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Univ. Of Maryland at College Park, ECE Department
ENEE324 Fall 2002 Instructor : M. Olfat
Solution to Assignment #: 5 Posted on: 10/09/
a)
Net reward
2
y = ax + bx − nd , where x is binomial.
[ ] ( )
( ) ( ) ( ) ( )
2 2 2
0 0 0 0
1
X X X X
x x x x
E y E ax bx nd ax bx nd P x a x P x b xP x nd P x
= = = =
=
∑ ∑ ∑ ∑
[ ] ( [ ]) ( ) [ ]
2
2
aE x bE x nd a E x var x bE x nd
( ) ( )
( )
2 2
= a n p + np 1 − p + bnp − nd = anp np + 1 − p + bnp − nd
b)
x
y = a
[ ] ( ) ( ) ( ) ( )
0 0
n
n n
n k k n k
x k k
k k
n n
E y E a a p p ap p ap p
k k
− −
= =
∑ ∑
by Khayyam-Newton formula
a) ( )
P Y < 1 = 0
b) ( ) ( )
Y
P Y ≤ = F =
c) ( ) ( ) ( )
Y
P Y > = − P Y ≤ = − F = − =
d) ( ) ( ) ( ) ( )
Y
P Y ≥ = − P Y < = − P Y ≤ = − F = − =
e) ( ) ( ) ( )
Y Y
P Y = = F − F = − =
f) ( ) ( ) ( )
Y Y
P Y = = F − F = − =
g) ( )
0 otherwise
Y
y
y
P y
y
Problem 2: [YAT]2.4.
Problem 1: [GAR]3.
[ ] ( )
0
x
k
E x kP k
∞
=
∑
. Instead of k we can write
1
k
l
k
=
∑
[ ] ( ) ( )
0 1 0 1
k k
X X
k l k l
E x P k P k
∞ ∞
= = = =
∑ ∑ ∑∑
We can change the order of sums:
[ ] ( )
0 1
k
X
k l
E x P k
∞
= =
∑∑
k = 0
k = 1
k = 2
k = 3
k = 4
( )
0 terms 0
X
P
1 terms ( 1 ) X
P
2 terms ( 2 )
X
P
3 terms ( 3 ) X
P
4 terms ( 4 ) X
P
So: [ ] ( ) ( ) ( ) ( )
0 1 0 1 0 0
k
X X X
k l l k l l
E x P k P k P x l P x l
∞ ∞ ∞ ∞ ∞
= = = = + =
∑∑ ∑ ∑ ∑ ∑
a)
( ) ( )
2
U Y
y u y
P u P y
=
∑
but in problem 2.4.1 : { }
y ∈1,2,
So
2
u = y has only one solution y = u
Problem 3: [YAT]2.5.
Problem 4: [YAT]2.6.
[ ]
( ) ( ) ( )
3 3 3 3 3 3 3
2 3
X X
x S x S
x x
E x x
∈ ∈
∑ ∑
b)
( ) ( )
( )
2 2
: , 0 X X
Z X X
x z x S x x S x z
P z P x P x
= ∈ ∈ − =
∑ ∑
( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
( )
( ) ( ) ( )
( )
( )
2
2
2
2
2
2
0 otherwise 9 3 3
Z X
Z X X
Z
Z X X
Z X X
P P
z
P P P
z
P z
P P P
z
P P P
c)
( ) ( [ ]) [ ]
2 1 2 9 1 8 81 1 16 81 98
Var 1 4 9 7
x E x E x E z
= − = = × + × + × = + + = = =
d)
( ) [ ] ( )
( ) ( )
4
2
2
Var
X X X
X X
x S x S x S
x
x x E x P x x P x
∈ ∈ ∈
∑ ∑ ∑
( ) ( ) ( )
4 4 4
4 4 4
e)
( )
4
2 2
X X
X
x S x S
x
E x x P x
∈ ∈
∑ ∑
( ) ( [ ])
2
2 2
Var x E x E x 7 0 7
Take x as the number of girls among natural children. Then it has a binomial distribution with
parameters 5 and
( )
5 5
0 otherwise
x x
X
x
P x x x
−
Problem 7:
If y is the number of girls among all children, then y = x + 2.
So:
( ) ( )
{ }
( )
2
Y X X
x x y
P y P x P y
∑
Note that the equation y = x + 2 has only one solution x = y − 2 and so we don’t need ∑
Also 0 ≤ y − 2 ≤ 5 ⇒ 2 ≤ y ≤ 7. So:
( )
5
0 otherwise
Y
y
P y y
