SOLUTIONS TO ASSIGNMENT 2 – ACTSC 431/831, FALL 2008
1. (a)
i.
E
(
Y
A
) =
1
2
(
E
(
X
1
) +
E
(
X
2
)) = 30 and
E
(
Y
B
) =
1
2
(
E
(
X
1
) +
E
(
X
2
)) = 30.
ii.
E
(
Y
2
A
) =
1
2
(
E
(
X
2
1
) +
E
(
X
2
2
)) = 1306
.
5 and thus
V ar
(
Y
A
) = 1306
.
5

30
2
=
406
.
5.
V ar
(
Y
B
) =
1
4
(
V ar
(
X
1
) +
V ar
(
X
2
)) = 13
/
4 = 3
.
25.
iii. Let the pdfs of
F
1
and
F
2
be
f
1
and
f
2
, respectively. Then, the mgf of
Y
A
is
given by
E
(
e
tY
A
) =
Z
∞
∞
e
ty
1
2
(
f
1
(
y
) +
f
2
(
y
))
dy
=
1
2
(
M
X
1
(
t
) +
M
X
2
(
t
))
=
1
2
(
e
10
t
+
5
2
t
2
+
e
50
t
+
8
2
t
2
)
.
If
Y
A
has a normal distribution, the mgf of
Y
A
must be identical to
e
30
t
+
406
.
5
2
t
2
.
Obviously,
e
30
t
+
406
.
5
2
t
2
is not identical to
1
2
(
e
10
t
+
5
2
t
2
+
e
50
t
+
8
2
t
2
). Hence, the
distribution of
Y
A
is not normal distribution.
iv. The mgf of
Y
B
is given by
M
Y
B
(
t
) =
M
X
1
2
(
t
)
M
X
2
2
(
t
) =
e
10
t
2
+
5
2
(
t
2
)
2
e
50
t
2
+
8
2
(
t
2
)
2
=
e
30
t
+
13
4
t
2
2
,
which is the mgf of the normal distribution
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 Fall '09
 david
 Probability theory, YA, yB

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