Mathematics Department, UCLA
T. Richthammer
spring 09, sheet 6
May 1, 2009
Homework assignments: Math 170B Probability, Sec. 1
52. Show that
E
(
X
) =
E
(
E
(
X

Y
)) in the case that
X, Y
have a joint PDF.
Answer:
Let
f
be the joint PDF, then
f
X
(
x

Y
=
y
) =
f
X,Y
(
x,y
)
f
Y
(
y
)
, so
E
(
X

Y
=
y
) =
R
dx
f
X,Y
(
x,y
)
f
Y
(
y
)
x
,
so
E
(
E
(
X

Y
)) =
R
dyf
Y
(
y
)
R
dx
f
X,Y
(
x,y
)
f
Y
(
y
)
x
=
R
dydxf
X,Y
(
x, y
)
x
=
E
(
X
).
53. Conditional covariance formula.
(a) Show that
Cov
(
X, Y
) =
E
((
Cov
(
X, Y

Z
)) +
Cov
(
E
(
X

Z
)
,
E
(
Y

Z
)).
(b) Let
X
1
, X
2
be chosen completely at random in the interval [0
, x
], where
X
=
x
is
randomly chosen with respect to a uniform distribution on [0
,
1]. Calculate
E
(
X
i
),
V
(
X
i
) and
Cov
(
X
1
, X
2
) by conditioning on
X
.
Answer:
(a)
E
((
Cov
(
X, Y

Z
)) =
E
(
E
(
XY

Z
)

E
(
X

Z
)
E
(
Y

Z
)) =
E
(
E
(
XY

Z
))

E
(
E
(
X

Z
)
E
(
Y

Z
))
and
Cov
(
E
(
X

Z
)
,
E
(
Y

Z
)) =
E
(
E
(
X

Z
)
E
(
Y

Z
))

E
(
E
(
X

Z
))
E
(
E
(
Y

Z
)). Using the power
property we thus get
E
((
Cov
(
X, Y

Z
)) +
Cov
(
E
(
X

Z
)
,
E
(
Y

Z
)) =
E
(
XY
)

E
(
X
)
E
(
Y
) =
Cov
(
X, Y
).
(b) We are given that
f
X
1
,X
2
(
x
1
, x
2

X
=
x
) =
1
x
1
[0
,x
]
(
x
1
)
1
x
1
[0
,x
]
(
x
2
) (independent uni
form distribution on an interval of length
x
).
Recall from 170A that a RV
Y
that
is uniformly distributed on [0
, a
] has
E
(
Y
) =
a
2
and
V
(
Y
) =
a
2
12
. So
E
(
X
i

X
) =
X
2
,
V
(
X
i

X
) =
X
2
12
and
Cov
(
X
1
, X
2

X
) = 0. So
E
(
X
i
) =
E
(
E
(
X
i

X
)) =
E
(
X
2
) =
1
2
E
(
X
) =
1
4
,
V
(
X
i
) =
E
(
V
(
X
i

X
)) +
V
(
E
(
X
i

X
)) =
E
(
X
2
12
) +
V
(
X
2
) =
1
12
1
3
+
1
4
1
12
=
7
144
.
Cov
(
X
1
, X
2
) =
E
((
Cov
(
X
1
, X
2

X
)) +
Cov
(
E
(
X
1

X
)
,
E
(
X
2

X
)) = 0 +
Cov
(
X
2
,
X
2
) =
1
4
V
(
X
) =
1
48
.
54. A bin contains
N
=
N
1
+
N
2
lightbulbs of 2 types (
N
i
of type
i
). The type
i
light bulbs
function for a random amount of time (with mean
μ
i
and standard deviation
σ
i
). Let
X
denote the lifetime of a bulb that is chosen at random. Calculate
E
(
X
) and
V
(
X
).
Answer:
Let
Y
denote the type of the chosen lightbulb. We are given that
E
(
X

Y
=
i
) =
μ
i
and
V
(
X

Y
=
i
) =
σ
2
i
. Using the tower property for the expectation we get
E
(
X
) =
∑
i
E
(
X

Y
=
i
)
P
(
Y
=
i
) =
μ
1
N
1
N
+
μ
2
N
2
N
=
μ
1
N
1
+
μ
2
N
2
N
. For the variance we have to cal
culate
E
(
V
(
X

Y
)) =
∑
i
σ
2
i
N
i
N
=
σ
2
1
N
1
+
σ
2
2
N
2
N
and
V
(
E
(
X

Y
)) =
∑
i
μ
2
i
N
i
N

(
μ
1
N
1
+
μ
2
N
2
N
)
2
=
(
μ
1

μ
2
)
2
N
1
N
2
N
2
, where we have used that
E
(
X

Y
) has the PMF
p
(
μ
i
) =
N
i
N
. Using the
conditional variance formula we get
V
(
X
) =
σ
2
1
N
1
+
σ
2
2
N
2
N
+
(
μ
1

μ
2
)
2
N
1
N
2
N
2
.
55. Conditional expectation, correlation and independence.