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1
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Chapter 2
Discrete Random Variables(R.V)
Part3
3
Expected value (the average) of a random variable:
is one number that summarizes an entire probability model.
After that we want to know : what are the chances of observing an
event far from the average ? ( variance)
Example exam average and variance
Refer to page.77
4
2
x
[ ] [( ) ]
Standard Deviation: = [ ]
xVar X E x
Var X
5
2 2 2
2 2
2 2
2 2
2 2
[ ] [( ) ] [ 2 ]
[ ] [ ] 2 [ ]
[ ] 2 [ ]
[ ] 2
[ ] [ ]
x x x
x x
x x
x x x
x
Var X E x E x x
E x E E x
E x E x
E x
Var X E x
2 2 2[ ] [( ) ] [ ]x xVar X E x E x
6
2
[ ] [ ]
[ ] [ ]
E aX b a E X b
Var aX b a Var X
2
2
2 2
2 22 2
2
[ ] [ ] [( ) ]
[ ]
= [ ] [ ]
[ ]
Y
x
Y aX b
Var aX b Var Y E Y
E aX b aE X b
E aX b aE X b E aX aE X
E a X E X a E X
a
[ ]Var X
7
[ ] [ ]Var X b Var X
shifting a random variable by a constant does not cahnge
the dispersion of outcomes around the expected value.
Dispersion
8
2 2
[ ] ( )
[ ] ( )
[ ] is a central moment of X
n n
X
X
E X X P x
E X X P x
Var X
ththe n moment :
the second moment :
9
[ ]
[ ]x x
E a a
E
10
Var[X]E[X]Families
p(1-p)pBernoulli
np(1-p)npBinomial
ααPoisson
(1-p)/p21/pGeometric
(l-k)(l-k+2)/12(k+l)/2Dis Uniform
k(1-p)/p2k/pPascal
11
Find:
0.2 0
0.7 1( )
0.1 2
0 otherwise
N
n
nP n
n
[N] (0.2)(0) (0.7)(1) (0.1)(2) 0.9E
(a) E[N]?
(c) Var[N]?2 2 2[ ] [ ] 1.1 (0.9 ) 0.29
DVar N E N
2 2 2 2[N ] (0.2)(0 ) (0.7)(1 ) (0.1)(2 ) 1.1E
(b) E[N2]?
(d) σN?
N = [ ] =0.538Var N
12
Let X have the Binomial PMF
Find:
44
( ) 0.5XP xx
44 4
0
4 4
4
4 4[X] ( ) 0 0.5 1 0.5
0 1
4 4 2 0.5 3 0.5
2 3
4 4 0.5 2
4
: [ ] 4*0.5 2
X
x
E xP x
or E X np
(a) The standard deviation σX?
13
44 42 2 2 2
0
4 42 2
42
2 2 2
4 4[X ] ( ) 0 0.5 1 0.5
0 1
4 4 2 0.5 3 0.5
2 3
4 4 0.5 5
4
[ ] [ ] 5 (2 ) 1
:
[ ] (1
X
X
x
E x P x
Var X E X
or
Var X np p
X
) 4*0.5*0.5 1
= [ ] 1Var X
14
(b) What is P[ µ X − σ X ≤ X ≤ µ X + σ X ] ?
4 4 4
[2 1 2 1] [1 3]
[ 1] [ 2] [ 3]
4 4 4 7 0.5 0.5 0.5
1 2 3 8
P x P x
P X P X P X
44
( ) 0.5XP xx
15
(c) What is P[ 1.38 ≤ X ≤ 3.62 ] ?
4 4
[1.38 3.62] [2 3]
[ 2] [ 3]
4 4 0.5 0.5
2 3
P x P x
P X P X
44
( ) 0.5XP xx
16
/
( )
[ ]( )
0 otherwise
X
X B
P xx B
P BP x
17
If B={Y<3} Find PY/B(y) , E[Y/B] , Var[Y/B] ?
1/ 4 1
1/ 4 2( )
1/ 2 3
0 otherwise
Y
y
yP y
y
/
{ 1} {1,2}
1 1 1[ ] [ 1] [ 2]
4 4 2
( ) ( ) {1,2}
[ ]( ) 0.5
0 otherwise0 otherwise
Y Y
Y B
B Y
P B P y P y
P y P yy B y
P BP y
18
/
/
1/ 4=0.5 1
1/ 2( )
{1,2} 1/ 4( ) =0.5 20.5
1/ 20 otherwise
0 otherwise
( ) (0.5)(1) (0.5)(2) 1.5
Y
Y B
Y B
y
P yy
P y y
E Y
19
If B={|X|>0} Find PX/B(y) ?
