Transcript Probability theory

Random Vector
Martina Litschmannová
[email protected]
K210
Random Vectors
 An k- dimensional random vector is a function X = (𝑋1, … , 𝑋𝑘 )𝑇 that
associates a vector of real numbers with each element is a random
variable.
For example:
 A semiconductor chip is divided into ‘M’ regions. For the random
experiment of finding the number of defects and their locations, let
𝑁𝑖 denote the number of defects in ith region. Then 𝑵 =
(𝑁1, … , 𝑁𝑀)𝑇 is a discrete random vector.
 In a random experiment of selecting a student’s name, let 𝐻𝑖 =
height of ith student in inches and 𝑊𝑖 = weight of ith student in
pounds. Then 𝑺 = (𝐻𝑖, 𝑊𝑖 )𝑇 is a continuous random vector.
2 – dimensional Random Vectors
 We're going to focus on 2-dimensional distributions (i.e. random
vector consists only of two random variables) but higher
dimensions (more than two variables) are also possible.
Joint Probability Distribution
 Joint distribution for random vector 𝑋, 𝑌 𝑇 defines the probability
of events defined in terms of both X and Y.
Joint cumulative distribution function for 𝑿 = 𝑋, 𝑌 𝑇 is given by
𝐹 𝑥, 𝑦 = 𝑃 𝑋 < 𝑥 ∧ 𝑌 < 𝑦 = 𝑃 𝑋 < 𝑥; 𝑌 < 𝑦 .
𝐹 𝑥, 𝑦
Joint cumulative distribution function
Joint cumulative distribution function for 𝑿 = 𝑋, 𝑌 𝑇 is given by
𝐹 𝑥, 𝑦 = 𝑃 𝑋 < 𝑥 ∧ 𝑌 < 𝑦 = 𝑃 𝑋 < 𝑥; 𝑌 < 𝑦 .
Properties of joint CDF:
1.
2.
3.
∀ 𝑥, 𝑦 ∈ ℝ2 : 0 ≤ 𝐹 𝑥, 𝑦 ≤ 1,
lim 𝐹 𝑥, 𝑦 = lim 𝐹 𝑥, 𝑦 = 0,
𝑥→−∞
lim
𝑥,𝑦 → ∞,∞
𝑦→−∞
𝐹 𝑥, 𝑦 = 1,
4. 𝐹 𝑥, 𝑦 is nondecreasing in each variable,
5. 𝐹 𝑥, 𝑦 is continous from the left in each variable.
6. 𝑃 𝑎 ≤ 𝑋 < 𝑏, 𝑐 ≤ 𝑌 < 𝑑 = 𝐹 𝑏, 𝑑 − 𝐹 𝑎, 𝑑 − 𝐹 𝑏, 𝑐 + 𝐹 𝑎, 𝑐
Discrete Joint Probability Distributions
The probability function, also known as the probability mass function
for a joint probability distribution is defined such that:
𝑝 𝑥𝑖 , 𝑦𝑗 = 𝑃 𝑋 = 𝑥𝑖 ∧ 𝑌 = 𝑦𝑗 = 𝑃 𝑋 = 𝑥𝑖 ; 𝑌 = 𝑦𝑗 .
𝑝 𝑥𝑖 , 𝑦𝑗 = P 𝑋 = 𝑥𝑖 |𝑌 = 𝑦𝑗 ∙ 𝑃 𝑌 = 𝑦𝑗 = P 𝑌 = 𝑦𝑗 |𝑋 = 𝑥𝑖 ∙ 𝑃 𝑋 = 𝑥𝑖
𝑛1 𝑛2
𝑝 𝑥𝑖 , 𝑦𝑗 = 1,
𝑖=1 𝑗=1
𝑛1
≥ 1, 𝑛2 ≥ 1
Discrete Joint Probability Distributions
Probability Mass Function for a Joint Probability Distribution :
𝑝 𝑥𝑖 , 𝑦𝑗 = 𝑃 𝑋 = 𝑥𝑖 ∧ 𝑌 = 𝑦𝑗 = 𝑃 𝑋 = 𝑥𝑖 ; 𝑌 = 𝑦𝑗
Properties of p.m.f.:
1. 𝑝 𝑥𝑖 , 𝑦𝑗 > 0 only for a finite or countable set of values 𝑥𝑖 , 𝑦𝑗
2. 0 ≤ 𝑝 𝑥𝑖 , 𝑦𝑗 ≤ 1,
𝑛1
𝑛2
3.
