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Lecture 4
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Random Variables (Discrete)
Real-valued functions defined on a sample space are random
vars.  determined by outcome of experiment, we can
assign probability to possible values of it (them).
Exs: Toss 3 fair coins, let Y be number of H appearing, then
Y is a random var. with possible values (0,1,2,3) with
probability
P{y = 0} = P(T, T, T) = 1/8
P(y = 1) = P(T, T, H), (T, H,T), (H, T, T) = 3/8
P(y = 2) = P(T, H, H), (H, T, H), (H, H, T) = 3/8
P(y = 3) = P(H, H, H) = 1/8
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Since y is between 0 and 3, 1  P ( { y  i}) 
P( y  i )

i 0

i 0
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Distribution functions
The c.d.f. or distribution function F of random variable x is
defined for all real numbers b,   b  
F(b) = P{x  b} denotes prob. that x takes on a value  b
Properties
1. F is a non-decreasing function, if a  b F(a) < F(b)
F (b)  1
2. blim

3. lim . . .= 0
b  
4. F is right continuous. For any b and any decreasing
sequence bn,
n  1 that converges to b,
lim F (bn)  F (b)
n 
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Exs: Distribution function of a random variable x is given
0
x 0
x/2
0 x 1
F(x) = 2/3
1 x  2
11/12 2  x  3
1
3 x
Calculate P{x  3}
 lim P{x  3  1 }  lim F (3  1 )  11
n
n
12
n
n
4
Calculate P{x = 1}
1 2
 P( x  1)  P( x  1)  F (1)  lim F (1  ) 
1 1
3
2
6
n
Calculate P{x  ½}
 P{x  1 }  1  P{x  1 }  1  F ( 1 )  3
2
2
2
4
Calculate P(2  x

4)
P(2  x  4)  F (4)  F (2)  1
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(draw graph)
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Discrete Random Variables
A random variable that can take on at most a countable
number of possible values.
Probability mass function P(a) = P{x = a}
If x can be x1, x2, x3, . . . then
P(xi)  0, i = 1, 2, . . .
P(x) = 0, else 
P ( xi)  1
and

i 1
If we have P(0) = ¼ P(1) = ½
P(2) = ¼
(draw graph)
The plot for a random var. of sum of 2 dice are:
(draw graph)
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Cumulative Distr. Function F can be expressed in terms of
mass function as F(a) =  p(x)
all x  a
If x is discrete random variable where x1  x2  x3  . . . then
distrib. function F is a step function. The value of F is
constant in intervals [xi-1, xi) and with a step (jump) of
size p(xi) at xi
Exs:
x has probability mass function
P(1) = ¼
p(2) = ½ p(3) = 1/8
P(4) = 1/8
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Then, the c.d.f is
0
a 1
¼
1  a 2
F(x) = ¾
2  a 3
7/8
3  a 4
1
4a
Step size at any of values 1, 2, 3, 4 is equal to the probability
that x assumes a particular value.
(draw graph)
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Expected Value: if x is a discrete random variable with probability
mass function p(x) then
expectation or expected value
E[ x]   xp( x)
x: p ( x )  0
This is a weighted average of possible values x can take on each
value, weighted by probability that x assumes it.
If p(0) = ½ = p(1)
E[x] = 0*1/2 + 1 ½ = ½ ordinary avg. of 2 possible values 0 and
1 x can assume
Or if p(0) = ½ p(1) = 2/3  E[x] = 0*(1/2) + 1*(2/3) = 2/3 where
value of 1 is given twice as much weight as value 0 so, p(1) =
2p(0)  2/3 = 2*(1/3)
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Frequency Interpretation if an infinite sequence of
independent replications of an experiment is done, then for
any event E, the proportion of time that E occurs is p(E).
So, if x can be x1, x2, x3, . . . with p(x1), p(x2), . . . then,
n
average expectation
E[ x]   xip ( xi )
i 1
Exs:
Find E[x] where x is outcome of a roll of fair die
 Since p(1) = p(2) = . . . p(6) = 1/6
Then, E[x] = 1(1/6) + 2(1/6) + 3(1/6) + . . . 6(1/6) = 7/2
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Exs
If I is a stock indicator for trading volume V then
I = 1 if V true
Find E[I]
0 if VC true
 Since p(1) = p(v), p(0) = 1 – p(A) we haven then E[I] = p(A)
Expected value of indicator is the probability that event
occurs.
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Expectation of a function of a Random Variable:
Having a discrete random variable and its probability mass
function, we need expected value of some function x, g(x).
One method:
*calculate the prob. mass function of g(x) since it is a discrete
random variable
*calculate E[g(x)] by using expected value definition
Exs X is a random variable with any value in –1, 0, 1 with
x -1
x 0
x 1
P(-1) = .2
P(0) = .5
P(1) = .3
Calculate E[x2] :
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Let y = x2, it follows that prob. mass function of y is
P(y = 1) = P(x = -1) + P{x = 1} = .5
P{y = 0} = P{x = 0} = .5
hence, E[x2] = E[y] = 1(.5) + 0(.5) = .5
note that .5 = E[x2]  (E[x])2  .01
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Method 2:
if x is discrete random variable that takes on values xi, i ≥ 1
with P(xi) then for any real-valued function g,
E[ g ( x)]   g ( xi ) p( xi )
i
Going back to previous example:
E[x2] = (-1)2 (.2) + 02 (.5) + 12 (.3)
= 1(.2 + .3) + 0 (.5)
= .5
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Corollary
if a and b are constants then E[aX + b] = a E[X] + b
Because E[aX + b] =
(ax  b) p( x)

