Probability and Random Processes

1
Probability and Random Processes

• Case Study Solution
• Module 2

2
Concept of Random Variable

• Problem 1 The two components in the system C1
  and C2, are tested and declared to be in one of
  the three possible state F functioning , R
  not functioning but repairable, and K kaput.
• (a) What is the sample space in this experiment?
• (b) What is the set corresponding to the event
  none of the component is kaput?
• Solution
• (a) S set of order pairs ( x1, x2 ) where xi
  specifies state of Ci
• (b) ( F, F), ( F, R), ( R, F), (R, R)

3
Concept of Random Variable

• Problem 2 A desk drawer contain five pens, three
  of which are dry.
• (a) The pens are selected at random one by one
  until a good pen is found.
• The sequence of result is noted. What is the
  sample space?
• (b) Suppose that the pens are selected one by one
  and tested until both good
• pens have been identified, and the sequence of
  the result is noted. What is the sample space?
• Solution
• (a) Each testing of a pen has two possible
  outcomes good pen (g) or bad pen (b). The
  experiment consist of testing pen until a good
  pen is found. Therefore each outcome of the
  experiment consist of a string of bs ended by
  a g. We assume that each pen is not put back in
  the drawer after being tested.
• So, S g, bg, bbg, bbbg
• Continue

4
Concept of Random Variable

• (b) The outcome now consist of a substituting of
  bs and one g in any order followed by a
  final g.
• S gg, bgg, gbg, gbbg, bbgg, gbbbg, bgbbg,
  bbgbg, bbbgg

5
Discrete probability distributions

• Problem 1
• The CDF of the random variable X is given by
• Fx( x) 1/3 (2/3)( x 1)2 -1 ? x ? 0
• 0 x lt -1
• find the probability of the events
• A X gt 1/3, B X 1, C X -1/3
  lt1, D X lt 0.
• Solution
• Continue

Fx( x)
1
1/3
0
-1
0
6
Discrete probability distributions

• Now from the above graph
• (a) P X gt 1/3 0
• (b) P X 1 P X -1 1/3
• (c) now,
• X 1/3 lt 1 -2/3 lt X lt 4/3
• P X 1/3 lt 1 FX( 4/3) FX (-2/3)
• (1- 2/3) (1/3)2
• 25/27
• (d) P X lt 0 FX( 0) 1

7
Discrete probability distributions

• Problem 2
• A random variable Y has cdf
• FY (y) 0 y lt 1
• 1 y n y 1, where n
  is a positive integer.
• (a) Plot the cdf of Y.
• (b) Find the probability P k lt Y k1 for a
  positive integer k.
• Solution
• (a)
• Continue

FY( y)
1
1 yn
1
y
8
Discrete probability distributions

• (b)
• P k ltY k1 FY ( k1) FY ( k)
• ( 1/ k) n ( 1/ k1)n .

9
Continuous Probability Distribution

• Problem 1
• A random variable X has pdf
• fX ( x) c x ( 1 x) ,-1 x 1
• 0 ,elsewhere
• (a) find c
• (b) find P 1/2 X 3/4
• (c) Find FX ( x)
• Solution (a) We use the fact that the pdf must
  integrate to one
• 1 0ò1 fX ( x) dx c 0ò1 x( 1-x)dx .
• c x2/2 x3/3 0 to 1
• c/6
• So, c 6

10
Continuous Probability Distribution

• (b)
• P 1/2 X 3/4 6 1/2ò3/4 x( 1-x) dx
• 6 x2/2 x3/3 1/2 to 3/4
• 0.34375
• (c)
• For x lt 0, FX ( x) 0 for x gt1, FX ( x) 1
• For 0 x 1
• FX ( x) 0òx fX ( x) dx 3x2 2x2

11
Continuous Probability Distribution

• Problem 2
• A random variable X has pdf
• fX ( x) c ( 1 x4) ,-1 x 1
• 0 ,elsewhere
• (a) find c
• (b) find the cdf of X
• (c) Find P X lt 1/2
• Solution (a)
• Now, 1 -1ò1fX ( x) dx c -1ò1 x(1 x4) dx
• c x x5/5-11
  8/5 c
• So, c 8/5

12
Continuous Probability Distribution

• (b)
• FX (x) 0 for x lt -1 FX (x) 1 for x gt1
• hence,
• for -1 x 1
• FX (x) -1òx 5/8 ( 1 x 4) dx ½ 1/8 (5x
  x5)
• (c)
• P X lt 1/2 P -1/2 lt X lt 1/2
• FX (1/2) FX ( -1/2)
• 79 /8(16)