𝑛1 ≥ 1, 𝑛2 ≥ 1,
𝑖=1 𝑗=1 𝑝 𝑥𝑖 , 𝑦𝑗 = 1,
4. 𝑝 𝑥𝑖 , 𝑦𝑗 = P 𝑋 = 𝑥𝑖 |𝑌 = 𝑦𝑗 ∙ 𝑃 𝑌 = 𝑦𝑗 = P 𝑌 = 𝑦𝑗 |𝑋 = 𝑥𝑖 ∙
𝑃 𝑋 = 𝑥𝑖 ,
5. If X, Y are independent 𝑝 𝑥𝑖 , 𝑦𝑗 = P 𝑋 = 𝑥𝑖 ∙ 𝑃 𝑌 = 𝑦𝑗
Table of joint probabilities
X\Y
x1
x2
y1
p(x1, y1)
p(x2, y1)
y2
p(x1, y2)
p(x2, y2)
xn1
p(xn1, y1) p(xn1, y2)
...
...
...
...
...
yn2
p(x1, yn2)
p(x2, yn2)
p(xn1, yn2)
1. A random experiment consists of tossing coin (X) and flipping die
(Y). Find probability mass function for a joint probability
distribution of a random vector 𝑋, 𝑌 .
X/Y
0 (head)
1 (tail)
1
1/12
1/12
2
1/12
1/12
3
1/12
1/12
4
1/12
1/12
5
1/12
1/12
6
1/12
1/12
1. A random experiment consists of tossing coin (X) and flipping die
(Y). Find probability mass function for a joint probability
distribution of a random vector 𝑋, 𝑌 .
X/Y
0 (head)
1 (tail)
1
1/12
1/12
2
1/12
1/12
3
1/12
1/12
4
1/12
1/12
5
1/12
1/12
6
1/12
1/12
1. A random experiment consists of tossing coin (X) and flipping die
(Y). Find probability mass function for a joint probability
distribution of a random vector 𝑋, 𝑌 .
X/Y
0 (head)
1 (tail)
1
1/12
1/12
2
1/12
1/12
3
1/12
1/12
4
1/12
1/12
5
1/12
1/12
6
1/12
1/12
1
control cell
2. Probability mass function for a joint probability distribution of a
random vector 𝑋, 𝑌 is given as:
X/Y
-1
0
1
-2
0,13
0
0,23
0
0,11
0,11
0,01
1
0,07
0
0
2
0
0,33
0,01
1
Find:
a) 𝑝 0; 1
b) 𝑝 1; 0
c) 𝑝 3; 1
d) 𝑃 𝑋 < 1,3; 𝑌 < −0,6
e) 𝑃 𝑋 < 1,3; 𝑌 > −0,6
f) 𝐹 0,7; 1,3
Continous Joint Probability Distributions
𝐹 𝑥, 𝑦 =
𝑦
𝑥
𝑓
−∞ −∞
𝑠, 𝑡 𝑑𝑠 𝑑𝑡,
where 𝑓 𝑥, 𝑦 is Joint Probability Density Function.
Properties of Joint Probability Density Function:
1. 𝑓 𝑥, 𝑦 ≥ 0,
2.
∞ ∞
𝑓
−∞ −∞
3. If
𝜕2 𝐹 𝑥,𝑦
𝜕𝑥𝜕𝑦
𝑥, 𝑦 𝑑𝑥 𝑑𝑦 = 1,
exists, pak 𝑓 𝑥, 𝑦 =
4. 𝑃 𝑎 ≤ 𝑋 < 𝑏, 𝑐 ≤ 𝑌 < 𝑑 =
𝜕2 𝐹 𝑥,𝑦
𝜕𝑥𝜕𝑦
𝑑 𝑏
𝑓
𝑐 𝑎
,
𝑥, 𝑦 𝑑𝑥 𝑑𝑦.
3.
Find the constant c so that function
𝑓 𝑥, 𝑦 =
𝑐 𝑥+𝑦
0
𝑥, 𝑦 ∈ 0; 1 × 0; 1
𝑥, 𝑦 ∉ 0; 1 × 0; 1
can be a joint probability density function of a random vector 𝑋, 𝑌 𝑇 .
