Brownian Bridge™
Learn
Practice
Pricing
For employers
About
Contact
Log in
Start practicing
Courses
/
Probability Theory
/
03 — Article
/
Quiz
Quiz: Conditional Expectation and the Tower Property
Module 3 of 5 · Hard
Quick Quiz
1.
The modern measure-theoretic definition characterises
E
[
X
∣
G
]
\mathbb{E}[X\mid\mathcal{G}]
E
[
X
∣
G
]
(for
X
∈
L
1
X\in L^1
X
∈
L
1
, sub-σ-algebra
G
⊆
F
\mathcal{G}\subseteq\mathcal{F}
G
⊆
F
) as:
The ratio
P
(
A
∩
G
)
/
P
(
G
)
\mathbb{P}(A\cap G)/\mathbb{P}(G)
P
(
A
∩
G
)
/
P
(
G
)
for the most probable
G
\mathcal{G}
G
-event
G
G
G
.
The unique (a.s.)
G
\mathcal{G}
G
-measurable
Z
Z
Z
satisfying
∫
G
Z
d
P
=
∫
G
X
d
P
\int_G Z\,d\mathbb{P}=\int_G X\,d\mathbb{P}
∫
G
Z
d
P
=
∫
G
X
d
P
for all
G
∈
G
G\in\mathcal{G}
G
∈
G
.
The Fourier projection of
X
X
X
onto the span of
G
\mathcal{G}
G
-atoms.
The pointwise limit of
E
[
X
∣
G
n
]
\mathbb{E}[X\mid\mathcal{G}_n]
E
[
X
∣
G
n
]
as
G
n
\mathcal{G}_n
G
n
increases to
G
\mathcal{G}
G
.
2.
Let
Ω
=
{
a
,
b
,
c
,
d
}
\Omega=\{a,b,c,d\}
Ω
=
{
a
,
b
,
c
,
d
}
with
P
\mathbb{P}
P
uniform,
G
=
σ
(
{
a
,
b
}
,
{
c
,
d
}
)
\mathcal{G}=\sigma(\{a,b\},\{c,d\})
G
=
σ
({
a
,
b
}
,
{
c
,
d
})
, and
X
(
a
)
=
1
,
X
(
b
)
=
3
,
X
(
c
)
=
0
,
X
(
d
)
=
4
X(a)=1,X(b)=3,X(c)=0,X(d)=4
X
(
a
)
=
1
,
X
(
b
)
=
3
,
X
(
c
)
=
0
,
X
(
d
)
=
4
. What is
E
[
X
∣
G
]
(
a
)
\mathbb{E}[X\mid\mathcal{G}](a)
E
[
X
∣
G
]
(
a
)
?
2
1
2.5
3
3.
Same setup as the previous question. Which correctly verifies
∫
{
c
,
d
}
E
[
X
∣
G
]
d
P
=
∫
{
c
,
d
}
X
d
P
\int_{\{c,d\}}\mathbb{E}[X\mid\mathcal{G}]\,d\mathbb{P}=\int_{\{c,d\}}X\,d\mathbb{P}
∫
{
c
,
d
}
E
[
X
∣
G
]
d
P
=
∫
{
c
,
d
}
X
d
P
?
E
[
X
∣
G
]
(
c
)
=
0
\mathbb{E}[X\mid\mathcal{G}](c)=0
E
[
X
∣
G
]
(
c
)
=
0
and
E
[
X
∣
G
]
(
d
)
=
4
\mathbb{E}[X\mid\mathcal{G}](d)=4
E
[
X
∣
G
]
(
d
)
=
4
, so
∫
=
(
0
+
4
)
/
4
=
1
\int=(0+4)/4=1
∫
=
(
0
+
4
)
/4
=
1
.
4
⋅
1
2
=
2
4\cdot\tfrac12=2
4
⋅
2
1
=
2
, and
0
⋅
1
4
+
4
⋅
1
4
=
1
0\cdot\tfrac14+4\cdot\tfrac14=1
0
⋅
4
1
+
4
⋅
4
1
=
1
— not equal.
2
⋅
1
2
=
1
2\cdot\tfrac12=1
2
⋅
2
1
=
1
, and
0
⋅
1
4
+
4
⋅
1
4
=
1
0\cdot\tfrac14+4\cdot\tfrac14=1
0
⋅
4
1
+
4
⋅
4
1
=
1
— equal.
2
⋅
1
4
=
1
2
2\cdot\tfrac14=\tfrac12
2
⋅
4
1
=
2
1
, and
0
⋅
1
4
+
4
⋅
1
4
=
1
0\cdot\tfrac14+4\cdot\tfrac14=1
0
⋅
4
1
+
4
⋅
4
1
=
1
— not equal.
4.
The tower property: if
H
⊆
G
⊆
F
\mathcal{H}\subseteq\mathcal{G}\subseteq\mathcal{F}
H
⊆
G
⊆
F
, then
E
[
E
[
X
∣
G
]
∣
H
]
=
E
[
X
∣
H
]
\mathbb{E}[\mathbb{E}[X\mid\mathcal{G}]\mid\mathcal{H}]=\mathbb{E}[X\mid\mathcal{H}]
E
[
E
[
X
∣
G
]
∣
H
]
=
E
[
X
∣
H
]
a.s. What is the key step in the proof?
The tower property is a corollary of Jensen's inequality for the identity function.
