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Stochastic Independence and Functions of Random Variables, Study notes of Mathematical Statistics

The concept of stochastic independence of random variables and how to determine if two random variables are stochastically independent. It also covers the concept of functions of random variables, including how to find the probability density function (pdf) of a function of a random variable. Examples are provided to illustrate these concepts.

Typology: Study notes

2012/2013

Uploaded on 04/20/2013

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Independent%Random%Variables%
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Intuitively, the rvs X and Y are independent
if the distribution of one rv does not in any
way influence distribution of the other rv.
Independence is a useful assumption for
hydrologic analysis in many situations.
However, there must be a sound physical
basis for the assumption.
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3

Independent Random Variables

  • Intuitively, the rvs X and Y are independent

if the distribution of one rv does not in any

way influence distribution of the other rv.

  • Independence is a useful assumption for

hydrologic analysis in many situations.

However, there must be a sound physical

basis for the assumption.

4

  • As an example, inflow to a reservoir (X) and the rainfall

in the command area (Y) may be taken as independent,

if the command area is far removed from the reservoir.

Inflow

Reservoir

Command area

Rainfall

Independent Random Variables

6

Independent Random Variables

  • When two rvs are independent, g(x/y)=g(x)
    • Distribution of X given Y is independent of Y and

hence the conditional pdf is equalt to the marginal pdf.

( )

f x y

g x y

h y

f x y

g x

h y

f x y g x h y

h(y) > 0

77

Independent Random Variables

  • The random variables X and Y are stochastically

independent if and only if their joint density is equal to

the product of their marginal densities.

  • Discrete case: the two r.v.s are independent if and only if

p(x

i

, y

j

) = p(x

i

). p(y

j

) v i,j

Example-1(contd.)

( )

1 1

0 0

1

2

0

f x y dy x y dy
y
xy x x

∫ ∫

g(x)
h(y)

( )

1 1

0 0

1

2

0

f x y dx x y dx
x
xy y y

∫ ∫

9

Example-1(contd.)

g x h y x y

× = + × +

Therefore X and Y are not stochastically
independent.

f ( , x y ) ≠ g x h y ( ). ( )

10

Example-2(contd.)

( )

0 0

0

x y

x y x

f x y dy e dy
e e dy e x

∞ ∞

− +

− − −

∫ ∫

g(x)
h(y)

( )

0 0

0

x y

y x y

f x y dx e dx
e e dx e y

∞ ∞

− +

− − −

∫ ∫

12

Example-2 (contd.)

( )

x y

x y

g x h y e e

e

− −

− +

× = ×

Therefore X and Y are stochastically
independent

f ( , x y ) = g x h y ( ). ( )

13

  • X: discrete; Y = H(X), a funcAon of X.
  • The pmf of X is known
  • Enumerate possible values of Y for the

discrete values of X

  • Then obtain the probabiliAes of the possible

values of Y from the probabiliAes of the

corresponding values of X.

Functions of Random Variable

Example for discrete case

p(x) = 60/77x ; x = 2, 3, 4, 5

discrete values

y = x

2

-7x+

Distribution of y:

p(Y=-7) = p(X=3)+p(X=4) = 20/77+ 15/77 = 35/77 = 5/

p(Y=-5) = p(X=2)+p(X=5) = 30/77+ 12/77 = 42/77 = 6/

x 2 3 4 5

y -­‐5 -­‐7 -­‐7 -­‐

p(x) 30/77 20/77 15/77 12/

16

b. Differentiate G(y) w.r.t y to get g(y)
c. Since g(y) must be non-negative, determine
those values of y over which g(y) > 0
and check,

General procedure for functions of continuous

random variables:

g y dy ( ) 1

−∞

18

The rv X has a pdf
f(x) = x/2 0 < x < 2
= 0 elsewhere
Let H(X) = 4X+
Find the pdf of Y=H(X)
a. Get the CDF of Y
G(y) = P [Y < y]
= P [4X+1 < y]

1

4

y

P X

⎡ ⎤

⎢ ⎥

⎣ ⎦

Example-

19

b. g(y) =

c. From 0 < x < 1, we get

( ) ( )

2

1

64

dG y y

d

dy dy

⎛ ⎞

⎜ ⎟

⎜ ⎟

⎝ ⎠

( )

1

( ) 1 9

32

y

g y y

= < <

Example-1(contd.)

( )

y
y

21

y=4x+

Example-1(contd.)

Check :

9

1

y

dy

9

9

2

1

1

2

( 1) 1 ( 1)

32 32 2

1

(8 0)

64

1

y y

dy

⎡ ⎤

− −

=

⎢ ⎥

⎣ ⎦

= −

=

22