CHAPTER 17
SOLUTIONS TO PROBLEMS
17.1
(i) Let
m
0
denote the number (not the percent) correctly predicted when
y
i
= 0 (so the
prediction is also zero) and let
m
1
be the number correctly predicted when
y
i
= 1.
Then the
proportion correctly predicted is (
m
0
+
m
1
)/
n
, where
n
is the sample size.
By simple algebra, we
can write this as (
n
0
/
n
)(
m
0
/n
0
) + (
n
1
/
n
)(
m
1
/
n
1
) = (1
−
y
)(
m
0
/
n
0
) +
y
(
m
1
/
n
1
), where we have used
the fact that
y
=
n
1
/
n
(the proportion of the sample with
y
i
= 1) and 1
−
y
= n
0
/
n
(the proportion
of the sample with
y
i
= 0).
But
m
0
/
n
0
is the proportion correctly predicted when
y
i
= 0, and
m
1
/
n
1
is the proportion correctly predicted when
y
i
= 1.
Therefore, we have
(
m
0
+
m
1
)/
n
=
(1
−
y
)(
m
0
/
n
0
) +
y
(
m
1
/n
1
).
If we multiply through by 100 we obtain
ˆ
p
=
(1
−
y
)
+
0
ˆ
q
y
⋅
,
1
ˆ
q
where we use the fact that, by definition,
ˆ
p
= 100[(
m
0
+
m
1
)/
n
],
= 100(
m
0
/
n
0
), and
=
100(
m
1
/
n
1
).
0
ˆ
q
1
ˆ
q
(ii) We just use the formula from part (i):
ˆ
p
= .30(80) + .70(40) = 52.
Therefore, overall we
correctly predict only 52% of the outcomes.
This is because, while 80% of the time we correctly
predict
y
= 0,
y
i
= 0 accounts for only 30 percent of the outcomes.
More weight (.70) is given to
the predictions when
y
i
= 1, and we do much less well predicting that outcome (getting it right
only 40% of the time).
17.3
(i) We use the chain rule and equation (17.23).
In particular, let
x
1
≡
log(
z
1
).
Then, by the
chain rule,
1
11
1
1
( 0
,)
,)1
,
x
Ey y
zx
z
x
1
z
∂
∂>
=⋅
∂∂
∂
∂
xx
x
where we use the fact that the derivative of log(
z
1
) is 1/
z
1
.
When we plug in (17.23) for
∂
E(
y

y
> 0,
x
)/
∂
x
1
, we obtain the answer.
(ii) As in part (i), we use the chain rule, which is now more complicated:
12
1
2
( 
0, )
( 
0, )
( 
0, )
,
1
x
x
z
x
z
∂
∂
+
∂
∂
x
⋅
∂
99