That the function 𝑓 𝑥, 𝑦 can be a joint probability density function of a
random vector 𝑋, 𝑌 𝑇 , it has be satisfy condition
∞ ∞
𝑓
−∞ −∞
1 1
𝑐
0 0
1 1
𝑐
0 0
𝑥, 𝑦 𝑑𝑥 𝑑𝑦 = 1.
𝑥 + 𝑦 𝑑𝑥 𝑑𝑦 = 1
𝑥 + 𝑦 𝑑𝑥 𝑑𝑦 = 𝑐
=𝑐
1 𝑥2
0 2
+ 𝑥𝑦
1 1
+
0 2
1
0
𝑑𝑦 = 𝑐
𝑦 𝑑𝑦 = 𝑐
1
𝑦
2
1 𝑥2
0 2
+
+ 𝑥𝑦
1
𝑦2
2 0
1
0
𝑑𝑦 =
=𝑐 ⇒ 𝑐=1
Marginal probability distributions
Obtained by summing or integrating the joint probability distribution
over the values of the other random variable.
 Discrete Random Vector
𝑃𝑋 𝑥𝑖 =
𝑦𝑗
𝑝 𝑥𝑖 , 𝑦𝑗 , 𝑖 ≥ 1,
𝑃𝑌 𝑦𝑗 =
𝑥𝑖
𝑝 𝑥𝑖 , 𝑦𝑗 , 𝑗 ≥ 1.
 Continous Random Vector
𝑓𝑋 𝑥 =
𝑓𝑌 𝑦 =
∞
𝑓
−∞
∞
𝑓
−∞
𝑥, 𝑦 𝑑𝑦, 𝑥 ∈ ℝ,
𝑥, 𝑦 𝑑𝑥 , 𝑦 ∈ ℝ.
4. A random experiment consists of tossing coin (X) and flipping die
(Y). Probability mass function for a joint probability distribution of
a random vector 𝑋, 𝑌 is given as:
X/Y
0 (head)
1 (tail)
1
1/12
1/12
2
1/12
1/12
3
1/12
1/12
4
1/12
1/12
5
1/12
1/12
6
1/12
1/12
1
Find Marginal Probability Mass Functions 𝑃𝑋 𝑥𝑖 and 𝑃𝑌 𝑦𝑗 .
4. A random experiment consists of tossing coin (X) and flipping die
(Y). Probability mass function for a joint probability distribution of
a random vector 𝑋, 𝑌 is given as:
X/Y
0 (head)
1 (tail)
𝑃𝑌 𝑦𝑗
1
1/12
1/12
2/12
2
1/12
1/12
2/12
3
1/12
1/12
2/12
4
1/12
1/12
2/12
5
1/12
1/12
2/12
6
𝑃𝑋 𝑥𝑖
1/12 6/12
1/12 6/12
2/12
1
Find Marginal Probability Mass Functions 𝑃𝑋 𝑥𝑖 and 𝑃𝑌 𝑦𝑗 .
4. A random experiment consists of tossing coin (X) and flipping die
(Y). Probability mass function for a joint probability distribution of
a random vector 𝑋, 𝑌 is given as:
X/Y
0 (head)
1 (tail)
𝑃𝑌 𝑦𝑗
1
1/12
1/12
2/12
2
1/12
1/12
2/12
3
1/12
1/12
2/12
4
1/12
1/12
2/12
5
1/12
1/12
2/12
6
𝑃𝑋 𝑥𝑖
1/12 6/12
1/12 6/12
2/12
1
Marginal Probability Mass Functions 𝑃𝑋 𝑥𝑖 and 𝑃𝑌 𝑦𝑗 .
Y
𝑃𝑌 𝑦𝑗
X
𝑃𝑋 𝑥𝑖
1
2/12
2
2/12
0 (head)
6/12
3
2/12
1 (tail)
6/12
4
2/12
5
2/12
6
2/12
4. A random experiment consists of tossing coin (X) and flipping die
(Y). Probability mass function for a joint probability distribution of
a random vector 𝑋, 𝑌 is given as:
X/Y
0 (head)
1 (tail)
𝑃𝑌 𝑦𝑗
1
1/12
1/12
1/6
2
1/12
1/12
1/6
3
1/12
1/12
1/6
4
1/12
1/12
1/6
5
1/12
1/12
1/6
6
𝑃𝑋 𝑥𝑖
1/12
1/2
1/12
1/2
1/6
1
Marginal Probability Mass Functions 𝑃𝑋 𝑥𝑖 and 𝑃𝑌 𝑦𝑗 .
