Uniform integrability

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Definition (Uniformly integrable)

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. A family [math](X_i)_{i\in I}[/math] of r.v.'s in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math], indexed by an arbitrary index set [math]I[/math], is called uniformly integrable (denoted by u.i.) if

[[math]] \lim_{a\to\infty}\sup_{i\in I}\E[\vert X_i\vert \one_{\{\vert X_i\vert \gt a\}}]=0. [[/math]]

A singel r.v. in [math]L^1(\Omega,\F,(\F_n)_{n\geq0},\p)[/math] is always u.i. (this follows from dominated convergence). If [math]\vert I\vert \lt \infty[/math], then using

[[math]] \vert X_i\vert \leq \sum_{j\in I}\vert X_j\vert, [[/math]]

we get

[[math]] \E[\vert X_i\vert \one_{\{\vert X_i\vert \gt a\}}]\leq \sum_{j\in I}\E[\vert X_j\vert\one_{\{ X_j\vert \gt a\}}]\xrightarrow{a\to\infty}0. [[/math]]

Let [math](X_i)_{i\in I}[/math] be u.i. For [math]a[/math] large enough, we then have

[[math]] \sup_{i\in I}\E[\vert X_i\vert \one_{\{\vert X_i\vert \gt a\}}]\leq 1. [[/math]]

Hence

[[math]] \sup_{i\in I}\E[\vert X_i\vert]=\sup_{i\in I}\E[\vert X_i\vert (\one_{\{\vert X_i\vert\leq a\}}+\one_{\{ \vert X_i\vert \gt a\}})]\leq 1+a \lt \infty, [[/math]]

which implies that [math](X_i)_{i\in I}[/math] is bounded in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math].

Example

Let [math]Z\in L^1(\Omega,\F,(\F_n)_{n\geq0},\p)[/math]. Then the family

[[math]] \Theta=\{X\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)\mid \vert X\vert\leq Z\} [[/math]]

is u.i. Indeed, we have

[[math]] \sup_{X\in\Theta}\E[\vert X\vert \one_{\{\vert X\vert \gt a\}}]\leq \E[Z\one_{\{Z \gt a\}}]\xrightarrow{a\to\infty}0. [[/math]]

Example

Let [math]\phi:\R_+\to\R_+[/math] be a measurable map, such that

[[math]] \frac{\phi(x)}{x}\xrightarrow{x\to\infty}\infty. [[/math]]

Then for all [math]C \gt 0[/math], the family

[[math]] \Theta_C=\{X\in L^1(\Omega,\F,(\F_n)_{n\geq0},\p)\mid \E[\phi(\vert X\vert)]\leq C\} [[/math]]

is u.i. Indeed, for [math]a[/math] large enough, we have

[[math]] \E[\vert X\vert \one_{\{\vert X\vert \gt a\}}]=\E\left[\frac{\vert X\vert}{\phi(\vert X\vert)}\phi(\vert X\vert)\one_{\{\vert X\vert \gt a\}}\right]\leq \sup_{x \gt a}\left(\frac{x}{\phi(x)}\right)\E[\phi(\vert X\vert)]\leq C\sup_{x \gt a}\left(\frac{x}{\phi(x)}\right)\xrightarrow{a\to\infty}0. [[/math]]

Thus

[[math]] \sup_{X\in\Theta}\E[\vert X\vert\one_{\{\vert X\vert \gt a\}}]\leq C\sup_{x \gt a}\left(\frac{x}{\phi(x)}\right) [[/math]]

Proposition

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_i)_{i\in I}[/math] be a family of r.v.'s bounded in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math], i.e. [math]\sup_{i\in I}\E[\vert X_i\vert] \lt \infty[/math]. Then [math](X_i)_{i\in I}[/math] is u.i. if and only if for all [math]\varepsilon \gt 0[/math] there is a [math]\delta \gt 0[/math] such that for all [math]\A\in \F[/math], if [math]\p[A] \lt \delta[/math] then [math]\sup_{i\in I}\E[\vert X_i\vert \one_A] \lt \varepsilon[/math].

Show Proof

For the right implication, let [math]\varepsilon \gt 0[/math]. Then there exists [math]a \gt 0[/math] such that

[[math]] \sup_{i\in I}\E[\vert X_i\vert \one_{\{\vert X_i\vert \gt a\}}] \lt \frac{\varepsilon}{2}. [[/math]]

Now let [math]\delta=\frac{\varepsilon}{2a}[/math] and [math]A\in\F[/math] such that [math]\p[A] \lt \delta[/math]. Then

[[math]] \E[\vert X_i\vert \one_A]=\E[\vert X_i\vert \one_A\one_{\{\vert X_i\vert \gt a\}}]+\E[\vert X_i\vert \one_A\one_{\vert X_i\vert \leq a\}}]\leq \E[\vert X_i\vert \one_{\{\vert X_i\vert \gt a\}}]+a\underbrace{\E[\one_A]}_{\p[A]} \lt \frac{\varepsilon}{2}+a\delta \lt \frac{\varepsilon}{2}+a\frac{\varepsilon}{2a}=\varepsilon. [[/math]]

For the left implication, let [math]C=\sup_{i\in I}\E[\vert X_i\vert] \lt \infty[/math]. From Markov's inequality,we get

[[math]] \p[\vert X_i\vert \gt a]\leq \frac{\E[\vert X_i\vert]}{a}\leq \frac{C}{a}. [[/math]]

Now let [math]\delta \gt 0[/math] such that [math](ii)[/math] holds. If [math]\frac{C}{a} \lt \delta[/math], then for all [math]i\in I[/math]

[[math]] \E[\vert X_i\vert \one_{\{\vert X_i\vert \gt a\}}] \lt \varepsilon. [[/math]]

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Corollary

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]X[/math] be a bounded r.v., i.e. [math]X\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math]. Then the family

[[math]] \{\E[X\mid\mathcal{G}]\mid \mathcal{G}\subset \F,\text{ $\mathcal{G}$ is a $\sigma$-Algebra}\} [[/math]]

is u.i.

