05 Discrete random variables examples - solutions -...

University of California, Los Angeles
Department of Statistics
Statistics 100A Instructor: Nicolas Christou
Examples - solutions
Example 1:
E ( N ) = E ( Cπr
2
) = CπEr
2
= Cπ ( σ
2
+ μ
2
) .
The mean and variance of the distribution of r are: E ( r ) = 23 . 5, var ( r ) = 1 . 54. Therefore,
E ( N ) = 8 π (1 . 54 + 23 . 4
2
) = 13800 . 39 ≈ 13801 .
Or simply compute E ( N ) as:
E ( N ) = E ( Cπr
2
) = Cπ
X
r
r
2
P ( r ) = 8 π 21
2
(0 . 05) + 22
2
(0 . 20) + . . . + 26
2
(0 . 05) = 13800 . 39 ≈ 13801 .
Example 2:
It is given that P ( X = i ) = cP ( X = i - 1) for i = 1 , 2. Let P(X=0)=p.
P ( X = 1) = cP ( X = 0) = cp , and P ( X = 2) = cP ( X = 1) = c
2
p . Also, P ( X = 0) + P ( X = 1) + P ( X = 2) = 1.
Or p + cp + c
2
p = 1 ⇒ p =
1
1+ c + c
2
. The expected value of X is:
E ( X ) = 0( p ) + 1( cp ) + 2( c
2
p ) =
c
1 + c + c
2
+
2 c
2
1 + c + c
2
=
c (1 + 2 c )
1 + c + c
2
.
Example 3:
There are 8 white, 4 black and 2 orange balls. Two balls are selected without replacement. For each black we win $2, for each white we lose $1.
We neither win nor we lose anything if we select an orange ball.
a.
Color X ($) P ( X )
W W - 2
8
14
7
13
=
56
182
W O or OW - 1 2
8
14
2
13
=
32
182
OO 0
2
14
1
13
=
2
182
BW or W B 1 2
4
14
8
13
=
64
182
BO or OB 2 2
4
14
2
13
=
16
182
BB 4
4
14
3
13
=
12
182
b.