PS_Basic Probability part 4_PPT_2020.pdf

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Two Dimensional Random Variable Two Dimensional Discrete Random Variable Two Dimensional Continuous Random Variable
Basic Probability Part 4
Dr M A Patel
Assistant Professor
Mathematics
Government Engineering College, Gandhinagar,
GUJARAT(INDIA).
September 18, 2024

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Plan of Presentation
1 Two Dimensional Random Variable
2 Two Dimensional Discrete Random Variable
3 Two Dimensional Continuous Random Variable

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Two Dimensional Random Variable
Definition: (Two Dimensional Random Variable)
Let S be the sample space of a random experiment and let X and Y be
two random variables defined on S. Thus, X = X(s) and Y = Y (s)
are two functions which assign real numbers x and y to each outcomes
s ∈ S. Then the pair, (X, Y ) is called a two dimensional random
variable.

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Two Dimensional Discrete Random Variable
Definition: (Joint Probability Mass Function)
If (X, Y ) is the two dimensional discrete random variable, then the
probability that X = xi and Y = yj is given by
pij = P(X = xi , Y = yj)
then pij is called as the joint probability mass function or joint proba-
bility function of (X, Y ) provided it satisfies the following conditions:
(1) pij ≥ 0, ∀i, j
(2)
X
j
X
i
pij = 1.

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NOTE:
(1) pij = P(xi , yj) gives the probability that the random variable X
takes value xi and the random variable Y takes value yj simultaneously
i.e.
P(xi , yj) = P(X = xi ∩ Y = yj).
(2) The set of triplets {xi , yj, pij}, i = 1, 2, . . . , m and j = 1, 2, . . . , n
is called the joint probability distribution of discrete random variable
(X, Y ).

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Definition: (Joint Cumulative Distribution Function)
If (X, Y ) is a two dimensional discrete random variable then the func-
tion,
F(x, y) = FXY (x, y) = P(X ≤ x, Y ≤ y)
= P(−∞ < X ≤ x, −∞ < Y ≤ y)
=
X
j
X
i
P(X = xi , Y = yj), for xi ≤ x, yj ≤ y
is called the joint c.d.f of (X, Y ).

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Properties of jcdf:
1 F(−∞, y) = 0 = F(x, −∞)
2 F(∞, ∞) = 1
3 P(a < X ≤ b, Y ≤ y) = F(b, y) − F(a, y)
4 P(X ≤ x, c < y ≤ d) = F(x, d) − F(x, c)
5 P(a < X ≤ b, c < y ≤ d) =
F(b, d) + F(a, c) − F(a, d) − F(b, c)

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Definition: (Marginal Probability Function)
For a two dimensional discrete r.v. (X, Y ), the marginal probability
function of X is given by
pi∗ = PX (xi ) = P(X = xi )
= P(X = xi , Y = y1) orP(X = xi , Y = y2) or . . .
= pi1 + pi2 + . . .
=
X
j
pij
The set {xi , pi∗}, i = 1, 2, . . . , m is called the marginal (probability)
distribution of X.

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Similarly, the marginal probability function of Y is given by
p∗j = PY (yj) = P(Y = yj)
= p1j + p2j + . . .
=
X
i
pij
The set {yj, p∗j}, j = 1, 2, . . . , n is called the marginal (probability)
distribution of Y .

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Definition: (Conditional Probability Distribution)
For a discrete r.v. (X, Y ), the conditional probability function of X
given Y = yj is given by
P(X = xi /Y = yj) =
P(X = xi
T
Y = yj)
P(Y = yj)
=
pij
p∗j
The set
n
xi , pij
p∗j
o
, i = 1, 2, . . . m is called the conditional probability
distribution of X given Y = yj.

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Definition: (Conditional Probability Distribution)
Similarly, the conditional probability function of Y given X = xi is
given by
P(Y = yj/X = xi ) =
P(X = xi
T
Y = yj)
P(X = xi )
=
pij
pi∗
The set
n
yj, pij
pi∗
o
, j = 1, 2, . . . n is called the conditional probability
distribution of Y given X = xi .
The necessary and sufficient condition for the discrete r.v. X and Y
to be independent is P(X = xi , Y = yj) = P(X = xi ) × P(Y = yj)
for all values of (xi , yj) of (X, Y ).

