑆𝐹 = 0,80 ∙ 0,20 ∙ 0,80 = 0,128
𝑃 𝐹𝐹𝑆 = 0,80 ∙ 0,80 ∙ 0,20 = 0,128
𝑃 𝐹𝐹𝐹 = 0,80 ∙ 0,80 ∙ 0,80 = 0,512
x
P(x)
0
0,512
1
2
3
0,384 0,096 0,008
2. A mutual fund sales person knows that there is 20% chance of
closing a sale on each call she makes. What is the probability
distribution of the number of sales if she plans to call three
customers?
 Another Approach: Tree Diagram
Sales Call 1
Sales Call 2
P(S)=,2
P(S)=,2
P(F)=,8
P(S)=,2
P(F)=,8
Sales Call 3
P(S)=,2
SSS
P(F)=,8
P(S)=,2
SSF
S FS
P(F)=,8
P(S)=,2
S FF
FS S
P(F)=,8
P(S)=,2
FS F
FFS
P(F)=,8
FFF
P(F)=,8
X
3
2
1
0
P(x)
,23 = .008
3(,032)=.096
3(,128)=.384
,83 = .512
Cumulative Distribution Function F(x)
CDF 𝐹 𝑥 is probability that random variable X is less than real
number x.
𝐹 𝑥 =𝑃 𝑋<𝑥
Cumulative Distribution Function F(x)
CDF 𝐹 𝑥 is probability that random variable X is less than real
number x.
𝐹 𝑥 =𝑃 𝑋<𝑥
Properties of the distribution function 𝑭 𝒙 :
 0 ≤ 𝐹 𝑥 ≤ 1,
 𝐹 𝑥 is nondecreasing,
 𝐹 𝑥 is continous from the left,
 𝐹 𝑥 → 0 𝑓𝑜𝑟 𝑥 → −∞ („start“ in 0),
 𝐹 𝑥 → 1 𝑓𝑜𝑟 𝑥 → ∞
(„finished“ in 1).
Relationship between p.d.f. P(x) and c.d.f. F(x)
P(x)
F(x)
0.8
0.95
0.7
0.6
0.75
0.5
0.4
0.55
0.3
0.35
0.2
0.15
0.1
0
0
1
2
x
3
4
-1
-0.05
0
1
2
x
3
4
Relationship between p.d.f. P(x) and c.d.f. F(x)
P(x)
F(x)
0.8
0.95
0.7
0.6
0.75
0.5
0.4
0.55
0.3
0.35
0.2
0.15
0.1
0
0
1
2
x
3
4
-1
-0.05
0
1
2
3
x
 Points of discontinuity of F(x) are points, where P(x) is nonzero.
4
Relationship between p.d.f. P(x) and c.d.f. F(x)
P(x)
F(x)
0.8
0.95
0.7
0.6
0.75
0.5
0.4
0.55
0.3
0.35
0.2
0.15
0.1
0
0
1
2
x
3
4
-1
-0.05
0
1
2
3
x
 Points of discontinuity of F(x) are points, where P(x) is nonzero.
4
Relationship between p.d.f. P(x) and c.d.f. F(x)
P(x)
F(x)
0.8
0.95
0.7
0.6
0.75
0.5
0.4
0.55
0.3
0.35
0.2
0.15
0.1
0
0
1
2
x
3
4
-1
-0.05
0
1
2
3
x
 Points of discontinuity of F(x) are points, where P(x) is nonzero.
4
Relationship between p.d.f. P(x) and c.d.f. F(x)
P(x)
F(x)
0.8
0.95
0.7
0.6
0.75
0.5
0.4
0.55
0.3
0.35
0.2
0.15
0.1
0
0
1
2
x
3
4
-1
-0.05
0
1
2
3
x
 Points of discontinuity of F(x) are points, where P(x) is nonzero.