Y
𝑃𝑌 𝑦𝑗
X
𝑃𝑋 𝑥𝑖
1
1/6
2
1/6
0 (head)
1/2
3
1/6
1 (tail)
1/2
4
1/6
5
1/6
6
1/6
5. A certain farm produces two kinds of eggs on any given day;
organic and non-organic. Let these two kinds of eggs be
represented by the random variables X and Y respectively. Given
that the joint probability density function of these variables is
given by
2
𝑓 𝑥; 𝑦 = 3 𝑥 + 2𝑦
0
0 ≤ 𝑥 ≤ 1; 0 ≤ 𝑦 ≤ 1
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒.
Find:
a) marginal density functions 𝑓𝑋 𝑥 and 𝑓𝑌 𝑦 .
∞
𝑓𝑋 𝑥 =
1
𝑓 𝑥; 𝑦 𝑑𝑦 =
−∞
∞
𝑓𝑌 𝑦 =
0
1
𝑓 𝑥; 𝑦 𝑑𝑥 =
−∞
0
2
2
𝑥 + 2𝑦 𝑑𝑦 = 𝑥𝑦 + 𝑦 2
3
3
1
0
1
2
2
2 𝑥
𝑥 + 2𝑦 𝑑𝑥 =
+ 2𝑥𝑦
3
3 2
2
= 𝑥+1
3
0
1
= 1 + 4𝑦
3
5. A certain farm produces two kinds of eggs on any given day;
organic and non-organic. Let these two kinds of eggs be
represented by the random variables X and Y respectively. Given
that the joint probability density function of these variables is
given by
2
𝑓 𝑥; 𝑦 = 3 𝑥 + 2𝑦
0
Find:
b) 𝑃 𝑋
𝑃 𝑋<
1
;𝑌
2
≤
1
1
< ;𝑌 ≤
2
2
0,5 0,5 2
1
2
=
0
0
3
0 ≤ 𝑥 ≤ 1; 0 ≤ 𝑦 ≤ 1
𝑥 + 2𝑦 𝑑𝑥𝑑𝑦 =
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒.
0,5 2 𝑥 2
0
3 2
+
5. A certain farm produces two kinds of eggs on any given day;
organic and non-organic. Let these two kinds of eggs be
represented by the random variables X and Y respectively. Given
that the joint probability density function of these variables is
given by
2
𝑓 𝑥; 𝑦 = 3 𝑥 + 2𝑦
0
Find:
c) 𝑃 𝑋 <
1
1
1
;𝑌
2
𝑃 𝑋 < 2;𝑌 > 4 =
=
>
0 ≤ 𝑥 ≤ 1; 0 ≤ 𝑦 ≤ 1
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒.
1
4
1
0,5 2
0,25 0 3
1 2 1
0,25 3 8
𝑥 + 2𝑦 𝑑𝑥𝑑𝑦 =
+ 𝑦 𝑑𝑦 =
1
2
0,25 3
2 1
𝑦
3 8
+
𝑥2
2
0,5
+ 2𝑥𝑦
1
𝑦2
2
0,25
=
𝑑𝑦 =
0
5
12
−
1
24
=
𝟑
𝟖
Conditional probability distributions
Conditional Probability Distributions arise from joint probability
distributions where by we need to know that probability of one event
given that the other event has happened, and the random variables
behind these events are joint.
 Discrete Random Vector
𝑃 𝑥𝑖 |𝑦𝑗
𝑝 𝑥𝑖 ,𝑦𝑗
= 𝑃 𝑦
𝑌 𝑗
𝑝 𝑥𝑖 ,𝑦𝑗
𝑃 𝑦𝑗 |𝑥𝑖 =
𝑃𝑋 𝑥𝑖
,
𝑃𝑌 𝑦𝑗 ≠0,
,
𝑃𝑋 𝑥𝑖 ≠ 0.
 Continous Random Vector
𝑓 𝑥|𝑦 =
𝑓 𝑦|𝑥 =
𝑓 𝑥,𝑦
𝑓𝑌 𝑦
𝑓 𝑥,𝑦
𝑓𝑋 𝑥
,
𝑓𝑌 𝑦 ≠0,
,
𝑓𝑋 𝑥 ≠ 0.