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Expectation for Two Dimensional Discrete
Random Variable
Definition: (Expectation)
If (X, Y ) is a two-dimensional discrete r.v. with joint probability mass
function pij = P(X = xi , Y = yj) then the expectation of a function
g(X, Y ) is
E[g(X, Y )] =
X
j
X
i
g(xi , yj)pij.
Properties of Expected Values:
(1) E(X) =
P
i
xi pi∗ and E(Y ) =
P
j
yjp∗j.
(2) E(aX + bY ) = aE(X) + bE(Y ).
(3) If X and Y are independent random variables then
E(XY ) = E(X)E(Y ).
(4) The converse of above result need not be true.

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Example on Two Dimensional Discrete
Random Variable
Example-1
For the following bivariate probability distribution of (X, Y ), find:
(1) P(X ≤ 1) (2) P(Y ≤ 3)
(3) P(X ≤ 1, Y ≤ 3) (4) P(X ≤ 1/Y ≤ 3)
(5) P(Y ≤ 3/X ≤ 1) (6) P(X + Y ≤ 4)
(7) E(Y − 2X)
HH
HHH
H
X
Y
1 2 3 4 5 6
0 0 0 1/32 2/32 2/32 3/32
1 1/16 1/16 1/8 1/8 1/8 1/8
2 1/32 1/32 1/64 1/64 0 2/64

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Random Variable
Solution: Given is the probability distribution of two dimensional
discrete r.v. (X, Y ). Hence, the given figures represent P(X =
i, Y = j), i = 0, 1, 2, j = 1, 2, . . . 6.
The marginal probability
P(X = 0) = p0∗ = P(X = 0, Y = j), j = 1, 2, . . . , 6
= P(X = 0, Y = 1) + . . . + P(X = 0, Y = 6)
= 0 + 0 +
1
32
+
2
32
+
2
32
+
3
32
=
8
32
=
1
4
.

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Random Variable
P(X = 1) = p1∗ = P(X = 1, Y = j), j = 1, 2, . . . , 6
= P(X = 1, Y = 1) + . . . + P(X = 1, Y = 6)
=
1
16
+
1
16
+
1
8
+
1
8
+
1
8
+
1
8
=
10
16
=
5
8
.
P(X = 2) = p2∗ = P(X = 2, Y = j), j = 1, 2, . . . , 6
= P(X = 2, Y = 1) + . . . + P(X = 2, Y = 6)
=
1
32
+
1
32
+
1
64
+
1
64
+ 0 +
2
64
=
1
8
.

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Random Variable
P(Y = 1) = p∗1 = P(X = i, Y = 1), i = 0, 1, 2
= P(X = 0, Y = 1) + P(X = 1, Y = 1) + P(X = 2, Y = 1)
= 0 +
1
16
+
1
32
=
3
32
.
P(Y = 2) = p∗2 = P(X = i, Y = 2), i = 0, 1, 2
= P(X = 0, Y = 2) + P(X = 1, Y = 2) + P(X = 2, Y = 2)
= 0 +
1
16
+
1
32
=
3
32
.

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Random Variable
P(Y = 3) = p∗3 = P(X = i, Y = 3), i = 0, 1, 2
= P(X = 0, Y = 3) + P(X = 1, Y = 3) + P(X = 2, Y = 3)
=
1
32
+
1
8
+
1
64
=
11
64
.
P(Y = 4) = p∗4 = P(X = i, Y = 4, i = 0, 1, 2
= P(X = 0, Y = 4) + P(X = 1, Y = 4) + P(X = 2, Y = 4)
=
2
32
+
1
8
+
1
64
=
13
64
.

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Random Variable
P(Y = 5) = p∗5 = P(X = i, Y = 5, i = 0, 1, 2
= P(X = 0, Y = 5) + P(X = 1, Y = 5) + P(X = 2, Y = 5)
=
2
32
+
1
8
+ 0
=
6
32
=
3
16
.
P(Y = 6) = p∗6 = P(X = i, Y = 6, i = 0, 1, 2
= P(X = 0, Y = 6) + P(X = 1, Y = 6) + P(X = 2, Y = 6)
=
3
32
+
1
8
+
2
64
=
16
64
=
1
4
.

