Mathematics Department, UCLA
T. Richthammer
spring 09, sheet 1
Mar 30, 2009
Homework assignments: Math 170B Probability, Sec. 1
01. Let
X
be a normal RV with parameters
μ, σ
2
, and
a, b
∈
R
.
(a) Show that
Y
=
aX
+
b
again is a normal RV (with which parameters?) if
a
±
= 0.
(Hint: consider the cases
a >
0 and
a <
0 separately.)
(b) What happens if
a
= 0?
Answer:
(a) For
a >
0 we have
F
Y
(
c
) =
P
(
Y
≤
c
) =
P
(
aX
+
b
≤
c
) =
P
(
X
≤
c

b
a
) =
F
X
(
c

b
a
), so
f
Y
(
c
) =
d
dc
F
X
(
c

b
a
) =
f
X
(
c

b
a
)
1
a
=
1
√
2
πσ
e

(
c

b
a

μ
)
2
/
2
σ
2
1
a
=
1
√
2
πaσ
e

(
c

b

aμ
)
2
/
2
a
2
σ
2
, so
Y
is
a normal RV with parameters
μ
±
=
aμ
+
b
and
σ
±
2
=
a
2
σ
2
.
Similarly for
a <
0 we have
F
Y
(
c
) =
P
(
aX
+
b
≤
c
) =
P
(
X
≥
c

b
a
) = 1

F
X
(
c

b
a
), so
f
Y
(
c
) =

d
dc
F
X
(
c

b
a
) =

f
X
(
c

b
a
)
1
a
=
1
√
2
π

a

σ
e

(
c

b

aμ
)
2
/
2
a
2
σ
2
, so again
Y
is a normal RV
with parameters
μ
±
=
aμ
+
b
and
σ
±
2
=
a
2
σ
2
.
(b) For
a
= 0 we have
Y
=
b
, so
Y
is a discrete RV with all the mass on the value
b
.
02. Let
X
1
, X
2
be independent exponential RVs with parameter 1. Determine the PDF of
(a)
Y
= max
{
X
1
, X
2
}
(b)
Z
= min
{
X
1
, X
2
}
.
Answer:
(a)
R
(
Y
) = (0
,
∞
) For
c >
0 we have
F
Y
(
c
) =
P
(
Y
≤
c
) =
P
(max
{
X
1
, X
2
} ≤
c
) =
P
(
X
1
≤
c, X
2
≤
c
) =
P
(
X
1
≤
c
)
P
(
X
2
≤
c
) = (1

e

c
)
2
. So
f
Y
(
c
) =
d
dc
F
Y
(
c
) =
2(1

e

c
)
e

c
, so
f
Y
(
y
) = 2(
e

y

e

2
y
)1
{
y>
0
}
.
(b)
R
(
Z
) = (0
,
∞
) For
c >
0 we have 1

F
Z
(
c
) =
P
(
Z > c
) =
P
(min
{
X
1
, X
2
}
> c
) =
P
(
X
1
> c, X
2
> c
) =
P
(
X
1
> c
)
P
(
X
2
> c
) =
e

2
c
. So
f
Z
(
c
) =
d
dc
F
Z
(
c
) = 2
e

2
c
, so
f
Z
(
z
) = 2
e

2
z
1
{
z>
0
}
.
03. The joint PDF of
X
and
Y
is given by
f
(
x, y
) = (
y

x
)
e

y
1
{
0
<x<y
}
.
(a) Check that
f
(
x, y
) is a density function.
(b) Find the PDFs of
X
and
Y
. Are
X
and
Y
independent?
(c) Calculate
E
(
Y
),
P
(
1
3
Y
≤
X
≤
2
3
Y
) and
P
(
X
+
Y
≤
1).
Answer:
(a) We have
f
(
x, y
)
≥
0 and
f
is normalized:
R
dy
R
dxf
(
x, y
) =
R
dy
R
dx
(
y

x
)
e

y
1
{
0
<x<y
}
=
R
∞
0
dye

y
R
y
0
dx
(
y

x
) =
R
∞
0
dye

y y
2
2
= 1.
(b)
f
Y
(
y
) =
R
dxf
(
x, y
) =
e

y y
2
2
1
{
y>
0
}
by the same calculation as in (a).
f
X
(
x
) =
R
dyf
(
x, y
) =
R
∞
x
dy
(
y

x
)
e

y
1
{
x>
0
}
=
R
∞
0
dy
±
y
±
e

y
±

x
1
{
x>
0
}
=
e

x
1
{
x>
0
}
, so
X
is an
exponential RV with parameter 1.
f
(
x, y
)
±
=
f
X
(
x
)
f
Y
(
y
), so
X
and
Y
are not independent.
(c)
E
(
Y
) =
R
dyyf
Y
(
y
) =
R
∞
0
dye

y y
3
2
= 3.
P
(
1
3
Y
≤
X
≤
2
3
Y
) =
R
∞
0
dye

y
R
2
y/
3
y/
3
dx
(
y

x
) =
R
∞
0
dye

y
1
6
y
2
=
1
3
.