6. Joint and marginal probability distribution of a random vector
𝑋, 𝑌 is given as:
X/Y
-1
0
1
𝑃𝑌 𝑦𝑗
Find:
a) 𝑃 𝑋 = 0|𝑌
b) 𝑃 𝑌 = 0|𝑋
c) 𝑃 𝑋 = 0|𝑌
d) 𝑃 𝑌 = 0|𝑋
e) 𝑃 𝑥|𝑦
f) 𝑃 𝑦|𝑥
=1
=1
=2
=2
-2
0,13
0
0,23
0,36
0
0,11
0,11
0,01
0,23
1
0,07
0
0
0,07
2
0
0,33
0,01
0,34
𝑃𝑋 𝑥𝑖
0,31
0,44
0,25
1
7. Joint probability density function of 𝑋, 𝑌 is given by
2
𝑓 𝑥; 𝑦 = 3 𝑥 + 2𝑦
0
0 ≤ 𝑥 ≤ 1; 0 ≤ 𝑦 ≤ 1
Find:
a) conditional density function 𝑓 𝑦|𝑥 ,
𝑓 𝑦|𝑥 =
𝑓 𝑥, 𝑦
𝑓𝑋 𝑥
2
𝑓𝑋 𝑥 = 3 𝑥 + 1
0
0≤𝑥≤1
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒.
7. Joint probability density function of 𝑋, 𝑌 is given by
2
𝑓 𝑥; 𝑦 = 3 𝑥 + 2𝑦
0
0 ≤ 𝑥 ≤ 1; 0 ≤ 𝑦 ≤ 1
Find:
b) conditional density function 𝑓 𝑥|𝑦 ,
𝑓 𝑥|𝑦 =
𝑓 𝑥, 𝑦
𝑓𝑌 𝑦
1
𝑓𝑌 𝑦 = 3 1 + 4𝑦
0
0≤𝑦≤1
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒.
Conditional expected value (expectation)
Conditional expectation is the expected value of a real random variable
with respect to a conditional probability distribution.
Discrete random vector:
 𝐸 𝑋|𝑌 = 𝑦 = (𝑥) 𝑥 ∙ 𝑃 𝑥|𝑌 = 𝑦
 𝐸 𝑌|𝑋 = 𝑥 =
(𝑦) 𝑦
∙ 𝑃 𝑦|𝑋 = 𝑥
Continous random vector:
 𝐸 𝑋|𝑌 = 𝑦 =
 𝐸 𝑌|𝑋 = 𝑥 =
∞
𝑥
−∞
∞
𝑦
−∞
∙ 𝑓 𝑥|𝑌 = 𝑦 𝑑𝑥
∙ 𝑓 𝑦|𝑋 = 𝑥 𝑑𝑦
Conditional variance
Conditional variance is the variance of a conditional probability
distribution.
𝐷 𝑋|𝑌 = 𝑦 = 𝐸
𝑋 − 𝐸 𝑋|𝑌 = 𝑦
𝐷 𝑌|𝑋 = 𝑥 = 𝐸
𝑌 − 𝐸 𝑌|𝑋 = 𝑥
2
|𝑌 = 𝑦 = 𝐸 𝑋 2 |𝑌 = 𝑦 − 𝐸 𝑋|𝑌 = 𝑦
2
2
|𝑋 = 𝑥 = 𝐸 𝑌 2 |𝑋 = 𝑥 − 𝐸 𝑌|𝑋 = 𝑥
2
8. Joint and marginal probability distribution of a random vector
𝑋, 𝑌 is given as:
X/Y
-1
0
1
𝑃𝑌 𝑦𝑗
Find:
a) 𝐸 𝑋|𝑌 = 1
b) 𝐷 𝑋|𝑌 = 1
-2
0,13
0
0,23
0,36
0
0,11
0,11
0,01
0,23
1
0,07
0
0
0,07
2
0
0,33
0,01
0,34
𝑃𝑋 𝑥𝑖
0,31
0,44
0,25
1
9. Joint probability density function of 𝑋, 𝑌 is given by
2
𝑓 𝑥; 𝑦 = 3 𝑥 + 2𝑦
0
Find:
a) E 𝑌|𝑋 = 0,5 ,
b) D 𝑌|𝑋 = 0,5 .
0 ≤ 𝑥 ≤ 1; 0 ≤ 𝑦 ≤ 1
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒.
Independence
Two random variables X and Y are independent if
 Discrete Random Variables
∀𝑖, 𝑗:
𝑝 𝑥𝑖 , 𝑦𝑗 = 𝑃𝑋 𝑥𝑖 ∙ 𝑃𝑌 𝑦𝑗
 Continous Random Vector
∀𝑥, 𝑦:
𝑓 𝑥, 𝑦 = 𝑓𝑋 𝑥 ∙ 𝑓𝑌 𝑦
.