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Hence, the marginal probability is
H
HHH
H
H
X
Y
1 2 3 4 5 6 P(X = i)
0 0 0 1/32 2/32 2/32 3/32 1/4
1 1/16 1/16 1/8 1/8 1/8 1/8 5/8
2 1/32 1/32 1/64 1/64 0 2/64 1/8
P(Y = j) 3/32 3/32 11/64 13/64 3/16 1/4
(1)
P(X ≤ 1) = P(X = 0 or 1) =
1
4
+
5
8
=
7
8
.

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(2)
P(Y ≤ 3) = P(Y = 1 or 2 or 3) =
3
32
+
3
32
+
11
64
=
23
64
.
(3)
P(X ≤ 1, Y ≤ 3) = P(X ≤ 1 ∩ Y ≤ 3)
=
3
X
j=1
P(X = 0, Y = j) +
3
X
j=1
P(X = 1, Y = j)
= 0 + 0 +
1
32
+
1
16
+
1
16
+
1
8
=
9
32
.

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(4)
P(X ≤ 1/Y ≤ 3) =
P(X ≤ 1 ∩ Y ≤ 3)
P(Y ≤ 3)
=
P(X ≤ 1, Y ≤ 3)
P(Y ≤ 3)
=
9
32
23
64
=
18
23
.
(5)
P(Y ≤ 3/X ≤ 1) =
P(Y ≤ 3, X ≤ 1)
P(X ≤ 1)
=
9
32
7
8
=
9
28
.

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(6)
P(X + Y ≤ 4)
=
4
X
j=1
P(X = 0, Y = j) +
3
X
j=1
P(X = 1, Y = j) +
2
X
j=1
P(X = 2, Y = j)
= 0 + 0 +
1
32
+
2
32
+
1
16
+
1
16
+
1
8
+
1
32
+
1
32
=
3
32
+
1
4
+
1
16
=
13
32
.

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(7) E(Y − 2X)
E(X) =
X
xi pi∗
=
X
xi P(X = xi )
= 0

1
4

+ 1

5
8

+ 2

1
8

=
7
8
= 0.875.
E(Y ) =
X
yjp∗j =
X
yjP(Y = yj)
= 1

3
32

+ 2

3
32

+ 3

11
64

+ 4

13
64

+ 5

3
16

+ 6

1
4

=
6
=
259
64
= 4.047.

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So,
E(Y − 2X) = E(Y ) − 2E(X)
= 4.047 − 2(0.875)
= 2.297.

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Two Dimensional Continuous Random
Variable
Definition: (Joint Probability Density Function (JPDF))
If X and Y are continuous r.v. then their joint probability density
function is fXY (x, y) or f (x, y) if
P
(
x −
dx
2
≤ X ≤ x +
dx
2
, y −
dy
2
≤ Y ≤ y +
dy
2
)
= f (x, y)dxdy
and provided f (x, y) satisfies the following conditions
(1) f (x, y) ≥ 0, ∀(x, y)
(2)
∞
R
−∞
∞
R
−∞
f (x, y)dxdy = 1.

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Variable
NOTE: The probability P(a ≤ X ≤ b, c ≤ Y ≤ d) is given by
P(a ≤ X ≤ b, c ≤ Y ≤ d) =
d
Z
y=c
b
Z
x=a
f (x, y)dxdy.

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Variable
Definition: (Joint Cumulative Distribution Function)
If (X, Y ) is a two dimensional continuous r.v. then the function
F(x, y) = FXY (x, y) = P(X ≤ x, Y ≤ y)
= P(−∞ X ≤ x, −∞ Y ≤ y)
=
Z x
−∞
Z y
−∞
f (x, y)dydx
is called the joint c.d.f of (X, Y ).

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Properties of jcdf:
1 F(−∞, y) = 0 = F(x, −∞)
2 F(∞, ∞) = 1
3 At points of continuity of f (x, y), f (x, y) = ∂2F(x,y)
∂x∂y
4 P(a x ≤ b, c y ≤ d) = F(b, d)+F(a, c)−F(a, d)−F(b, c)

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Variable
For a two dimensional continuous r.v. (X, Y ), then the probability
distribution of X is given by
fX (x) =
∞
Z
−∞
f (x, y)dy
and it is called the marginal (probability) density function of X.
Similarly, the marginal probability density function of Y is
fY (y) =
∞
Z
−∞
f (x, y)dx.