0.2 1
0.5 0( )
0.3 1
0 otherwise
X
x
xP x
x
/
{| | 0} {1,1}
1 and x= 1 satisfy
[ ] [ 1] [ 1] 0.2 0.3 0.5
( ) ( ) { 1,1}
[ ]( ) 0.5
0 otherwise0 otherwise
X X
X B
B X
x
P B P x P x
P x P xx B x
P BP x
20
/
0.2 1
0.5 0.4 10.3
( ) = 1 0.6 10.5
0 otherwise0 otherwise
X B
x
x
P x x x
21
4
X/B
4( ) 0.5 , B={X 0}
Find P (x) & E[X/B] ?
XP xx
4 4 4 4
4
{0,1,2,3,4}
{1,2,3,4}
[ ] [ 1] [ 2] [ 3] [ 4]
4 4 4 4 15 = 0.5 0.5 0.5 0.5
1 2 3 4 16
4 15: [ ] 1 [ 0] 1 0.5
0 16
x
B
P B P X P X P X P X
or P B P X
22
/
4
/
( ) ( ) {1,2,3,4}
[ ]( ) 15 /16
0 otherwise0 otherwise
40.5
( ) = 1, 2,3,415 /16
0 other
X X
X B
X B
P x P xx B x
P BP x
xP x x
4
/
1
4 1 1, 2,3,4
15
0 otherwisewise
4 4 4 41 1 1 1 32[ / ] ( ) 1 2 3 4
1 2 3 415 15 15 15 15X B
x
xx
E X B xP x
23
Test a sequence of integrated circuits until you find the first
failure and then stop.
Let N be the number of tests.
The probability of failure p=0.1
Consider the condition B={N≥20}
1(1 ) n=1,2,3,.....( ) , Geometric PMF
0 otherwise
n
N
p pP n
(a) Find the PMF PN(n)
24
(b) Find the conditional PMF of N given that there have
been 20 consecutiveعلى التوالي tests without a failure .
191 1
20 1
1
1
191 19
1
19
[ ] (1 ) 1 (1 )
(1 ) 1
1 1
( )(1 ) 1 (1 )
[ ] 1 1 (1 ) (1 )
n n
n n
nj n
j
x
x
n
P B p p p p
x x x
x p x p
p p p
P B p p
1
1
(1 ) 1n
j n
j
x x x
25
/
1
19
20
( )
[ ]( )
0 otherwise
(1 ) 20,21,22,......
(1 )
0 otherwise
(1 ) 20,
N
N B
n
n
P nn B
P BP n
p pn
p
p p n
21,22,......
0 otherwise
26
On the internet, data is transmitted in packets the number of
packets N needed to transmit a Web page depends on
whether the page has graphic images.
Event I ,the page has images then N between 150
Event T , the page is just text then N between 15
The page has images with probability ¼
Find:
/(a) The conditional probability ( )?N IP n
/
1 3(I) (T)
4 4
1 1,2,3,....,50
( ) 50
0 otherwiseN I
P = , P =
nP n
27
/(b) The conditional probability ( )?N TP n
/
1 1, 2,3,.4,5
( ) 5
0 otherwiseN T
nP n
28
N(c) The PMF for packets P (n)?
/
/
/
( ) 1, 2,3,....,50
[ ]( )
0 otherwise
1 1( ) ( ) ( ) 0.005
4 50
( ) 1, 2,3,.4,5
[ ]( )
0 otherwise
(
N
N I
N N I
N
N T
N
P nn
P IP n
P n P I P n
P nn
P TP n
P n
/
3 1) ( ) ( ) 0.15
4 5N IP T P n
29
0.005 0.15 0.155 1, 2,3, 4,5
( ) 0.005 6,7,.....,50
0 otherwise
N
n
P n n
30
(d) (n) ? N / N 10 P
/ 10
10
1
/ 10
( )( )
( 10)
( 10) ( ) (0.155)(5) (0.005)(5) 0.8
0.155 1, 2,3, 4,5
0.8
0.005( ) 6,7,8,9,10
0.8
0 otherwise
NN N
n
N N
P nP n
P N
P N P N
n
P n n
31
(e) [ ] ? E N / N 10
5 10
1 6
0.155 0.005[ ] 3.16
0.8 0.8n n
E N / N 10 n n
(f) [ ] ? Var N / N 10
5 102 2 2
1 6
2
0.155 0.005[ ] 12.7
0.8 0.8
[ ] 12.7 3.16 2.7
n n
E N / N 10 n n
Var N / N 10
