User:Caliburn/s/prob/Characterization of Uniformly Integrable Family of Random Variables
Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\sequence {X_\alpha}_{\alpha \in I}$ be an $I$-indexed uniformly integrable family of random variables.
Then the following hold:
- $\ds (1) \quad \sup_{\alpha \in I} \expect {\size {X_\alpha} } < \infty$
- $\ds (2) \quad \sup_{\alpha \in I} \map \Pr {\size {X_\alpha} > K} \to 0$ as $K \to \infty$
- $\ds (3) \quad$ for all $\epsilon > 0$ there exists $\delta > 0$ such that whenever $A \in \Sigma$ has $\map \Pr A < \delta$ implies that $\expect {\size {X_\alpha} \cdot \chi_A} < \epsilon$ for all $\alpha \in I$.
Conversely, if $(1)$ and $(3)$ simultaneously hold, or $(2)$ and $(3)$ simultaneously hold, then $\sequence {X_\alpha}_{\alpha \in I}$ is uniformly integrable.
Proof
Proof that UI implies $(1)$
Suppose that $\sequence {X_\alpha}_{\alpha \in I}$ is uniformly integrable.
Then there exists $K^\ast > 0$ such that:
- $\ds \sup_{\alpha \in I} \expect {\size {X_\alpha} \cdot \chi_{\set {\size {X_\alpha} > K^\ast} } } \le 1$
Now take $\alpha \in I$, we have:
\(\ds \expect {\size {X_\alpha} }\) | \(=\) | \(\ds \expect {\size {X_\alpha} \cdot \chi_{\set {\size {X_\alpha} \le K^\ast} } } + \expect {\size {X_\alpha} \cdot \chi_{\set {\size {X_\alpha} > K^\ast} } }\) | Expectation is Linear | |||||||||||
\(\ds \) | \(\le\) | \(\ds K^\ast + 1\) | Expectation is Monotone |
So we have:
- $\ds \sup_{\alpha \in I} \expect {\size {X_\alpha} } \le K^\ast + 1$
$\Box$
Proof that UI implies $(3)$
Suppose that $\sequence {X_\alpha}_{\alpha \in I}$ is uniformly integrable.
Let $A \in \Sigma$ and $\alpha \in I$.
We have:
\(\ds \expect {\size {X_\alpha} \cdot \chi_A}\) | \(=\) | \(\ds \expect {\size {X_\alpha} \chi_{A \cap \set {\size {X_\alpha} > K} } } + \expect {\size {X_\alpha} \chi_{A \cap \set {\size {X_\alpha} \le K} } }\) | Expectation is Linear, Characteristic Function of Disjoint Union | |||||||||||
\(\ds \) | \(\le\) | \(\ds \expect {\size {X_\alpha} \cdot \chi_{\set {\size {X_\alpha} > K} } } + K \expect {\chi_{\set {\size {X_\alpha} > K} \cap A} }\) | Characteristic Function of Subset, Expectation is Monotone | |||||||||||
\(\ds \) | \(=\) | \(\ds \expect {\size {X_\alpha} \cdot \chi_{\set {\size {X_\alpha} > K} } } + K \map \Pr {A \cap \set {\size {X_\alpha > K} } }\) | Integral of Characteristic Function | |||||||||||
\(\ds \) | \(\le\) | \(\ds \expect {\size {X_\alpha} \cdot \chi_{\set {\size {X_\alpha} > K} } } + K \map \Pr A\) | Measure is Monotone |
Let $\epsilon > 0$.
Fix $K^\ast > 0$ such that:
- $\ds \expect {\size {X_\alpha} \cdot \chi_{\set {\size {X_\alpha} > K^\ast} } } < \frac \epsilon 2$
for all $\alpha \in I$.
Now take:
- $\ds \map \Pr A < \frac \epsilon {2 K^\ast}$
Then we have:
- $\expect {\size {X_\alpha} \cdot \chi_{\set {\size {X_\alpha} > K^\ast} } } + K^\ast \map \Pr A < \epsilon$
for all $\alpha \in I$.
So, if $A \in \Sigma$ is such that:
- $\ds \map \Pr A < \frac \epsilon {2 K^\ast}$
we have:
- $\ds \expect {\size {X_\alpha} \cdot \chi_A} < \epsilon$ for all $\alpha \in I$.
$\Box$
Proof that $(1)$ implies $(2)$
We show that $(1)$ implies $(2)$.
Suppose that:
- $\ds \sup_{\alpha \in I} \expect {\size {X_\alpha} } < \infty$
Let:
- $\ds M = \sup_{\alpha \in I} \expect {\size {X_\alpha} }$
Then by Markov's Inequality, we have for each $\alpha \in I$ and $K > 0$:
- $\ds \map \Pr {\size {X_\alpha} > K} \le \frac 1 K \expect {\size {X_\alpha} }$
so that:
\(\ds \sup_{\alpha \in I} \map \Pr {\size {X_\alpha} > K}\) | \(\le\) | \(\ds \frac 1 K \sup_{\alpha \in I} \expect {\size {X_\alpha} }\) | Multiple of Supremum | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac M K\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds 0\) | as $K \to \infty$ |
$\Box$
Proof that UI implies $(2)$
Suppose that $\sequence {X_\alpha}_{\alpha \in I}$ is uniformly integrable.
Then $(1)$ holds.
We know that $(1)$ implies $(2)$, so $(2)$ holds.
$\Box$
Proof that $(2)$ and $(3)$ holding simultaneously implies UI
Suppose that for $\sequence {X_\alpha}_{\alpha \in I}$, $(2)$ and $(3)$ simultaneously hold.
Let $\epsilon > 0$.
From $(3)$, we can pick $\delta > 0$ such that for all $A \in \Sigma$ with $\map \Pr A < \delta$, we have:
- $\ds \expect {\size {X_\alpha} \cdot \chi_A} < \epsilon$
From $(2)$, there exists $K^\ast > 0$ such that:
- $\map \Pr {\size {X_\alpha} > K^\ast} < \delta$ for all $\alpha \in I$.
So we have, setting:
- $A = \set {\size {X_\alpha} > K^\ast}$
we have:
- $\expect {\size {X_\alpha} \cdot \chi_A} < \epsilon$ for all $\alpha \in I$.
$\Box$
Proof that $(1)$ and $(3)$ holding simultaneously implies UI
Suppose that $(1)$ and $(3)$ simultaneously hold for $\sequence {X_\alpha}_{\alpha \in I}$.
Since $(1)$ implies $(2)$, $(2)$ and $(3)$ simultaneously hold.
So $\sequence {X_\alpha}_{\alpha \in I}$ is uniformly integrable.