Pages 375377
36
Set
X
=
∑
n
i
=1
X
i
,
Y
=
∑
n
i
=1
Y
i
, where
X
i
=
I
(
i
th roll results in 1) and
Y
i
=
I
(
i
th roll results in 2). Then
Cov
(
X, Y
) =
i
j
Cov
(
X
i
Y
j
) =
i
Cov
(
X
i
, Y
i
) =

n
1
36
The above used
Cov
(
X
i
Y
j
) = 0, whenever
i
=
j
(due to the independence of
X
i
and
Y
j
),
and
Cov
(
X
i
, Y
i
) =
E
(
X
i
Y
j
)

E
(
X
i
)
E
(
Y
j
) = 0

E
(
X
i
)
E
(
Y
j
) =

(1
/
6)
2
.
37
Cov
(
X
1
+
X
2
, X
1

X
2
) =
Cov
(
X
1
, X
1
)

Cov
(
X
1
, X
2
) +
Cov
(
X
2
, X
1
)

Cov
(
X
2
, X
2
)
=
Cov
(
X
1
, X
1
)

Cov
(
X
2
, X
2
) =
V ar
(
X
1
)

V ar
(
X
2
) = 0, where
X
i
= outcome of roll i.
55
Let
V
i
=
I
(
i
th duck gets hit), and
N
the number of ducks in a flock.
Thus, the
number of ducks that get hit is
∑
N
i
=1
V
i
. Use
E
(
Y
) =
E
(
E
(
Y

X
)) with
Y
=
∑
N
i
=1
V
i
and
X
=
N
to get that the expected number of ducks that are hit can be found from
E
N
i
=1
V
i
=
E
E
N
i
=1
V
i
N
.
Since any duck is equally likely to get hit (no matter what
N
is) we have
E
N
i
=1
V
i
N
=
NE
(
V
1
N
)
Thus, we need to find
E
(
V
1
N
)
, as a function of
N
, and then take the expected value of
NE
(
V
1
N
)
with the information that
N
∼
Poisson(
λ
= 6).
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 '08
 G.JOGESHBABU
 Probability, Probability theory, probability density function

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