Mathematics Department, UCLA
T. Richthammer
spring 09, sheet 7
May 8, 2009
Homework assignments: Math 170B Probability, Sec. 1
65. Here we will show that
E
(
1
X
+1
) =
1

(1

p
)
n
+1
(
n
+1)
p
for a binomial RV
X
with parameters
n, p
:
Let
X
1
, . . . , X
n
+1
be a Bernoulli sequence with success rate
p
,
X
±
=
X
1
+
. . .
+
X
n
+1
and
X
=
X
2
+
. . .
+
X
n
+1
. Calculate
E
(
X
1
X
±
1
{
X
±
>
0
}
) in the following two ways:
(a) by using symmetry and
P
(
X
±
>
0)
(b) by conditioning on the value of
X
1
.
Answer:
(a) By symmetry we have
E
(
X
1
X
±
1
{
X
±
>
0
}
) =
E
(
X
i
X
±
1
{
X
±
>
0
}
), which implies
n
+1
∑
i
=1
E
(
X
i
X
±
1
{
X
±
>
0
}
) =
(
n
+ 1)
E
(
X
1
1
{
X
±
>
0
}
). On the other hand
n
+1
∑
i
=1
E
(
X
i
X
±
1
{
X
±
>
0
}
) =
E
(
X
1
+
...
+
X
n
+1
X
±
)1
{
X
±
>
0
}
) =
E
(
X
±
X
±
1
{
X
±
>
0
}
) =
P
(
X
±
>
0) = 1

q
n
+1
, and thus
E
(
X
1
X
±
1
{
X
±
>
0
}
) =
1

q
n
+1
n
+1
.
(b)
X
=
X
±

X
1
is a binomial RV with parameters
n, p
that is independent from
X
1
. We
have
E
(
X
1
X
±
1
{
X
±
>
0
}

X
1
= 1) =
E
(
1
1+
X
) and
E
(
X
1
X
±
1
{
X
±
>
0
}

X
1
= 0) = 0, so
E
(
X
1
1
{
X
±
>
0
}
) =
p
E
(
1
1+
X
). Combining (a) and (b) gives the desired result.
66.* In the lecture we considered
N
x
:= min
{
n
:
X
1
+
. . .
+
X
n
≥
x
}
, where
X
1
, X
2
, . . .
are
independent RVs with uniform distribution on [0
,
1].
For
m
(
x
) :=
E
(
N
x
) we showed
m
(
x
) =
e
x
for 0
≤
x
≤
1 by conditioning on the value of
X
1
.
(a) Calculate
m
(
x
) for 1
≤
x
≤
2 by conditioning, and describe a procedure how to
obtain
m
(
x
) for larger
x
.
(b) Calculate
m
(
x
) for 0
≤
x
≤
1 using indicators.
Answer:
(a) Let
x
≥
1. By conditioning on
X
1
we get.
E
(
N
x

X
1
=
x
1
) =
E
(
N
x

x
1
) + 1 =
m
(
x

x
1
) + 1. So
m
(
x
) =
E
(
N
x
) =
R
1
0
dx
1
m
(
x

x
1
) + 1. This recursive equation can
be used to calculate
m
(
x
) for
x
∈
[
n, n
+ 1) if you already know
m
(
x
) for
x
∈
[
n

1
, n
).
If
m
n
(
x
) =
m
(
n
+
x
) for
x
∈
[0
,
1] we get
m
n
(
x
) =
R
x
0
dx
1
m
n
(
x

x
1
) +
R
1
x
dx
1
m
n

1
(1 +
x

x
1
) + 1 =
R
x
0
m
n
(
u
)
du
+
R
1
x
m
n

1
(
u
)
du
+ 1. So
m
±
n
(
x
) =
m
n
(
x
)

m
n

1
(
x
). This gives
diﬀerential equations for
m
n
(
x
) that can be solved recursively, using the initial conditions
m
n
(0) =
m
n

1
(1) and
m
0
(
x
) =
e
x
. For instance for
m
1
we get
m
±
1
(
x
) =
m
1
(
x
)

e
x
and
m
1
(0) =
e
. Solving this diﬀerential equation by means of variation of the constant we get
m
1
(
x
) = (
e

x
)
e
x
, which gives
m
(
x
) = (
e

x
+ 1)
e
x

1
for 1
≤
x
≤
2.
(b) We have
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