H
⊆
G
\mathcal{H}\subseteq\mathcal{G}
H
⊆
G
means every
H
∈
H
H\in\mathcal{H}
H
∈
H
is also in
G
\mathcal{G}
G
, so the defining property of
E
[
X
∣
G
]
\mathbb{E}[X\mid\mathcal{G}]
E
[
X
∣
G
]
applies on each
H
\mathcal{H}
H
-set.
H
\mathcal{H}
H
and
G
\mathcal{G}
G
generate the same σ-algebra, so their conditional expectations coincide.
G
⊆
H
\mathcal{G}\subseteq\mathcal{H}
G
⊆
H
means every
G
∈
G
G\in\mathcal{G}
G
∈
G
is in
H
\mathcal{H}
H
, so
E
[
X
∣
H
]
\mathbb{E}[X\mid\mathcal{H}]
E
[
X
∣
H
]
is
G
\mathcal{G}
G
-measurable.
5.
In the
L
2
L^2
L
2
geometric interpretation,
E
[
X
∣
G
]
\mathbb{E}[X\mid\mathcal{G}]
E
[
X
∣
G
]
is the orthogonal projection of
X
X
X
onto
L
2
(
Ω
,
G
,
P
)
L^2(\Omega,\mathcal{G},\mathbb{P})
L
2
(
Ω
,
G
,
P
)
. The orthogonality condition states:
E
[
(
X
−
E
[
X
∣
G
]
)
2
]
=
0
\mathbb{E}[(X-\mathbb{E}[X\mid\mathcal{G}])^2]=0
E
[(
X
−
E
[
X
∣
G
]
)
2
]
=
0
.
E
[
X
∣
G
]
\mathbb{E}[X\mid\mathcal{G}]
E
[
X
∣
G
]
minimises the
L
1
L^1
L
1
norm
E
[
∣
X
−
Z
∣
]
\mathbb{E}[|X-Z|]
E
[
∣
X
−
Z
∣
]
over
G
\mathcal{G}
G
-measurable
Z
Z
Z
.
E
[
X
⋅
E
[
X
∣
G
]
]
=
(
E
[
X
]
)
2
\mathbb{E}[X\cdot\mathbb{E}[X\mid\mathcal{G}]]=(\mathbb{E}[X])^2
E
[
X
⋅
E
[
X
∣
G
]]
=
(
E
[
X
]
)
2
.
E
[
(
X
−
E
[
X
∣
G
]
)
⋅
Z
]
=
0
\mathbb{E}[(X-\mathbb{E}[X\mid\mathcal{G}])\cdot Z]=0
E
[(
X
−
E
[
X
∣
G
])
⋅
Z
]
=
0
for every
G
\mathcal{G}
G
-measurable
Z
∈
L
2
Z\in L^2
Z
∈
L
2
.
6.
If
X
X
X
is independent of the σ-algebra
G
\mathcal{G}
G
, what is
E
[
X
∣
G
]
\mathbb{E}[X\mid\mathcal{G}]
E
[
X
∣
G
]
?
E
[
X
∣
F
]
=
X
\mathbb{E}[X\mid\mathcal{F}]=X
E
[
X
∣
F
]
=
X
a.s. — full information recovers
X
X
X
.
X
X
X
a.s. — independence leaves
X
X
X
unchanged by conditioning.
E
[
X
]
\mathbb{E}[X]
E
[
X
]
a.s. — the unconditional mean, a constant.
$0$ a.s. — independent random variables have zero correlation.
7.
Let
(
X
,
Y
)
(X,Y)
(
X
,
Y
)
be jointly Gaussian with means
(
μ
X
,
μ
Y
)
(\mu_X,\mu_Y)
(
μ
X
,
μ
Y
)
, standard deviations
(
σ
X
,
σ
Y
)
(\sigma_X,\sigma_Y)
(
σ
X
,
σ
Y
)
, and correlation
ρ
\rho
ρ
. What is
E
[
X
∣
Y
=
y
]
\mathbb{E}[X\mid Y=y]
E
[
X
∣
Y
=
y
]
?
μ
X
+
ρ
σ
X
σ
Y
(
y
−
μ
Y
)
\mu_X+\rho\dfrac{\sigma_X}{\sigma_Y}(y-\mu_Y)
μ
X
+
ρ
σ
Y
σ
X
(
y
−
μ
Y
)
.
μ
X
+
ρ
σ
Y
σ
X
(
y
−
μ
Y
)
\mu_X+\rho\dfrac{\sigma_Y}{\sigma_X}(y-\mu_Y)
μ
X
+
ρ
σ
X
σ
Y
(
y
−
μ
Y
)
.
μ
X
+
σ
X
σ
Y
(
y
−
μ
Y
)
\mu_X+\dfrac{\sigma_X}{\sigma_Y}(y-\mu_Y)
μ
X
+
σ
Y
σ
X
(
y
−
μ
Y
)
.
μ
X
\mu_X
μ
X
(a constant, independent of
y
y
y
).
8.
A risk quant prices a Bermudan swaption with LSMC, regressing on three state variables, but the true
E
[
continuation
∣
F
t
]
\mathbb{E}[\text{continuation}\mid\mathcal{F}_t]
E
[
continuation
∣
F
t
]
depends on a fourth omitted variable. Which limitation does this illustrate?
Tower-property failure: a coarser filtration breaks time-consistency.
L
1
L^1
L
1
vs
L
2
L^2
L
2
gap: the continuation value is not square-integrable.
Misspecified conditioning set: the regression targets the wrong σ-algebra.
a.s.-uniqueness: her CE version disagrees with the true CE on a null set.
Submit
← Back to article
Open Notebook
Next: Filtrations, Adapted Processes, and Martingales →