x: p ( x )  0
=
a

xp( x)  b
x: p ( x )  0
 p( x)  aE[ x]  b
x: p ( x )  0
The expected value of a random variable x, E[x] is also
referred to as the mean or 1st moment of x. The quantity
E[xn], n  1 is the nth moment of x and, E[xn] =
x
n
p( x)
x: p ( x )  0
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Variance
Given x with distribution function F and E[x]. E[x] is
weighted average of all possible values for x. We need the
variation or spread of these values. If x has a mean μ, then
variance of x, Var(x) = E[(x – μ)2]
Or 
( x  ) 2 p( x)

x
  ( x 2  2x   2 ) p( x)
x
  x 2 p( x)  2  xp( x)   2  p( x)
x
x
x
 E[ x 2 ]  2 2   2  E[ x 2 ]   2
That is also Var (x) = E[x2] – (E[x])2
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Exs
Calculate Var(x) if x is outcome of a fair die rolled
 from previous example, E[x] = 7/2 and
E[x2] = 12(1/6) + 22 (1/6) + 32(1/6) + . . . 62 (1/6) = 91/6
Hence, Var(x) = 91/6 – (7/2)2 = 35/12
Identity: for any constants a and b Var(ax + b) = a2 Var(x)
*note that analogous to the mean being center of gravity of a
distrib. of mass, Variance represents moment of inertia
*The SQRT of Var(x) is the STD. deviation of x denoted by SD(x)
=
Var (x)
Discrete random variables are classified according to their prob.
mass function.
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Exs:
52-card deck is well-shuffled and then cards are turned face
up one by one til Ace appears. Find expect number of
cards that are face up
 Let x be number cards face up til ace appears
A = { no ace among first i – 1 cards turned up }
B = { ith card is ace }
(i481 )
4
P(x = i) = P(AB) = P(B/A) P(A) =
.
52  (i  1) (52
i 1 )
P(x) =
49
48
i 1
i ( )4
 10.6  11

