Why Conditioning Preserves Tail Control
Conditional expectation replaces a random variable by its average given specified information. Averaging can reduce variation, but the information being conditioned on may differ from one member of a family to another. A useful question is whether such averaging can turn a family with uniformly controlled tails into one with large values that are no longer controlled. The answer is no: a conditional Jensen inequality gives an explicit tail bound, uniform over all the conditioning sigma-algebras.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A family \(\{X_\alpha:\alpha\in I\}\) of integrable real-valued random variables is uniformly integrable if
The tail event is strict: it is \(\{|X_\alpha|>K\}\), not \(\{|X_\alpha|\ge K\}\). The definition controls the expected magnitude contributed by values beyond a common threshold, not just the probability of those values. For each \(\alpha\), let \(\mathcal{G}_\alpha\) be a sub-sigma-algebra of \(\mathcal{F}\), and define \(Y_\alpha=\mathbb{E}[X_\alpha\mid\mathcal{G}_\alpha]\). Conditional expectations are defined up to almost-sure equality; changing a version on a null set does not change any tail expectation below.
Proof. Since \(|X|+X\geq0\) and \(|X|-X\geq0\), the order-preserving property of conditional expectation gives \(\mathbb{E}[|X|\mid\mathcal{G}]+\mathbb{E}[X\mid\mathcal{G}]\geq0\) and \(\mathbb{E}[|X|\mid\mathcal{G}]-\mathbb{E}[X\mid\mathcal{G}]\geq0\), almost surely. Thus \(-\mathbb{E}[|X|\mid\mathcal{G}]\leq\mathbb{E}[X\mid\mathcal{G}]\leq\mathbb{E}[|X|\mid\mathcal{G}]\). The middle quantity is therefore bounded in absolute value by the nonnegative right-hand side. Integrability follows from \(\mathbb{E}|\mathbb{E}[X\mid\mathcal{G}]|\leq\mathbb{E}|X|<\infty\), using the displayed inequality and the defining expectation property of conditional expectation. \(\square\)
The Uniform Tail Bound
The factor of two and the half-threshold come from separating each \(X_\alpha\) into a part bounded by \(K/2\) and a part above \(K/2\). The conditioning may vary with \(\alpha\); the estimate uses no common sigma-algebra.
Proof. Fix \(\alpha\) and \(K>0\), and abbreviate \(X=X_\alpha\), \(\mathcal{G}=\mathcal{G}_\alpha\), \(Y=Y_\alpha\), and \(U=|X|\). Define the nonnegative conditional expectations \[ A=\mathbb{E}[U\mathbf{1}_{\{U\leq K/2\}}\mid\mathcal{G}], \qquad V=\mathbb{E}[U\mathbf{1}_{\{U>K/2\}}\mid\mathcal{G}], \qquad Z=\mathbb{E}[U\mid\mathcal{G}]. \] By splitting \(U\) over the disjoint events \(\{U\leq K/2\}\) and \(\{U>K/2\}\), linearity gives \(Z=A+V\) almost surely. Since \(U\mathbf{1}_{\{U\leq K/2\}}\leq K/2\), monotonicity of conditional expectation gives \(A\leq K/2\) almost surely. Conditional Jensen gives \(|Y|\leq Z\). Thus \(D=\{|Y|>K\}\) is contained, up to a null set, in \(C=\{Z>K\}\).
On \(C\), the identity \(Z=A+V\) and the bound \(A\leq K/2\) imply \(V>K/2\). Consequently, using nonnegativity and then this event inclusion,
For the penultimate inequality, the pointwise bound \((K/2)\mathbf{1}_{\{V>K/2\}}\leq V\mathbf{1}_{\{V>K/2\}}\) gives \((K/2)\mathbb{P}(V>K/2)\leq\mathbb{E}[V\mathbf{1}_{\{V>K/2\}}]\). The final equality is the expectation property of conditional expectation. This proves the stated bound. Taking the supremum over \(\alpha\), its right-hand side tends to zero as \(K\to\infty\): uniform integrability of the \(X_\alpha\) applies at the threshold \(K/2\), which also tends to infinity. The \(Y_\alpha\) are integrable by conditional Jensen, so the definition of uniform integrability applies to them. \(\square\)
The argument proves more than preservation for a fixed conditioning sigma-algebra. At each threshold \(K\), every member’s tail is controlled by the corresponding original tail at the same threshold \(K/2\). Taking the supremum over the family happens only after this member-by-member estimate, so there is no need to choose a sigma-algebra uniformly across indices.
