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05 — Article
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Quiz
Quiz: Lp Spaces and Modes of Convergence
Module 5 of 5 · Medium
Quick Quiz
1.
For
1
≤
p
<
∞
1\le p<\infty
1
≤
p
<
∞
, the
L
p
L^p
L
p
norm of a random variable
X
X
X
is defined as:
∥
X
∥
p
=
(
E
[
∣
X
∣
p
]
)
1
/
p
\|X\|_p=(\mathbb{E}[|X|^p])^{1/p}
∥
X
∥
p
=
(
E
[
∣
X
∣
p
]
)
1/
p
.
∥
X
∥
p
=
E
[
∣
X
∣
p
]
\|X\|_p=\mathbb{E}[|X|^p]
∥
X
∥
p
=
E
[
∣
X
∣
p
]
.
∥
X
∥
p
=
ess sup
∣
X
∣
1
/
p
\|X\|_p=\operatorname{ess\,sup}|X|^{1/p}
∥
X
∥
p
=
ess
sup
∣
X
∣
1/
p
.
∥
X
∥
p
=
E
[
∣
X
∣
]
p
\|X\|_p=\mathbb{E}[|X|]^p
∥
X
∥
p
=
E
[
∣
X
∣
]
p
.
2.
On a probability space
(
Ω
,
F
,
P
)
(\Omega,\mathcal{F},\mathbb{P})
(
Ω
,
F
,
P
)
, which inclusion holds for
1
≤
p
≤
q
≤
∞
1\le p\le q\le\infty
1
≤
p
≤
q
≤
∞
?
L
q
⊆
L
p
L^q\subseteq L^p
L
q
⊆
L
p
(the higher-integrability space sits inside the lower).
L
p
=
L
q
L^p=L^q
L
p
=
L
q
for all
p
,
q
p,q
p
,
q
(the spaces coincide).
Neither inclusion holds in general.
L
p
⊆
L
q
L^p\subseteq L^q
L
p
⊆
L
q
(the lower-exponent space sits inside the higher).
3.
Let
X
=
1
X=1
X
=
1
with probability
1
2
\tfrac12
2
1
and
X
=
3
X=3
X
=
3
with probability
1
2
\tfrac12
2
1
. Compute
∥
X
∥
2
\|X\|_2
∥
X
∥
2
.
2
1
5
≈
2.236
\sqrt5\approx2.236
5
≈
2.236
5
4.
Which implication among modes of convergence holds for every sequence
(
X
n
)
(X_n)
(
X
n
)
on a probability space?
Convergence in probability
⇒
\Rightarrow
⇒
convergence a.s.
Convergence in distribution
⇒
\Rightarrow
⇒
convergence in probability.
Convergence in
L
p
L^p
L
p
⇒
\Rightarrow
⇒
convergence in probability.
Convergence in
L
1
L^1
L
1
⇒
\Rightarrow
⇒
convergence a.s.
5.
The Riesz–Fischer theorem states that
L
p
L^p
L
p
(for
1
≤
p
≤
∞
1\le p\le\infty
1
≤
p
≤
∞
) is:
separable but not complete whenever the exponent
p
<
∞
p<\infty
p
<
∞
.
finite-dimensional whenever the sample space is infinite.
complete: every Cauchy sequence has a limit in
L
p
L^p
L
p
.
a Hilbert space for every exponent
p
≥
1
p\ge1
p
≥
1
, not merely
p
=
2
p=2
p
=
2
.
6.
Hölder's inequality, for conjugate exponents
1
p
+
1
q
=
1
\tfrac1p+\tfrac1q=1
p
1
+
q
1
=
1
, bounds
E
[
∣
X
Y
∣
]
\mathbb{E}[|XY|]
E
[
∣
X
Y
∣
]
by:
∥
X
∥
p
∥
Y
∥
q
\|X\|_p\,\|Y\|_q
∥
X
∥
p
∥
Y
∥
q
.
∥
X
∥
p
+
∥
Y
∥
q
\|X\|_p+\|Y\|_q
∥
X
∥
p
+
∥
Y
∥
q
.
it provides only a lower bound,
E
[
∣
X
Y
∣
]
≥
∥
X
∥
p
∥
Y
∥
q
\mathbb{E}[|XY|]\ge\|X\|_p\|Y\|_q
E
[
∣
X
Y
∣
]
≥
∥
X
∥
p
∥
Y
∥
q
.
∥
X
∥
p
p
∥
Y
∥
q
q
\|X\|_p^p\,\|Y\|_q^q
∥
X
∥
p
p
∥
Y
∥
q
q
.
7.
If
X
n
→
X
X_n\to X
X
n
→
X
in probability, which additional condition upgrades the convergence to
L
1
L^1
L
1
(Vitali's convergence theorem)?
Each
X
n
X_n
X
n
has the same distribution.
The sequence
(
X
n
)
(X_n)
(
X
n
)
is monotone increasing.
The sequence
(
X
n
)
(X_n)
(
X
n
)
is uniformly integrable.
The sequence
(
X
n
)
(X_n)
(
X
n
)
is bounded in
L
2
L^2
L
2
.
8.
A quant's Monte-Carlo estimator of
E
[
X
]
\mathbb{E}[X]
E
[
X
]
converges a.s. by the strong law, but its sample variance never stabilises. The payoff satisfies
E
[
∣
X
∣
]
<
∞
\mathbb{E}[|X|]<\infty
E
[
∣
X
∣
]
<
∞
yet
E
[
X
2
]
=
∞
\mathbb{E}[X^2]=\infty
E
[
X
2
]
=
∞
. What does this reveal?
X
∈
L
1
∖
L
2
X\in L^1\setminus L^2
X
∈
L
1
∖
L
2
: the mean exists, but the standard
n
\sqrt n
n
CLT error bars are invalid.
The estimator is biased; a.s. convergence requires
L
2
L^2
L
2
integrability.
The strong law of large numbers has failed.
The estimator converges in distribution but not in probability.
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