52
i 1 ( i 1 )(52  i )
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Exs:
What are Expected number, Var., and S.D. of the number of
spades in a poker hand? (P.H. = set of 5 cards randomly
picked from 52)
 A = pick any 12 spades
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5
(13
)(
i
5 i )
E[ x]  i 52
 1.25
B = rest of cards

i 0
5
E[ x ]   i
2
i 0
(5 )
13
2 i
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5i
52
5
( )( )
 2.430
( )
Thus, Var(x) = 2.43 - (1.25)2 = 0.864 and σx = .864  .93
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Exs
A professor made 30 exams, 8 tough, 12 medium, 10 easy;
exams are mixed up and 4 are selected at random. How
many will be difficult?
 x  number of difficult ones
We need E(x), so
x can be 0, 1, 2, 3, 4 and its probability function is
8
i
22
4 i
30
4
( )( )
p(i )  p( x  i) 
, i  0,1,2,3,4
( )
Values of all p(i.s)  i
0
1
2 3
4
p(i)
0.27 .45 .24 .04 .003
 E(x) = 0(.27) + 1(.45) + 2(.24) + 3(.04) + 4(.003) = 1.06
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Continuous Random Variables
Set of possible values is uncountable (lifetime of a transistor,
mars rovers, etc)
X is continuous random variable if there exists a nonnegative
function f, defined for all real x  (, ) having the
property that any set B of real numbers:
P{x  B}   f ( x)dx
B
Probability density function of x
(states that probability that x will be in B may be obtained by
integrating the p.d.f. over set B) it must also satisfy

1  P{x  (, )}   f ( x)dx

(all probability stmts about x can be answered in terms of f)
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If B = [a, b] then
b
P{a  x  b}   f ( x)dx
a
if we let a = b then
b
P{x  a}   f ( x)dx  0
a
(means that probability of a continuous random variable will
assume any fixed value is zero).
a
 P{x  a}  P{x  a}  F (a)   f ( x)dx

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Exs:
Suppose x is continuous random variable whose p.d.f. is
f ( x) 
C ( 4 x 2 x 2 ) 0  x  2
0
else
What is value of C?
 Since f is a p.d.f., we have
and thus, C 2 (4 x  2 x 2 )dx  1



f ( x)dx  1

0
3
2
x
C[2 x 2 
]
3
x2
x 0
1 C  3
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Find P(x  1)?

 P( x  1)   f ( x)dx 
1
3
8

2
1
(4 x  2 x 2 )dx  1
2
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Exs
Lifetime in days of a MEMS wireless transceiver is a random
variable having p.d.f. given by
f(s) 0
100/x2
x  100
x > 100
What’s probability that 2 of 5 such devices needs replacing
within 150 days of continuous operation?
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Exs
 Assume that events Ei, i = 1, 2, 3, 4, 5
ith transceiver will need replacement within the 150 days are
independent
150
P( Ei)  
0
150
f ( x)dx  100 x dx  1
100
2
3
thus, from independence of events Ei, the probability is
P(E)
P(EC)
3
1 2 2 3 80
0 ( )( 3 ) ( 3 )  243
5
2
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Exs Loss in a stock option, in thousands of dollars, is a
continuous random var. x with density function:
f(x) = k(2x – 3x2) -1 < x < 0
0
else
Calculate k and find prob. the loss is at most $500
Since f is a p.d.f. 
,but






f ( x)dx  1
0
0
f ( x)dx   (2 x  3 x )dx  k  (2 x  3 x 2 )dx
 k[ x 2  x 3
2
1
0
1
1
 2k
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so,
 2k  1  k   1
2
and, loss at most $500 iff x ≥ -1/2, thus
0
1
P( x   )   1  1 (2 x  3x 2 )dx   1 [ x 2  x 3 ]0 1  3
2
2
2
16

2
2
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Relationship between c.d.f. F and p.d.f. f is
a
F (a)  P{x  (, a)}   f ( x)dx

Differentiating both sides, d F (a)  f (a)
da
Density is the derivative of cumulative distr. function. Also, we
can say that, from
b
P{a  x  b}   f ( x)dx
a
P{a 


a 
 x  a  }    2 f ( x)dx  f (a )
a
2
2
2
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Where  is small and when f is continuous at x = a. In other
words, probability that x will be contained in an interval of
length  around point a is roughly  f(a). f(a) is a measure
of how likely it is the random variable will be near “a”
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