 𝑃 𝑥 = lim 𝐹(𝑥) − 𝐹 𝑎
𝑥→𝑎+
4
Relationship between p.d.f. P(x) and c.d.f. F(x)
P(x)
F(x)
0.8
0.95
0.7
0.6
0.75
0.5
0.4
0.55
0.3
0.35
0.2
0.15
0.1
0
0
1
2
x
3
4
-1
-0.05
0
1
2
3
x
 Points of discontinuity of F(x) are points, where P(x) is nonzero.
 𝑃 𝑥 = lim 𝐹(𝑥) − 𝐹 𝑎
𝑥→𝑎+
4
Relationship between P(x) and F(x)
𝑃 𝑋 = 𝑎 = lim 𝐹 𝑥 − 𝐹 𝑎
𝑥→𝑎+
Relationship between P(x) and F(x)
 𝑃 𝑋<𝑎 =𝐹 𝑎 =
𝑥𝑖 <𝑎 𝑃(𝑥𝑖 )
 𝑃 𝑋 ≥𝑎 =1−𝑃 𝑋 <𝑎 =1−𝐹 𝑎 =
𝑥𝑖 ≥𝑎 𝑃(𝑥𝑖 )
 𝑃 𝑎 ≤𝑋 <𝑏 =𝑃 𝑋 <𝑏 −𝑃 𝑋 <𝑎 =𝐹 𝑎 −𝐹 𝑏 =
𝑎≤𝑥𝑖 <𝑏 𝑃(𝑥𝑖 )
 𝑃 𝑋 = 𝑎 = lim 𝐹 𝑥 − 𝐹(𝑎)
𝑥→𝑎+
 Probability distribution function P(x) is nonzero if Cumulative distribution
function is not continous.
3. If a coin is tossed three times, the number of heads obtained can
be 0, 1, 2 or 3. The probabilities of each of these possibilities can
be tabulated as shown:
Number of Heads
Probability
0
1/8
1
3/8
Specify Cumulative Distribution Function.
2
3/8
3
1/8
4. Cumulative distribution function of random variable X is:.
0
 0.2

F  x   0.52
 0.93

 1
Specify Probability Mass Function.
x0
0 x2
2 x4
4 x8
x 8
Probability Distribution
of Continous Random Variable
Probability Density Function
 Unlike a discrete random variable, a continuous random variable is
one that can assume an uncountable number of values.
 We cannot list the possible values because there is an infinite
number of them.
 Because there is an infinite number of values, the probability of
each individual value is virtually 0.
X … continous random variable ⇒ ∀𝑥 ∈ ℝ: 𝑃 𝑋 = 𝑥 = 0
8.35
Point Probabilities are Zero
 because there is an infinite number of values, the probability of
each individual value is virtually 0.
 Thus, we can determine the probability of a range of values only.
 E.g. with a discrete random variable like tossing a die, it is
meaningful to talk about P(X=5), say.
 In a continuous setting (e.g. with time as a random variable), the
probability the random variable of interest, say task length, takes
exactly 5 minutes is infinitesimally small, hence P(X=5) = 0.
It is meaningful to talk about P(X ≤ 5).
Probability Density Function f(x)
A function f(x) is called a probability density function (over the range
a ≤ x ≤ b if it meets the following requirements:
 f(x) ≥ 0 for all x between a and b, and
f(x)
a
area=1
b
x
 The total area under the curve between a and b is 1.0
Relationship between probability density
function f(x) and distribution function F(x)
𝑥
𝐹 𝑥 =
𝑓 𝑡 𝑑𝑡
−∞
Relationship between probability density
function f(x) and distribution function F(x)
𝑥
𝐹 𝑥 =
𝑓 𝑡 𝑑𝑡
−∞
Relationship between probability density
function f(x) and distribution function F(x)
𝑥
𝐹 𝑥 =
𝑓 𝑡 𝑑𝑡
−∞
5. X is a continuous random variable with probability density function
given by f(x) = 2x for 0 ≤ x ≤ 1.
Specify cumulative distribution function.
6. X is a continuous random variable with cumulative distribution
function given by:
0
𝑥2
𝐹 𝑥 =
4
1
𝑥<0
0≤𝑥≤2
𝑥>2
Specify probability density function f(x).