Measures of Linear Independence
Covariance:
𝜎𝑋𝑌 = 𝑐𝑜𝑣 𝑋, 𝑌 = 𝐸 𝑋 − 𝐸𝑋 𝑌 − 𝐸𝑌
 𝜎𝑋𝑌 ∈ −∞, ∞
Correlation coefficient:
𝜌𝑋𝑌
𝑐𝑜𝑣 𝑋, 𝑌
=
𝜎𝑋 𝜎𝑌
 𝜌𝑋𝑌 ∈ −1; 1
 𝜌𝑋𝑌 is a scaled version of covariance
= 𝐸 𝑋𝑌 − 𝐸𝑋 ∙ 𝐸𝑌
Covariance
Covariance:
𝜎𝑋𝑌 = 𝑐𝑜𝑣 𝑋, 𝑌 = 𝐸 𝑋 − 𝐸𝑋 𝑌 − 𝐸𝑌
 𝜎𝑋𝑌 ∈ −∞, ∞
Covariance matrix:
𝐷𝑋
𝑐𝑜𝑣 𝑋, 𝑌
𝑐𝑜𝑣 𝑋, 𝑌
𝐷𝑌
= 𝐸 𝑋𝑌 − 𝐸𝑋 ∙ 𝐸𝑌
Correlation
Correlation:
𝜌𝑋𝑌




𝜌𝑋𝑌
𝜌𝑋𝑌
𝜌𝑋𝑌
𝜌𝑋𝑌
𝑐𝑜𝑣 𝑋, 𝑌
=
𝜎𝑋 𝜎𝑌
∈ −1; 1
> 0 … 𝑋, 𝑌 are positively correlated
< 0 … 𝑋, 𝑌 are negatively correlated
= 0 … 𝑋, 𝑌 are uncorrelated
Correlation matrix:
1
𝜌𝑋𝑌
𝜌𝑋𝑌
1
Correlation
𝜌 𝑋, 𝑌
=1,000
𝜌 𝑋, 𝑌
=0,967
𝜌 𝑋, 𝑌
𝜌 𝑋, 𝑌
= -1,000
𝜌 𝑋, 𝑌
=0,000
𝜌 𝑋, 𝑌
=0,934
=0,857
𝜌 𝑋, 𝑌
=-0,143
𝜌 𝑋, 𝑌
=0,608
10. Joint and marginal probability distribution of a random vector
𝑋, 𝑌 is given as:
X/Y
-1
0
1
𝑃𝑌 𝑦𝑗
-2
0,13
0
0,23
0,36
0
0,11
0,11
0,01
0,23
1
0,07
0
0
0,07
2
0
0,33
0,01
0,34
Find:
a) 𝐸 𝑋 , 𝐸 𝑌
b) 𝐷 𝑋 , 𝐷 𝑌
c) 𝜎 𝑋 , 𝜎 𝑌
d) 𝑐𝑜𝑣(𝑋, 𝑌)
e) 𝜌 𝑋, 𝑌
f) Are random variable X, Y independent?
g) Are random variable X, Y linear independent?
𝑃𝑋 𝑥𝑖
0,31
0,44
0,25
1
11. Joint probability density function of 𝑋, 𝑌 is given by
2
𝑓 𝑥; 𝑦 = 3 𝑥 + 2𝑦
0
Marginal density functions are:
2
𝑓𝑋 𝑥 = 3 𝑥 + 1
0
0≤𝑥≤1 ,
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
0 ≤ 𝑥 ≤ 1; 0 ≤ 𝑦 ≤ 1
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒.
1
𝑓𝑌 𝑦 = 3 1 + 4𝑦
0
Find:
a) 𝐸 𝑋 , 𝐸 𝑌
b) 𝐷 𝑋 , 𝐷 𝑌
c) 𝜎 𝑋 , 𝜎 𝑌
d) 𝑐𝑜𝑣(𝑋, 𝑌)
e) 𝜌 𝑋, 𝑌
f) Are random variable X, Y independent?
g) Are random variable X, Y linear independent?
0≤𝑦≤1
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
Study materials :
 http://homel.vsb.cz/~bri10/Teaching/Bris%20Prob%20&%20Stat.pdf
(p. 64 - p.70)