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Variable
Definition: (Conditional Probability Function)
For a two dimensional continuous r.v. (X, Y ), the conditional proba-
bility density function of X given Y = y is defined by
f (X/Y = y) = fX/Y (x/y) = f (x/y) =
f (x, y)
fY (y)
.
Similarly, the conditional probability density function of Y given X =
x is defined by
f (Y /X = x) = fY /X (y/x) = f (y/x) =
f (x, y)
fX (x)
.
A necessary and sufficient condition for the continuous r.v. X and
Y to be independent is
f (x, y) = fX (x) · fY (y).

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Variable
Note:
1.
P(a ≤ X ≤ b) =
Z b
x=a
Z ∞
−∞
f (x, y)dydx =
Z b
x=a
fX (x)dx
and P(c ≤ Y ≤ d) =
Z d
y=c
Z ∞
−∞
f (x, y)dxdy =
Z d
y=c
fY (y)dy.
2.
P(a ≤ X ≤ b/Y = y) =
Z b
x=a
f (X/Y = y)dx =
Z b
x=a
f (x/y)dx
and P(c ≤ Y ≤ d/X = x) =
Z d
y=c
f (Y /X = x)dy =
Z d
y=c
f (y/x)dy.

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Definition: (Expectation)
If (X, Y ) is a two-dimensional continuous r.v. with joint probability
density function f (x, y) then the expectation of a function g(X, Y )
is
E[g(X, Y )] =
Z ∞
−∞
Z ∞
−∞
g(x, y)f (x, y)dxdy.
Properties of Expected Values:
(1) E[g(X)] =
R ∞
−∞ g(x)fX (x)dx and E[h(Y )] =
R ∞
−∞ h(y)fY (y)dy.
(2) E(aX + bY ) = aE(X) + bE(Y ).
(3) If X and Y are independent random variables then
E(g(X)h(Y )) = E(g(X))E(h(Y )).
(4) The converse of above result need not be true.

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Example on Two Dimensional Continuous
Random Variable
Example-2
Given the following joint probability density function
f (x, y) =





8
k
xy, 0 ≤ x ≤ y ≤ 2
0, otherwise
(1) Compute the value of k.
(2) Find the marginal densities.
(3) Find the conditional density functions.
(4) Are X and Y independent?
(5) Find P(0 ≤ X ≤ 1, Y ≤ 3/2), (6) P(X ≤ 1/Y 3/2),
(7) P(X ≥ 1), (8) P(X + Y ≤ 1),
(9) E(X), E(XY ), E(X/Y = 1).

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Random Variable
Solution: (1) Since given f (x, y) is the jpdf of two dimensional
continuous r.v. (X, Y ), we have
ZZ
R
f (x, y)dxdy = 1
∴
2
Z
0
2
Z
x
f (x, y)dydx = 1
∴
2
Z
0
2
Z
x
8
k
xydydx = 1

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∴
8
k
2
Z
0

y2
2
#2
x
xdx = 1
∴
8
k
2
Z
0

2 −
x2
2
#
xdx = 1
∴
8
k

x2
−
x4
8
#2
0
= 1
∴
8
k

4 −
16
8

= 1
∴ k = 16.

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So the jpdf is,
f (x, y) =





xy
2
, 0 ≤ x ≤ y ≤ 2
0, otherwise
(2) The marginal density of X is
fX (x) =
∞
Z
−∞
f (x, y)dy
=
2
Z
x
xy
2
dy =

xy2
4
#2
x
= x −
x3
4
, 0 ≤ x ≤ 2.

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Similarly, the marginal density of Y is,
fY (y) =
∞
Z
−∞
f (x, y)dx
=
y
Z
0
xy
2
dx =

x2
y
4
#y
0
=
y3
4
, 0 ≤ y ≤ 2.
(3) The conditional density function of X given Y is
fX/Y (x/y) = f (x/y) =
f (x, y)
fY (y)
=
xy
2
y3
4
, 0 ≤ x ≤ y ≤ 2.

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i.e.,
fX/Y (x/y) =
2x
y2
, 0 ≤ x ≤ y ≤ 2.
Similarly, the conditional density function of Y given X is,
fY /X (y/x) = f (y/x) =
f (x, y)
fX (x)
=
xy
2
x − x3
4
, 0 ≤ x ≤ y ≤ 2.
i.e.,
fY /X (y/x) =
2xy
4x − x3
=
2y
4 − x2
, 0 ≤ x ≤ y ≤ 2.