Relationship between probability,
p.d.f. f(x) and c.d.f. F(x)
 𝑃 𝑋<𝑎 =𝐹 𝑎 =
𝑎
𝑓
−∞
𝑥 𝑑𝑥
 𝑃 𝑋 ≥𝑎 =1−𝑃 𝑋 <𝑎 =1−𝐹 𝑎 =1−
𝑎
𝑓
−∞
𝑥 𝑑𝑥 =
∞
𝑓
𝑎
𝑥 𝑑𝑥
 𝑃 𝑎 ≤𝑋 <𝑏 =𝑃 𝑋 <𝑏 −𝑃 𝑋 <𝑎 =𝐹 𝑎 −𝐹 𝑏 =
=
𝑏
𝑓
−∞
𝑥 𝑑𝑥 −
𝑎
𝑓
−∞
𝑥 𝑑𝑥 =
𝑏
𝑓
𝑎
𝑥 𝑑𝑥
 𝑃 𝑋 = 𝑎 = lim 𝐹 𝑥 − 𝐹 𝑎 = 0
𝑥→𝑎+
 If the random variable X is continuous, it is possible in the above inequality
interchanged symbols sharp and unsharp inequality.
Numerical Characteristics of
Random Variable
Expected Value E(X)
 The expected value of a random variable X is the weighted average
of all possible values that this random variable can take on.
 The weights used in computing this average correspond to the
probabilities in case of a discrete random variable, or densities in
case of a continuous random variable.
Discrete Random Variable:
E 𝑋 =𝜇=
𝑖
𝑥𝑖 ∙ 𝑃 𝑥𝑖
Continous Random Variable:
E 𝑋 =𝜇=
∞
𝑥
−∞
∙ 𝑓 𝑥 𝑑𝑥
Laws of Expected Value
 𝐸 𝑐 =𝑐
The expected value of a constant (c) is just the value of the
constant.
 𝐸 𝑋+𝑐 =𝐸 𝑋 +𝑐
The expected value of a random variable plus a constant is the
expected value of the random variable plus the constant
 𝐸 𝑐𝑋 = 𝑐𝐸 𝑋
The expected value of a constant times a random variable is the
constant times the expected value of the random variable.
Laws of Expected Value
 𝐸 𝑐1 𝑋1 + 𝑐2 𝑋2 + ⋯ + 𝑐𝑛 𝑋𝑛 = 𝑐1 𝐸 𝑋1 + 𝑐2 𝐸 𝑋2 +…+𝑐𝑛 𝐸 𝑋𝑛
7. If a coin is tossed three times, the number of heads obtained can
be 0, 1, 2 or 3. The probabilities of each of these possibilities can
be tabulated as shown:
Number of Heads
Probability
0
1/8
1
3/8
2
3/8
3
1/8
Calculate expected value of random variable X.
x
P(x)
x.P(x)
0
1/8
0
1
3/8
3/8
E 𝑋 =
𝑥𝑖 ∙ 𝑃 𝑥𝑖
𝑖
2
3/8
6/8
3
1/8
3/8
12 3
=
= = 𝟏, 𝟓
8
2
Total
1
12/8
8. X is a continuous random variable with probability density
function given by f(x) = 2x for 0 ≤ x ≤ 1. Calculate expected value
of random variable X.
∞
E 𝑋 =
1
𝑥 ∙ 𝑓 𝑥 𝑑𝑥 =
−∞
0
2𝑥 3
𝑥 ∙ 2𝑥𝑑𝑥 =
3
1
2
𝟐
= −0=
3
𝟑
Variance (Dispersion) D(X)
 The variance is a measure of how far a set of numbers is spread out.
It is one of several descriptors of a probability distribution,
describing how far the numbers lie from the mean (expected value).
