For
x >
1
,
F
X
(
x
) = 1
since the dart always falls within the unit circle, and for
x <
0
,
F
X
(
x
) = 0
because
X
is
nonnegative (a distance is always nonnegative).
3. (b) The random variable
X
is continuous because its CDF has no jumps. The PDF of
X
can be found by differentiating
the CDF. For
x
∈
[0
,
1]
, we have
f
X
(
x
) =
dF
X
(
x
)
dx
= 2
x.
For
x <
0
and
x >
1
,
f
X
(
x
) = 0
.
3. (c) Note first that when
y <
0
, the CDF of
F
Y
(
y
) = 0
since the
Y
is nonnegative. For
y
≥
1
,
F
Y
(
y
) = 1
since the
maximum value
Y
can take is 1. Notice also that
Y
= 0
if
X
≤
1
2
and that
1
4
≤
Y
≤
1
when
X
∈
(1
/
2
,
1]
.
Consequently, for
y
∈
[0
,
1
/
4]
, we have
F
Y
(
y
) = Pr(
Y
≤
y
) = Pr(
Y
= 0) = Pr
X
≤
1
2
=
F
X
1
2
=
1
4
.
For
y
∈
[1
/
4
,
1]
, we have, from Total Probability,
F
Y
(
y
) = Pr(
Y
≤
y
)
= Pr
Y
≤
y, X
≤
1
2
+ Pr
Y
≤
y, X >
1
2
= Pr
0
≤
y, X
≤
1
2
+ Pr
X
2
≤
y, X >
1
2
= Pr
X
≤
1
2
+ Pr
1
2
< X
≤
p
(
y
)
=
F
X
1
2
+
F
X
p
(
y
)

F
X
1
2
=
y.
2
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3. (d) The random variable
Y
is neither continuous nor discrete.
Y
takes a continuum of values over the interval
[1
/
4
,
1]
and
is therefore not discrete. Moreover, it takes the value 0 with positive probability so it is not continuous either.
4. (a) The function
F
is a valid CDF because it satisfies all the properties of a CDF (refer to Figure 4). Indeed,
lim
x
→∞
F
(
x
) =
1
and
lim
x
→∞
F
(
x
) = 0
. Moreover we also have
0
≤
F
(
x
)
≤
1
. The function
F
is also nondecreasing and right
continuous. This is true because F is a linear combination with nonnegative coefficients of
u
(
x
)
and
u
(
x

2)
, which
are both nondecreasing and rightcontinuous.
4. (b) The function
F
is a CDF of a discrete random variable (we will call it
X
) because it increases only by jumps and is
constant between any two jumps. The CDF of
X
has jumps at
x
= 0
and
x
= 2
and is constant otherwise. This means
that
X
can take two values 0 and 2 and
P
(
X
= 2) =
P
(
X
≤
2)

P
(
X <
2) =
F
(2)

lim
x
→
2

F
(
x
) = 1

1
3
=
2
3
.
Similarly,
P
(
X
= 0) =
F
(0)

lim
x
→
0

F
(
x
) =
1
3

0 =
1
3
.
Figure 5 depicts this PMF.
4. (c)
Pr(
X
∈
[0
,
2)) = Pr(
X
= 0) =
1
3
.
Another way to establish this is the following:
P
(
X
∈
[0
,
2)) =
P
(
X <
2)

P
(
X <
0) =
P
(
X <
2) = lim
x
→
2

F
(
x
) =
1
3
.
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 Fall '05
 HAAS
 Probability theory, CDF, ﬁrst bit

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