Worked Examples
Worked Example: A Finite Conditional Average
Let \(\Omega=\{\omega_1,\omega_2,\omega_3\}\), with probabilities \(1/8,4/8,3/8\), respectively. Let \(X(\omega_1)=-4\), \(X(\omega_2)=0\), and \(X(\omega_3)=2\). Take \(\mathcal{G}\) to be generated by the partition \(\{\{\omega_1,\omega_2\},\{\omega_3\}\}\). Compute \(Y=\mathbb{E}[X\mid\mathcal{G}]\) and compare its tail with the theorem’s bound at \(K=1\).
The two partition cells have probabilities \(5/8\) and \(3/8\), both positive, so their conditional averages are defined by dividing by those probabilities. On the first cell, the conditional average is \[ \frac{(-4)(1/8)+(0)(4/8)}{5/8} =\frac{-4/8}{5/8} =-\frac45. \] On the second cell, \(X\) is constantly \(2\), so its conditional average is \(2\). Therefore \(Y=-4/5\) on the first cell and \(Y=2\) on the second.
With the strict tail event \(\{|Y|>1\}\), only the second cell is counted, because \(4/5<1\) and \(2>1\). Hence \[ \mathbb{E}[|Y|\mathbf{1}_{\{|Y|>1\}}]=2\cdot\frac38=\frac34. \] For the original variable, the theorem uses the strict threshold \(K/2=1/2\). The values exceeding \(1/2\) in absolute value are \(-4\) and \(2\); the value \(0\) does not exceed \(1/2\) and contributes nothing. Thus \[ \mathbb{E}[|X|\mathbf{1}_{\{|X|>1/2\}}] =4\cdot\frac18+2\cdot\frac38 =\frac48+\frac68 =\frac54. \] The bound is \(2(5/4)=5/2\), and \(3/4\leq5/2\), as required. The conditional average has a smaller tail contribution here, but the theorem does not claim equality.
Worked Example: A Uniformly Integrable Family Averaged to Constants
On \(\Omega=[0,1]\) with Lebesgue probability, let \(X_n=n\mathbf{1}_{\{U\leq1/n^2\}}\), where \(U(\omega)=\omega\), for integers \(n\geq1\). Let every \(\mathcal{G}_n\) be the trivial sigma-algebra \(\{\varnothing,\Omega\}\). Verify uniform integrability and find the conditional expectations.
The event \(\{U\leq1/n^2\}\) has probability \(1/n^2\), including probability \(1\) when \(n=1\). For \(K>0\), the strict event \(\{|X_n|>K\}\) is the rare event when \(n>K\), and is empty when \(n\leq K\). Therefore \[ \mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}] =\begin{cases} 1/n,&n>K,\\ 0,&n\leq K. \end{cases} \] The supremum over positive integers \(n\) is \(1/(\lfloor K\rfloor+1)\): the smallest integer strictly greater than \(K\) is \(\lfloor K\rfloor+1\), and \(1/n\) decreases with \(n\). This supremum tends to zero, so the family is uniformly integrable.
Conditioning on the trivial sigma-algebra gives the constant expectation: \[ Y_n=\mathbb{E}[X_n\mid\mathcal{G}_n] =\mathbb{E}[X_n] =n\cdot\frac1{n^2} =\frac1n. \] For every \(K\geq1\), \(0<1/n\leq1\leq K\), so the strict event \(\{|Y_n|>K\}\) is empty for every \(n\). Its tail expectations are all zero. As a numerical check of the theorem at \(K=2\), the left side is zero, while \[ 2\sup_n\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>1\}}] =2\sup_{n>1}\frac1n =2\cdot\frac12 =1. \] Thus the stated bound holds, although it is not sharp in this example.