Show Proof

Let [math]\varepsilon \gt 0[/math]. Then there exists a [math]\delta \gt 0[/math] such that for all [math]A\in \F[/math], if [math]\p[A] \lt \delta[/math] then [math]\E[\vert X\vert \one_A] \lt \varepsilon[/math]. Then for all [math]a \gt 0[/math], we have

[[math]] \p[\E[X\mid \mathcal{G}] \gt a]\leq \frac{\E[\vert \E[X\mid \mathcal{G}]\vert]}{a}\leq \frac{\E[\vert X\vert]}{a}. [[/math]]

For [math]a[/math] large enough, i.e. [math]\frac{\E[\vert X\vert]}{a} \lt \delta[/math], we have

[[math]] \E[\vert \E[X\mid \mathcal{G}]\vert \one_{\{ \vert \E[X\mid \mathcal{G}]\vert \gt a\}}]\leq \E[\E[\vert X\vert \one_{\vert \E[X\mid \mathcal{G}]\vert \gt a\}}\mid \mathcal{G}]]\leq \E[\vert X\vert \one_{\{\E[X\mid \mathcal{G}]\vert \gt a\}}] \lt \varepsilon. [[/math]]

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Theorem

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a sequence of r.v.'s in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math], which converges in probability to [math]X_\infty[/math]. Then [math]X_n\xrightarrow{n\to\infty\atop L^1}X_\infty[/math] if and only if [math](X_n)_{n\geq 0}[/math] is u.i.

Show Proof

For the right implication, we first note that [math](X_n)_{n\geq 0}[/math] is bounded in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] since it converges in [math]L^1[/math]. For [math]\varepsilon \gt 0[/math], there exists [math]N\in\N[/math] such that for [math]n\geq N[/math] we get

[[math]] \E[\vert X_N-X_n\vert] \lt \frac{\varepsilon}{2}. [[/math]]

Next we note that [math]\{X_0,X_1,...,X_N\}[/math] is u.i. since it is a finite family of bounded r.v.'s. Therefore, there exists a [math]\delta \gt 0[/math], such that for all [math]A\in\F[/math], if [math]\p[A] \lt \delta[/math] then [math]\E[\vert X_n\vert\one_A] \lt \frac{\varepsilon}{2}[/math] for all [math]n\in\{0,1,...,N\}[/math]. Finally for [math]n\geq N[/math], we get

[[math]] \E[\vert X_n\vert \one_A]\leq \E[\vert X_N-X_n\vert \one_A]+\E[\vert X_N\vert \one_A] \lt \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. [[/math]]

Thus [math](X_n)_{n\geq 0}[/math] is u.i.

For the left implication, we note that if [math](X_n)_{n\geq 0}[/math] is u.i., then the family [math](X_n-X_m)_{(n,m)\in\N^2}[/math] is also u.i. since

[[math]] \E[\vert X_n-X_m\vert \one_A]\leq \E[\vert X_n\vert \one_A]+\E[X_m\vert \one_A] [[/math]]

for all [math]A\in\F[/math]. Now for [math]\varepsilon \gt 0[/math], there exists [math]a[/math] sufficiently large, such that

[[math]] \E[\vert X_n-X_m\vert \one_{\{\vert X_n-X_m\vert \gt a\}}] \lt \varepsilon. [[/math]]

Moreover, we note that

[[math]] \E[\vert X_m-X_m\vert]\leq \E[\vert X_n-X_m\vert \one_{\{ \vert X_n-X_m\vert \lt \varepsilon\}}]+\E[\vert X_n-X_m\vert \one_{\{\vert X_n-X_m\vert\geq \varepsilon\}}] [[/math]]

[[math]] \leq \varepsilon +\E[\vert X_n-X_m\vert\one_{\{\vert X_n-X_m\vert \geq\varepsilon\}}\one_{\{\vert X_n-X_m\vert \leq a\}}]+\E[\vert X_n-X_m\vert\one_{\{\vert X_n-X_m\vert\geq \varepsilon\}}\one_{\{\vert X_n-X_m\vert \gt a\}}] [[/math]]

[[math]] \leq \varepsilon+a\p[\vert X_n-X_m\vert\geq\varepsilon]+\E[\vert X_n-X_m\vert \one_{\{\vert X_n-X_m\vert \gt a\}}]. [[/math]]

Then, using that [math]\lim_{n\to\infty}\p[\vert X_n-X_m\vert\geq\varepsilon]=0[/math], we can show that the right hand side converges to zero for [math]a[/math] large enough, which implies that [math](X_n)_{n\geq 0}[/math] is a Cauchy sequence [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] and hence converges in [math]L^1[/math].

■

Combining all our previous results, we have; if [math](X_n)_{n\geq 0}[/math] is a martingale, then the following are equivalent

• [math](X_n)_{n\geq 0}[/math] converges a.s. and in [math]L^1[/math].
• [math](X_n)_{n\geq 0}[/math] is u.i.
• [math](X_n)_{n\geq 0}[/math] is regular and [math]X_n=\E[X_\infty\mid \F_n][/math] a.s.

General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].