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 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:

So we have:


 * $\ds \sup_{\alpha \in I} \expect {\size {X_\alpha} } \le K^\ast + 1$

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:

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 $\alpha \in I$.

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:

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.

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 $\alpha \in I$.

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.