This preview shows page 1. Sign up to view the full content.
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)
²
= 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:
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/14/2011 for the course STATS 100A taught by Professor Wu during the Fall '07 term at UCLA.
 Fall '07
 Wu
 Statistics, Variance

Click to edit the document details