2
𝐷 𝑋 =𝜎 =𝐸 𝑋−𝐸 𝑋
2
= E 𝑋2 − 𝐸 𝑋
Discrete Random Variable:
D 𝑋 =
𝑖
𝑥𝑖 − 𝐸 𝑋
Continous Random Variable:
D 𝑋 =
∞
−∞
𝑥−𝐸 𝑋
2
2
2
∙ 𝑃 𝑥𝑖
∙ 𝑓 𝑥 𝑑𝑥
Laws of Variance
 𝐷 𝑐 =0
The variance of a constant (c) is zero.
 𝐷 𝑋+𝑐 =𝐷 𝑋
The variance of a random variable and a constant is just the
variance of the random variable.
 𝐷 𝑐𝑋 = 𝑐 2 𝐷 𝑋
The variance of a random variable and a constant coefficient is the
coefficient squared times the variance of the random variable.
Laws of Variance
 𝐷 𝑐1 𝑋1 + 𝑐2 𝑋2 + ⋯ + 𝑐𝑛 𝑋𝑛 = 𝑐1 2 𝐷 𝑋1 + 𝑐2 2 𝐷 𝑋2 + ⋯ + 𝑐𝑛 2 𝐷 𝑋𝑛
if 𝑋1 , 𝑋2 , …, 𝑋𝑛 are independent variable
9. You weight all 30 000 students.
 Random Variable: X = students weight
 𝐸 𝑋 = 160 𝑙𝑏𝑠
 𝐷 𝑋 = 900 𝑙𝑏𝑠 2
----------------------------------------------------------------------You now discover that the scales reported a student’s weight 5 lbs too
heavy. The student’s real weights (Y) should have been Y = X – 5.
What are the mean and variance of the student’s REAL weights?
 𝐸 𝑌 = 𝐸 𝑋 − 5 = 𝐸 𝑋 − 5 = 160 𝑙𝑏𝑠
 𝐷 𝑌 = 𝐷 𝑋 − 5 = 𝐷 𝑋 = 900 𝑙𝑏𝑠 2
10. You measure the height of all 30 000 students.
 Random Variable: X = students height in “Feet”
 𝐸 𝑋 = 5,8 𝑓𝑒𝑒𝑡
 𝐷 𝑋 = 0,09 𝑓𝑒𝑒𝑡 2
----------------------------------------------------------------------You now discover that the President wanted to measure student’s
heights in “Inches” and not “Feet”. The student’s height in “Inches”
(Y) should have been Y = 12*X . What are the mean and variance of
the student’s heights in Inches?
 𝐸 𝑌 = 𝐸 12𝑋 = 12𝐸 𝑋 = 69,6 𝑖𝑛𝑐ℎ
 𝐷 𝑌 = 𝐷 12𝑋 = 122 𝐷 𝑋 = 12,96 𝑖𝑛𝑐ℎ2
11. X = weight of right shoes: E(X) = 0,5 lbs and D(X) = 0,0004 lbs2
Y = weight of left shoes: E(Y) = 0,5 lbs and D(Y) = 0,0004 lbs2
What is the mean and variance of a “Pair” of shoes?
 NOTE: WEIGHTS OF RIGHT AND LEFT SHOE INDEPENDENT.
 𝑃 =𝑋+𝑌
 𝐸 𝑃 = 𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸 𝑌 = 1,0 𝑙𝑏𝑠
 𝐷 𝑃 = 𝐷 𝑋 + 𝑌 = 𝐷 𝑋 + 𝐷 𝑌 = 0,0008 𝑙𝑏𝑠 2
Median of Continous Random Variable
 Median 𝑥0,5
- information about the “middle” value of the random variable
𝑃 𝑋 < 𝑥0,5 = 𝐹 𝑥0,5 = 0,5
 Symmetric Random Variable
If a continuous random variable is symmetric about a point 𝜇, then
both the median and the expectation of the random variable are
equal to 𝜇.
Quantiles of Continous Random Variable
 The p-th quantile of a random variable X = 𝑥𝑝 .
A probability of p that the random variable takes a value less
than the p-th quantile.
