14 Answers and Solutions
73. As N
i
is a binomial random variable with para-
meters
(n, P
i
), we have (i) E[N
i
]=nP
ji
(ii) Va r( X
i
)
=
nP
i
=(1 − P
i
); (iii) for i = j, the covariance of
N
i
and N
j
can be computed as:
Cov
(N
i
, N
j
)=Cov
∑
k
X
k
,
∑
k
Y
k
,
where X
k
(Y
k
) is 1 or 0, depending upon whether
or not outcome k is type i
(j). Hence,
Cov
(N
i
, N
j
)=
∑
k
∑
Cov(X
k
, Y
).
Now for k
= , Cov(X
k
, Y
)=0 by independence
of trials and so
Cov
(N
i
, N
j
)=
∑
k
Cov(X
k
, Y
k
)
=
∑
k
(E[X
k
Y
k
] − E[X
k
]E[Y
k
])
= −
∑
k
E[X
k
]E[Y
k
] (since X
k
Y
k
= 0)
= −
∑
k
P
i
P
j
= −nP
i
P
j
.
(iv) Letting
Y
i
=
1, if no type i’s occur
0, otherwise,
we have that the number of outcomes that never
occur is equal to
r
∑
1
Y
i
and thus,
E
r
∑
1
Y
i
=
r
∑
1
E[Y
i
]
=
r
∑
1
P{outcomes i does not occur}
=
r
∑
1
(
1 − P
i
)
n
.
74. (i) As the random variables are independent,
identically distributed, and continuous, it fol-
lows that, with probability 1, they will all
have different values. Hence the largest of
X
1
,...,X
n
is equally likely to be either X
1
or
X
2
...orX
n
. Hence, as there is a record at time
n when X
n
is the largest value, it follows that
P
{a record occurs at n} =
1
n
.
(ii) Let I
j
=
1, if a record occurs at j
0, otherwise.
Then
E
n
∑
1
I
j
=
n
∑
1
E[I
j
]=
n
∑
1
1
j
.
(iii) It is easy to see that the random variables
I
1
, I
2
,...,I
n
are independent. For instance, for
j
< k
P
{I
j
= 1/I
k
= 1} = P{I
j
= 1},
since knowing that X
k
is the largest of
X
1
,...,X
j
,...,X
k
clearly tells us nothing
about whether or not X
j
is the largest of
X
1
,...,X
j
. Hence,s
Var
n
∑
1
I
j
=
n
∑
1
Var(I
j
)=
n
∑
j=1
1
j
j
−1
j
.
(iv) P
{N > n}
=
P{X
1
is the largest of X
1
,...,X
n
} =
1
n
.
Hence,
E
[N]=
∞
∑
n=1
P{N > n} =
∞
∑
n=1
1
n
= ∞.
75. (i) Knowing the values of N
1
,...,N
j
is equiva-
lent to knowing the relative ordering of the
elements a
1
,...,a
1
. For instance, if N
1
= 0,
N
2
= 1, N
3
= 1 then in the random permu-
tation a
2
is before a
3
, which is before a
1
. The
independence result follows for clearly the
number of a
1
,...,a
1
that follow a
i+1
does not
probabilistically depend on the relative order-
ing of a
1
,...,a
i
.
(ii) P
{N
i
= k} =
1
i
, k
= 0,1,...,i − 1
which follows since of the elements
a
1
,...,a
i+1
the element a
i+1
is equally
likely to be first or second or ...or
(i + 1)
st
.
(iii) E
[N
i
]=
1
i
i−1
∑
k=0
k =
i −1
2
E
[N
2
i
]=
1
i
i−1
∑
k=0
k
2
=
(
i −1)(2i −1)
6
and so
Var
(N
i
)=
(
i −1)(2i −1)
6
−
(
i −1)
2
4
=
i
2
−1
12
.
76. E
[XY]=µ
x
µ
y
E[( XY )
2
]=(µ
2
x
+
σ
2
x
)(
µ
2
y
+
σ
2
y
)
Var(XY )=E[(XY )
2
] − (E[XY])
2