6.262 Discrete Stochastic Processes, Spring 2010
Problem Set 1 — Solutions
due: Friday, February 12, 2010
Problem 1 (Exercise 1.11)
a We have two IID random variables
X
1
and
X
2
. One is strictly smaller than the other with
probability 1 because P(
X
1
=
X
2
) = 0. By symmetry, each variable is equally likely to be
the smallest, so
P
(
X
1
< X
2
) = 1
/
2
.
b The answer is very similar to that of the previous problem, but let’s be a little more sys
tematic. With probability 1, the random variables
X
1
,X
2
,...,X
n
can be ordered to form
a strictly increasing ﬁnite sequence since the random variables are continuous. It follows
that (
X
1
,X
2
,...,X
n
) can be put in correspondence with all possible orderings of the set
{
1
,
2
,...,n
}
. For instance, for
n
= 3, the event
{
X
3
> X
1
> X
2
}
then corresponds to (2
,
1
,
3).
Now it’s easy: since the random variables are IID, all the
n
! orderings of
{
1
,
2
,...,n
}
are
equally likely. Since the events of interest correspond to having
n
in the
n
th
position, it follows
that the desired probability is (
n

1)!
/n
! = 1
/n
.
c Again, let’s be systematic. Let
I
n
denote the indicator associated with the event that
X
n
is
the record to date. Then, E[
∑
m
n
=1
I
n
] yields the expected number of records in the sequences
X
1
,X
2
,...,X
m
. Since expectation is linear, the quantity of interest becomes
∑
m
n
=1
E[
I
n
] =
∑
m
n
=1
n

1
. As
m
→ ∞
, we obtain the (divergent) harmonic series, which tells us that the
expected number of recordstodate in an inﬁnite sequence is itself inﬁnite.
Problem 2 (Exercise 1.23)
(a) At a suitably abstracted level, this problem is identical to the previous one. Can you
see why the answer is (
n

1)!
/n
! = 1
/n
?
(b) Let
I
k
denote the indicator associated with the event that user
k
receives the correct pass
word. Then, E[
∑
n
k
=1
I
k
] yields the expected number of users who are able to continue
using the system. Again, since expectation is linear, the quantity of interest becomes
∑
n
k
=1
E[
I
k
] =
∑
n
k
=1
n

1
=
n/n
= 1.
Challenge question: can you compute the probability that exactly
k
users receive correct
passwords?
Problem 3 (Exercise 1.31)
(a) We are interested in calculating the probability that
S
n
is contained within an interval
of ﬁxed width 2
m
+ 1 centered at the mean,
nδ
. As
n
increases, the standard deviation
of
S
n
increases as well (as
√
n
, in fact), and we expect the probability mass associated
with the interval to vanish in the limit.
1
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View Full DocumentIntuition is useful, but intuition can be wrong, so let’s do this precisely. Let
Y
n
denote
the zeromean, unitvariance version of
S
n
, that is,
Y
n
= (
S
n

nδ
)
/
√
nσ
. (Note that
σ
=
p
var(
X
1
) =
p
δ
(1

δ
) does not vary with
n
.) Then,
X
nδ

m
≤
i
≤
nδ
+
m
P
(
S
n
=
i
) =
F
S
n
(
nδ
+
m
)

F
S
n
(
nδ

m
)

=
F
Y
n
±
m
√
nσ
²

F
Y
n
±

m
√
nσ
²

Note 1:
To clarify some notation,
F
S
n
(
a
) = P(
S
n
≤
a
), while
F
S
n
(
a
)

= P(
S
n
< a
). Can you
see why
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 Spring '08
 Moon,J
 Central Limit Theorem, Normal Distribution, lim

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