𝑃 𝑋 < 𝑥𝑝 = 𝐹 𝑥𝑝 = 𝑝
 Upper quartile 𝑥0,75
The 75th percentile of the distribution
 Lower quartile 𝑥0,25
The 25th percentile of the distribution
 Interquartile range 𝐼𝑄𝑅 = 𝑥0,75-𝑥0,25
The distance between the two quartiles
12. X is a continuous random variable with probability density
function given by f(x) = 2x for 0 ≤ x ≤ 1. Calculate lower quartile
𝑥0,25 .
𝐹 𝑥0,25 = 0,25
0
𝑥
𝐹 𝑥 =
𝑥
𝑓 𝑡 𝑑𝑡 =
−∞
𝑥≤0
2𝑡𝑑𝑡 = 𝑥 2 ,
0
1
𝐹 𝑥0,25 = 0,25 ⇒ 𝑥0,25 ∈ 0; 1
𝑥0,25 2 = 0,25
𝑥0,25 = ±0,5
𝑥0,25 = 𝟎, 𝟓
0<𝑥≤1
𝑥>1
Linear Functions of Random Variables
 If X is a random variable and 𝑌 = 𝑎𝑋 + 𝑏 for some numbers 𝑎, 𝑏 ∈
ℝ then 𝐸 𝑌 = 𝑎𝐸 𝑋 + 𝑏 and 𝐷 𝑌 = 𝑎2 𝐷 𝑋 .
 𝐹𝑌 𝑦 = 𝑃 𝑌 < 𝑦 = 𝑃 𝑎𝑋 + 𝑏 < 𝑦 = 𝑃 𝑋 <
𝑦−𝑏
𝑎
= 𝐹𝑋
 If X is a discrete random variable then
𝑃 𝑌 = 𝑦 = 𝑃 𝑎𝑋 + 𝑏 = 𝑦 = 𝑃 𝑋 =
𝑦−𝑏
𝑎
=
𝑦−𝑏
𝑃𝑋
𝑎
𝑦−𝑏
𝑎
13. Random variable X is the number of heads if a coin is tossed three
times. The probability mass function of X can be tabulated as
shown:
x
P(x)
0
1/8
1
3/8
2
3/8
3
1/8
𝑌 = 1 − 2𝑋. Calculate 𝑃𝑌 𝑦 , 𝐹𝑌 𝑦 , 𝐸 𝑌 and 𝐷 𝑌 .
Total
1
13. X is a continuous random variable with cumulative
distribution function given by:
0
𝐹𝑋 𝑥 = 𝑥
1
𝑥<0
0≤𝑥≤1
𝑥>1
𝑌 = 1 − 2𝑋. Calculate 𝐹𝑌 𝑦 , 𝑓𝑌 𝑦 , 𝐸 𝑌 and 𝐷 𝑌 .
Nonlinear Functions of Random Variables
 A nonlinear function of a random variable X is another random
variable Y=g(X) for some nonlinear function g.
𝑌 = 𝑋2,
𝑌 = 𝑋,
𝑌 = 𝑒 𝑋, …
 There are no general results that relate the expected value and
variance of the random variable Y to the expectation and variance of
the random variable X.
14. Random variable X is the number of heads if a coin is tossed three
times. The probability mass function of X can be tabulated as
shown:
x
P(x)
-1
1/8
0
3/8
1
3/8
𝑌 = 𝑋 2 . Calculate 𝑃𝑌 𝑦 , 𝐹𝑌 𝑦 , 𝐸 𝑌 and 𝐷 𝑌 .
2
1/8
Total
1
15. X is a continuous random variable with cumulative distribution
function given by:
0
𝐹𝑋 𝑥 = 𝑥
1
𝑥<0
0≤𝑥≤1
𝑥>1
𝑌 = 𝑋 2 . Calculate 𝐹𝑌 𝑦 , 𝑓𝑌 𝑦 , 𝐸 𝑌 and 𝐷 𝑌 .
Study materials :
 http://homel.vsb.cz/~bri10/Teaching/Bris%20Prob%20&%20Stat.pdf
(p. 55 - p.63)
 http://stattrek.com/random-variable/combination.aspx?tutorial=ap
(Random Variable)