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(4)
fX (x)fY (y) = x −
x3
4
!
y3
4
=
xy3
16
(4 − x2
), 0 ≤ x ≤ y ≤ 2
̸= f (x, y) =
xy
2
∴ X and Y are not independent random variables.
(5) P

0 ≤ X ≤ 1, Y ≤ 3
2

=
1
Z
0
3/2
Z
0
f (x, y)dydx (∵ y ≥ 0)

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Random Variable
=
1
Z
0
3/2
Z
0
xy
2
dydx
=
1
Z
0

xy2
4
#3/2
0
dx
=
9
16

x2
2
#1
0
=
9
32
.

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(6)
P

X 1/Y
3
2

=
P

X 1, Y 3
2

P

Y 3
2
=
9
32
P

Y 3
2

(∵ Y is a continuous r.v.)
Now,
P

Y
3
2

=
3/2
Z
0
fY (y)dy =
3/2
Z
0
y3
4
dy
=

y4
16
#3/2
0
=
81
256

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∴ P

X 1/Y
3
2

=
9/32
81/256
=
8
9
.
(7)
P(X ≥ 1) = P(1 ≤ X ≤ 2)
=
2
Z
1
fX (x)dx
=
2
Z
1
x −
x3
4
!
dx

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Random Variable
=

x2
2
−
x4
16
#2
1
= 2 − 1 −

1
2
−
1
16

= 1 −
7
16
=
9
16

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(8)
P(X + Y ≤ 1) = P(Y ≤ 1 − X)
Here,
0 ≤ y ≤ 1 − x
=⇒ 0 ≤ 1 − x ≤ 2
or 0 ≤ 2 ≤ 1 − x (∵ 0 ≤ y ≤ 2)
If
0 ≤ 1 − x ≤ 2
then
−1 ≤ −x ≤ 1
i.e. 1 ≥ x ≥ −1
i.e. − 1 ≤ x ≤ 1

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A
P
a
t
e
l
Random Variable
But 0 ≤ x ≤ 2 is given.
∴ Common range is 0 ≤ x ≤ 1.
and if
0 ≤ 2 ≤ 1 − x
∴ 1 ≤ −x
i.e. − 1 ≥ x
i.e. x ≤ −1
But 0 ≤ x ≤ 2 is given.
∴ There is no common range.
Hence, we must have 0 ≤ x ≤ 1.

D
r
M
A
P
a
t
e
l
Random Variable
∴ P(X + Y ≤ 1) = P(0 ≤ X ≤ 1, 0 ≤ Y ≤ 1 − X)
=
1
Z
0
1−x
Z
0
f (x, y)dydx
=
1
Z
0
1−x
Z
0
xy
2
dydx
=
1
Z
0

xy2
4
#1−x
0
dx

D
r
M
A
P
a
t
e
l
Random Variable
=
1
Z
0
x(1 − x)2
4
dx
=
1
4
1
Z
0
(x − 2x2
+ x3
)dx
=
1
4

x2
2
−
2x3
3
+
x4
4
#1
0
=
1
4

1
2
−
2
3
+
1
4

=
1
48
.

D
r
M
A
P
a
t
e
l
Random Variable
(9)
E(Y ) =
∞
Z
−∞
yfY (y)dy
=
2
Z
0
y
y3
4
dy
=

y5
20
#2
0
=
8
5
.

D
r
M
A
P
a
t
e
l
Random Variable
E(XY ) =
∞
Z
−∞
∞
Z
−∞
xyf (x, y)dxdy
=
2
Z
0
2
Z
x
xy
xy
2
dydx
=
2
Z
0
x2
2

y3
3
#2
x
dx
=
2
Z
0
x2
6
h
8 − x3
i
dx

D
r
M
A
P
a
t
e
l
Random Variable
=
2
Z
0

4x2
3
−
x5
6
#
dx
=

4x3
9
−
x6
36
#2
0
=
32
9
−
64
36
=
16
9
.

D
r
M
A
P
a
t
e
l
Random Variable
E(X/Y = 1) =
∞
Z
−∞
xf (x/y = 1)dx
=
1
Z
0
x(2x)dx ∵ f (x/y) =
2x
y2
, 0 ≤ x ≤ y ≤ 2y = 1
!
=

2x3
3
#1
0
=
2
3
.

PS_Basic Probability part 4_PPT_2020.pdf

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