13
Joint Probability Distribution

• Problem 1 Let X and Y denote the amplitude of
  noise signals at two antennas. The random vector
  (X, Y) has the joint pdf
• f ( x, y) axe ax2/2 bye by2/2 xgt0, ygt0,
  agt0, bgt0
• a. find the joint pdf.
• b. Find P XgtY
• c. Find the marginal pdfs
• Solution
• (a) For xgt0 and ygt0
• FXY ( x, y) 0òx 0òy axe ax2/2 bye by2/2 dx
  dy
• (1- e ax2/2 ) ( 1-e by2/2)

14
Joint Probability Distribution

• (b)
• P X gtY 0ò 0òx axe ax2/2 bye by2/2 dy
  dx
• 0ò axe ax2/2 (1 - e bx2/2 ) dx
• 1 a/ a b
• (c)
• FX( x) limy tp FX( x,y) 1 e ax2/2 x
  gt0
• so, fX( x) d/ dx FX( x) axe
  ax2/2 x gt0
• Similarly
• f y( y) bye by2/2

15
Joint Probability Distribution

• Problem 2 A point (X, Y, Z) is selected at
  random inside the unit sphere.
• a. Find the marginal joint pdf of X and Y.
• b. Find marginal pdf of X.
• c. Find the conditional joint pdf of X and Y
  given Z.
• Solution
• (a) f X, Y ( x, y) - ò f X, Y, Z ( x, y, z)
  dz
• -(1- x2 y 2 )1/2ò(1- x2- y2)1/2 3/4p dz
• 3/2p (1- x2 y 2 )1/2
• (b) f X ( x) - ò - ò f X, Y, Z ( x, y, z)
  dy dz

16
Joint Probability Distribution

• -(1- x2 )1/2ò(1- x2)1/2 -(1- x2 y 2
  )1/2ò(1- x2- y2)1/2 3/4p dy dz
• -(1- x2 )1/2ò(1- x2)1/2 3/2p (1- x2 y 2
  )1/2 dy
• Let a2 1- x2
• 0 òa dt 0 òp/2 a cos ua cos u
  du (a sin u)
• 1/2 a2 0 òp/2 (1 cos 2u) du
• 1/2 a2 ( p/2 ½ sin2u 0p/2 )

a2 t2
17
Joint Probability Distribution

• ¼ p a2.
• f X ( x) 3/2p 2 1/4p a2 3/4 (1- x2)
• (c) f ( x, y z) f (x, y, z) / f (z)
• (3/4p) / (3/4 (1- z2))
• 1/ p (1- z2 )

18
Joint Probability Distribution

• Problem 3 Let X cos F and Y sin F, where F
  is an angle that is uniformly distributed in
  interval (0, 2p ).
• Find f( y/ x).
• Solution X cos F and Y sin F,

1 x2
Y
1 x2
19
Joint Probability Distribution

• Now,
• X2 Y2 1
• \ Y
• So,
• fY ( y x) 1/2d ( y ) 1/2d (
  y )

1 x2
1 x2
1 x2
20
Statistical Independence

• Problem 1 Let X and Y be random variables that
  takes on values from the set -1, 0 ,1.
• (a). find the joint probability mass assignment
  from which X and Y are independent, and confirm
  that X2 and Y2 are then also independent.
• (b). find a joint pmf assignment from which X and
  Y are not independent, for the which X2 and Y2
  are independent.
• Solution
• Here we show probability mass assignment in
  tabular form for better understanding of the
  problem.
• (a) Probability mass assignment is shown in below
  tables

21
Statistical Independence

• So,
• Here in both table no value is zero so X and Y
  and X2 and Y2 are independent

22
Statistical Independence

• (b)
• Here for X and Y table has zero values so X and Y
  are not Independent but for X2 and Y2 there is no
  zero value so X2 and Y2 are independent.

23
Statistical Independence

• Problem 2 Let X and Y be independent random
  variables that are uniformly distributed in the
  0, 1. Find the probability of the following
  events
• (1). P X2 lt 1/2, Y-1 lt 1/2 .
• (2). P X/2 lt1 , Y gt0 .
• (3). P XY lt 1/2 .
• (4). P min (X, Y) gt 1/3 .
• Solution
• (1) P X2 lt 1/2, Y-1 lt 1/2 P X2 lt ½
  P Y-1 lt 1/2
• P X lt 1/ 2 P Y gt 1/2
• 1/2 1/ 2

24
Statistical Independence

• (2) P X/2 lt 1, Y gt0
• P X lt 2 P Y gt 0
• 1
• (3)
• f ( x, y) f (x) f (y) 1

Y
1
X
0
1
25
Statistical Independence

• So,
• P XY lt 1/2 1/2 1/2 ò1 0 ò1/2x 1 dy dx
• 1/2 1/2 ò1 1/2x dx
• 1/2 1/2 ln x 1/21
• 0.85
• (4) P min (X, Y) gt 1/3 P X gt 1/3 P Y gt
  1/3
• (2/3)2 4/9