Problem Set 12 — Solutions
Not due.
Problem 1
(Exercise 7.28)
a First note that
E(
Y
n

Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
) = E(
Z
n
+
m

Z
m

Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
)
= E(
Z
n
+
m

Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
)

E(
Z
m

Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
)
by the linearity of the expectation, since, as
Z
n
is a martingale, both terms are necessarily
ﬁnite. (Essentially,
Z
n
being a martingale rules out the fact that
E(
Z
n
+
m

Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
) = E(
Z
m

Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
) =
∞
,
in which case E(
Z
n
+
m

Z
m

Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
) may exist but the diﬀerence of the
two expectations is undeﬁned.)
Since E(
Z
n
+
m

Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
) =
Z
n
+
m

1
(as
Z
n
is a martingale) and E(
Z
m

Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
) =
Z
m
(as
Z
m
is included in the condition on
Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
),
it follows that
E(
Y
n

Z
n
+
m

1
,Z
n
+
m

2
,...,Z
1
) =
Z
n
+
m

1

Z
m
.
b First note that specifying
Z
n
+
m

1
,...,Z
1
entirely speciﬁes
Y
n

1
,...,Y
1
but not viceversa
(why?). However, we can see that specifying
Z
n
+
m

1
,...,Z
1
is equivalent to specifying
Y
n

1
,...,Y
1
,Z
m
,Z
m

1
,...,Z
1
. By Problem 2a), we therefore have
E(
Y
n

Y
n

1
,...,Y
1
) = E(E(
Y
n

Y
n

1
,...,Y
1
,Z
m
,Z
m

1
,...,Z
1
)

Y
n

1
,...,Y
1
)
= E(E(
Y
n

Z
n
+
m

1
,Z
n
+
m

2
,...,Y
m
+1
,Z
m
,Z
m

1
,...,Z
1
)

Y
n

1
,...,Y
1
)
= E(
Z
n
+
m

1

Z
m

Y
n

1
,...,Y
1
)
by part a)
= E(
Y
n

1

Y
n

1
,...,Y
1
)
by deﬁnition
=
Y
n

1
c E(

Y
n

) = E(

Z
n
+
m

Z
m

)
≤
E(

Z
n
+
m

+

Z
m

). Since
Z
n
is a martingale, it follows that
E(

Z
k

<
∞
for all
k
, and so E(

Y
n

)
≤
E(

Z
n
+
m

) + E(

Z
m

)
<
∞
.
Problem 2
a We show the tower property assuming
X,Y
and
Z
have densities. The proof adapts easily
to the case where the three random variables are discrete by considering the corresponding
probability mass functions. The most general case is discussed afterwards.
1