Mathematics Department, UCLA
T. Richthammer
spring 09, sheet 3
Apr 10, 2009
Homework assignments: Math 170B Probability, Sec. 1
19. Calculate the expectation and the variance for the sum of 12 rolls of a fair die.
Answer:
X
=
X
1
+
. . .
+
X
12
, where the
X
i
are the values of the 6 rolls. The
X
i
are independent
and we can calculate
E
(
X
i
) =
1
6
(1 +
. . .
+ 6) =
7
2
and
E
(
X
2
i
) =
1
6
(1
2
+
. . .
+ 6
2
) =
91
6
, so
V
(
X
i
) =
35
12
. So
E
(
X
) = 12
E
(
X
i
) = 42 and
V
(
X
) = 12
V
(
X
i
) = 35.
20. Let
X
n
be the time of the
n
th success in independent trials with success rate
p
.
(a) Calculate
E
(
X
n
) and
V
(
X
n
). (Hint: Use 08(c))
(b) Explain brieﬂy why you would expect
ρ
(
X
n
, X
n
+
k
)
≥
0, then calculate it.
Answer:
(a) By 08(c) we have that
X
n
=
Y
1
+
. . .
+
Y
n
, where the
Y
i
are independent geometric
RVs. Since
E
(
Y
i
) =
1
p
and
V
(
Y
i
) =
1

p
p
2
for all
i
, we get
E
(
X
n
) =
E
(
Y
1
) +
. . .
+
E
(
Y
n
) =
n
p
and
V
(
X
n
) =
V
(
Y
1
) +
. . .
+
V
(
Y
n
) =
n
(1

p
)
p
2
.
(b) If
X
n
is known to be large, then
X
n
+
k
is more likely to be large than if
X
n
is known to
be small. Thus
X
n
, X
n
+
k
should be positively correlated. Using 08(c),
Cov
(
X
n
, X
n
+
k
) =
Cov
(
Y
1
+
. . .
+
Y
n
, Y
1
+
. . .
+
Y
n
+
k
) =
n
∑
i
=1
n
+
k
∑
j
=1
Cov
(
Y
i
, Y
j
) =
n
∑
i
=1
Cov
(
Y
i
, Y
i
)+
∑
i
±
=
j
Cov
(
Y
i
, Y
j
) =
n
∑
i
=1
V
(
Y
i
) + 0 =
n
(1

p
)
p
2
, where we have used that independent RVs are uncorrelated. So
ρ
(
X
n
, X
n
+
k
) =
Cov
(
X
n
,X
n
+
k
)
√
V
(
X
n
)
V
(
X
n
+
k
)
=
q
n
n
+
k
.
21. Suppose there are two methods (A, B) for measuring the distance from the earth to the
moon. A scientist using method A gets the values 402, 405, 399, 398, 399 (in 1000 km).
A second scientist using method A gets the values 407, 399, 400, 398, 399, 400, 397, 402,
401, 397. A third scientist using method B gets the values 399, 400, 402, 399, 402, 399.
(a) What estimate does each scientist get for the true value of the distance?
(b) Explain brieﬂy why you would expect the estimate of the second scientist to be better
than the estimate of the ﬁrst scientist.
(c) Which method seems to be better? Conﬁrm your answer by a calculation.
Answer:
(a) The estimate for the true value is the sample mean
¯
X
=
X
1
+
...
+
X
n
n
. Here we get
400
.
6
,
400 and 400
.
2 respectively.
(b) The average deviation of
¯
X
from the true value is given by
V
(
¯
X
). By the lecture
V
(
¯
X
) =
1
n
V
(
X
i
). So a larger
n
gives a smaller deviation.
(c) The variance of the measurement can be estimated by the sample variance
¯
V
=
(
X
1

¯
X
)
2
+
...
+(
X
n

¯
X
)
2
n

1
. Here we get 8
.
3
,
8
.
7
,
2
.
2 respectively. So the second method is better.