Worked Example: Conditioning Can Cancel Large Values
On the same probability space \([0,1]\) with Lebesgue probability, define, for \(n\geq1\), \[ A_n=\{U\leq1/n^2\},\quad B_n=\{U\leq1/(2n^2)\},\quad C_n=\{1/(2n^2)<U\leq1/n^2\}. \] Let \(X_n=n\) on \(B_n\), \(X_n=-n\) on \(C_n\), and \(X_n=0\) outside \(A_n\). Let \(\mathcal{G}_n\) be generated by \(A_n\) and its complement. Find \(Y_n=\mathbb{E}[X_n\mid\mathcal{G}_n]\) and check the tail estimate.
The two subevents \(B_n\) and \(C_n\) each have probability \(1/(2n^2)\), while \(A_n\) has probability \(1/n^2>0\). On \(A_n\), the conditional average is \[ \frac{n(1/(2n^2))+(-n)(1/(2n^2))}{1/n^2} =\frac{1/(2n)-1/(2n)}{1/n^2}=0. \] Outside \(A_n\), \(X_n=0\), so the conditional average there is also zero whenever that cell has positive probability. When \(n=1\), the complement is null and does not affect the conditional expectation. Hence \(Y_n=0\) almost surely for every \(n\), and every tail expectation of \(Y_n\) is zero.
For \(X_n\), the absolute value is \(n\) on all of \(A_n\) and zero outside it. At threshold \(K/2\), its tail expectation is \(n(1/n^2)=1/n\) if \(n>K/2\), and zero otherwise. In particular, for \(K=4\), only integers \(n>2\) contribute; the largest contribution is \(1/3\), at \(n=3\). The theorem’s right-hand side for this \(n\) is \(2/3\), while its left-hand side is zero. The example shows why conditional Jensen gives an inequality rather than an equality: opposite signs can cancel under conditioning.
A Common Mistake and Its Correction
A tempting but generally false shortcut is to compare the conditional tail at threshold \(K\) directly with the original tail at the same threshold \(K\). Consider a two-point probability space with each point having probability \(1/2\). Let \(X=9/10\) at the first point and \(X=3\) at the second, and condition on the trivial sigma-algebra. Then \(Y=\mathbb{E}X=(9/10+3)/2=39/20\). At the strict threshold \(K=1\), \(Y\) exceeds \(1\), so \[ \mathbb{E}[|Y|\mathbf{1}_{\{|Y|>1\}}]=\frac{39}{20}. \] For \(X\), only the value \(3\) exceeds \(1\), giving \[ \mathbb{E}[|X|\mathbf{1}_{\{|X|>1\}}]=3\cdot\frac12=\frac32. \] Thus the proposed same-threshold inequality would assert \(39/20\leq3/2\), which is false since \(39/20>30/20\). The proved estimate instead uses \(K/2=1/2\). Both values of \(|X|\) exceed \(1/2\), so the original tail expectation at that threshold is \(39/20\), and twice this value is \(39/10\), which does bound the conditional tail \(39/20\).
The half-threshold is not an arbitrary technical detail. The bounded portion of \(|X|\), after conditioning, contributes at most \(K/2\). If the conditional average’s magnitude exceeds \(K\), the conditional contribution from the part of \(|X|\) above \(K/2\) must therefore exceed \(K/2\) as well. That implication is what makes the tail estimate work.
This result is useful whenever a sequence or family of random variables is replaced by conditional expectations given available information. In particular, martingale sequences are built from conditional expectations. The estimate shows that uniform integrability is retained under such conditioning, even if the information changes with the index. It does not assert that each conditional expectation has the same distribution as the variable being conditioned, or that its tail at a fixed threshold is bounded by the original tail at that identical threshold.
Check Your Understanding
Use strict tail events throughout, and justify each comparison with the conditional tail bound.
- State the definition of uniform integrability used here. Why does the definition involve tail expectations rather than only tail probabilities?
- For \(K>0\), explain why \(\mathbb{E}[|X|\mathbf{1}_{\{|X|\leq K/2\}}\mid\mathcal{G}]\leq K/2\) almost surely.
- In the finite conditional-average example, which values are counted in the original tail at threshold \(1/2\), and what is the exact tail expectation?
- For the family \(X_n=n\mathbf{1}_{\{U\leq1/n^2\}}\), compute the tail expectation at a fixed \(K>0\) when \(n>K\).
- Why can the theorem allow \(\mathcal{G}_\alpha\) to vary with \(\alpha\)? Identify where the proof uses a bound independent of that sigma-algebra.