Stat 310A/Math 230A Theory of Probability
Homework 4 Solutions
Andrea Montanari
Due on October 25, 2010
Exercises on independent random variables and product measures
Exercise [1.4.18]
We will use the following fact repeatedly: the probability that
X
is divisible by
j
is
P
(
D
j
) =
∞
X
k
=1
(
kj
)

s
/ζ
(
s
) =
j

s
.
1. Let
p
i
be an enumeration of distinct primes. By deﬁnition, it suﬃces to show that for any ﬁnite
subcollection
{
p
j
}
n
j
=1
and any
n <
∞
,
P
(
n
\
j
=1
D
p
j
) =
n
Y
j
=1
P
(
D
p
j
)
.
Indeed, all the primes
p
j
in the subcollection divide
k
if and only if
k
is divisible by their product. So
the LHS is just (
Q
n
i
=1
p
j
)

s
which by the fact we derived before, equals the RHS.
2. Euler’s formula is the statement that
{
X
= 1
}
if and only if
X
is not divisible by any prime number.
Indeed, as the latter event is
T
p
D
c
p
, by continuity from below of
P
(
·
) we have that
1
ζ
(
s
)
=
P
(
X
= 1) = lim
n
→∞
P
(
n
\
j
=1
D
c
p
j
)
.
In part (a) we veriﬁed the mutual independence of the collections
{
D
p
}
,
p
prime, each of which is
trivially a
π
system. This implies the mutual independence of
σ
(
D
p
),
p
prime (see Corollary 1.4.7),
hence that of the events
D
c
p
. With
P
(
D
p
) =
p

s
, we get that
P
(
T
n
j
=1
D
c
p
j
) =
Q
n
j
=1
(1

p

s
j
) leading
to Euler’s formula when taking
n
→ ∞
.
3. The event that no perfect square other than 1 divides
X
, is precisely the event that
p
2
does not divide
X
for every prime
p
, which is
T
p
D
c
p
2
. Similarly to part (a), it is not hard to verify that
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 '11
 MONTANARI,A.
 Probability, Probability theory, Tn, Xn, xn ∈Bn

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