CS648
Randomized Algorithms
Lecture notes : 2
1
Some probability problems
Lemma 1.1
For a nonnegative integer valued random variable
X
,
E
[
X
] =
∞
X
t
=0
Pr
[
X > t
]
Proof:
Since
X
is a nonnegative integer values random variable, the following equality holds.
Pr
[
X > t
] =
∞
X
j
=
t
+1
Pr
[
X
=
j
]
(1)
If we plug Equation 1 in the expression
∑
∞
t
=0
Pr
[
X > t
]
, a term
Pr
[
X
=
j
]
will appear
j
times in the sum
for each integer
j >
0
, hence
∞
X
t
=0
Pr
[
X > t
] =
∞
X
t
=0
t
Pr
[
X
=
t
] =
E
[
X
]
2
Lemma 1.2
If we select
k
integers uniformly from set
{
1
, ..., l
}
, the expected value of the smallest of them
is
(
l
+ 1)
/
(
k
+ 1)
.
We provide below a “calculative” way to prove the lemma above. We shall later present a simple and
calculationfree proof. Let
X
be the random variable for the smallest of the
k
numbers selected from
{
1
, ..., l
}
. Using Lemma 1.1,
E
[
X
]
=
∞
X
i
=0
Pr
[
X > i
]
=
∞
X
i
=0
(
n

i
k
)
(
n
k
)
=
1
(
n
k
)
i
=
n

k
X
i
=0
±
n

i
k
!
=
1
(
n
k
)
"±
n
k
!
+
±
n

1
k
!
+
...
+
±
k
k
!#
=
1
(
n
k
)
"±±
n
+ 1
k
+ 1
!

±
n
k
+ 1
!!
+
±±
n
k
+ 1
!

±
n

1
k
+ 1
!!
+
· · ·
+
±±
k
+ 1
k
+ 1
!

±
k
k
+ 1
!!#
=
1
(
n
k
)
±
n
+ 1
k
+ 1
!
=
n
+ 1
k
+ 1
1
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View Full DocumentIn the ﬁfth equality, we have used the well known binomial equality
(
m

1
r

1
)
=
(
m
r
)

(
m

1
r
)
for each term.
Calculation free proof for Lemma 1.2 :
First we state a few simple observations.
We say that two random variables
X
and
Y
have same probability distribution if for all
t
,
Pr
[
X
=
t
] =
Pr
[
Y
=
t
]
. It is easy to see that if two random variables have the same probability distribution, then their
expected values is also the same.
Consider a circle with
l
+ 1
nodes connected with edges of unit length each. All the nodes are colored
white initially. If we select
k
+ 1
nodes uniformly randomly from the set of nodes forming the circle, we get
k
+ 1
segments. Each segment is the sequence of one of more edges with black nodes at the boundaries and
all internal nodes colored white. Let us assign number 1 to the segment containing the top most node, and
going in an anticlockwise direction, assign numbers to the all other segments from 2 to
k
+ 1
. In case the
topmost node is a black node, then the ﬁrst segment is the segment with topmost node as its right boundary
node.
Problem 1.1
Consider selecting
k
+1
nodes uniformly from
l
+1
nodes of a circle and coloring them black.
What is expected length of
i
th segment ?
Solution:
The crucial point to be noted is that the nodes are selected uniformly and irrespective of their
positions. Therefore, due to this uniform sampling and symmetry, lengths of each of the
(
k
+ 1)
segments
formed have identical probability distribution though they are not independent (since they all sum up to
l
+ 1
). This implies the following assertion : Let
X
i
denote the length of
i
th segment, then
E
[
X
1
] =
E
[
X
2
] =
E
[
X
3
] =
...
=
E
[
X
k
+1
]
Also note that
∑
i
X
i
=
l
+ 1
. So
(
k
+ 1)
E
[
X
i
] =
X
j
E
[
X
j
] =
E
[
X
j
X
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 Spring '08
 SurenderBaswana
 Algorithms, Probability, actual size

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