# If for every $x>0,P(|X_k|>x) \leq P(Y>x)$ then $\lim_k E[X_k|\mathcal{G}]=E[X|\mathcal{G}]$ a.s.?

Consider on a probability space $$(\Omega,\mathcal{F},P),$$ a sub-$$\sigma$$-algebra $$\mathcal{G}$$ and a sequence of random variables $$(X_k)_k$$ converging a.s. to $$X$$. Let $$Y \in L^1.$$

1. Prove that if $$|X_k| \leq Y$$ then $$E[X_k|\mathcal{G}]$$ converges a.s. to $$E[X|\mathcal{G}]$$.

2. Does it follow that $$\lim_k E[X_k|\mathcal{G}]=E[X|\mathcal{G}]$$ a.s.

a. if for every $$x>0,P(|X_k|>x) \leq P(Y>x)?$$

b. if for every $$x>0,P(|X_k|>x|\mathcal{G}) \leq P(Y>x|\mathcal{G})$$ a.s. ?

Attempt:

1. The result follows by applying Fatou's lemma: $$E[\liminf_k(2Y-|X_k-X|)|\mathcal{G}] \leq \liminf_k E[2Y-|Y_k-Y||\mathcal{G}]$$ a.s. which implies that $$2E[Y|\mathcal{G}]\leq 2 E[Y|\mathcal{G}]-\limsup_kE[|X_k-X||\mathcal{G}]$$ a.s. so that $$\limsup_kE[|X_k-X||\mathcal{G}]=0$$ a.s. concluding the proof.

Any ideas for part 2. are welcomed!

Let $$(X_n)_{n\geqslant 1}$$ be a sequence of random variables on a probability space $$(\Omega,\mathcal F,\mathbb P)$$ that converges almost surely to a random variable $$Y$$. Let $$\mathcal G$$ be a sub-$$\sigma$$-algebra of $$\mathcal F$$. Consider the following four conditions:

• $$(C1)$$: there exists an integrable random variable $$Y$$ such that $$\lvert X_n\rvert\leqslant Y$$ almost surely.
• $$(C2)$$: there exists an integrable random variable $$Y$$ such that for each $$x>0$$, $$\mathbb P\left(\lvert X_n\rvert>x\mid\mathcal G\right)\leqslant \mathbb P\left(Y>x\mid\mathcal G\right)$$ almost surely.
• $$(C3)$$: there exists an integrable random variable $$Y$$ such that for each $$x>0$$, $$\mathbb P\left(\lvert X_n\rvert>x\right)\leqslant \mathbb P\left(Y>x\right)$$.
• $$(C4)$$: the sequence $$\left(X_n\right)_{n\geqslant 1}$$ is uniformly integrable.

The chain of implication $$(C1)\Rightarrow (C2)\Rightarrow (C3) \Rightarrow (C4)$$ takes places. Now the question is: which conditions among $$(Ci)$$, $$1\leqslant i\leqslant 4$$ guarantees that $$\tag{*}\mathbb E\left[X_n\mid\mathcal G\right]\to \mathbb E\left[X\mid\mathcal G\right].$$

As it was shown in the opening post, $$(C1)$$ is sufficient and it was established in this thread that $$(C4)$$ is not sufficient in general.

Interestingly, the example provided in this thread satisfies also the stochastic domination in $$(C3)$$. To this this, let write explicitly the counter-example. Let $$\left(A_n\right)_{n\geqslant 1}$$ and $$\left(B_n\right)_{n\geqslant 1}$$ be two mutually independent sequences of events, both sequences consisting of independent events, and such that $$\mathbb P(A_n)=1/n$$, $$\mathbb P(B_n)=1/n^2$$. Let $$Y_n=\mathbf{1}_{A_n}$$, $$Z_n=n^2\mathbf{1}_{B_n}$$ and $$X_n=Y_nZ_n$$. Let $$\mathcal G:= \sigma(A_k,k\geqslant 1)$$.

• By the first Borel-Cantelli lemma, $$\mathbb P(\limsup_n B_n)=0$$ hence for almost every $$\omega$$, there exists an $$n(\omega)$$ such that $$X_n(\omega)=0$$ for $$n\geqslant n(\omega)$$. Therefore, $$X_n\to X:=0$$ almost surely.
• For a positive $$x$$, $$\mathbb P(\lvert X_n\rvert>x)=\mathbf 1_{(0,n^2]}(x)\mathbb P(A_n\cap B_n)=\mathbf 1_{(0,n^2)}(x)n^{-3}$$ hence $$\sup_{n\geqslant 1}\mathbb P(\lvert X_n\rvert>x)\leqslant \sup_{n\geqslant 1}\mathbf 1_{(0,n^2)}(x)n^{-3},$$ which is reached at $$n=\lfloor \sqrt{x}\rfloor+1$$ hence any random variable $$Y$$ whose tail behaves as $$(\lfloor \sqrt{x}\rfloor+1)^{-3}$$ would do the job, as such a random variable would be integrable.
• $$\mathbb E\left[X_n\mid\mathcal G\right]=n^2\mathbf{1}_{A_n}\mathbb E\left[\mathbf{1}_{B_n}\mid\mathcal G\right]=\mathbf{1}_{A_n}$$ and by the second Borel-Cantelli lemma, $$\mathbb E\left[X_n\mid\mathcal G\right]$$ does not converge to $$0$$ almost surely.

It remains to see whether $$(C2)$$ is sufficient. For $$R>0$$, let $$\Phi_R\colon\mathbb R\to \mathbb R$$ be the function defined by $$\Phi_R(t)=-R$$ if $$t<-R$$; $$\Phi_R(t)=t$$ if $$-R\leqslant t\leqslant R$$ and $$\Phi_R(t)=R$$ if $$t>R$$. Since $$\left( \Phi_R(X_n) \right)_{n\geqslant 1}$$ converges almost surely to $$\Phi_R(X)$$ and $$\lvert \Phi_R(X_n)\rvert\leqslant R$$, we derive that $$\mathbb E\left[ \Phi_R(X_n)\mid \mathcal G\right]\to \mathbb E\left[ \Phi_R(X)\mid \mathcal G\right]$$. Moreover, we can show that $$\tag{C5} \sup_{n\geqslant 1} \mathbb E\left[\lvert X_n\rvert\mathbb{1}_{\{\lvert X_n\rvert>2^k\}}\mid\mathcal G\right]\to 0\mbox{ a.s. as }k\to \infty$$ by applying the inequality in $$(C2)$$ with $$x=2^j$$ and sum over $$j\geqslant k$$. Since $$\left\lvert \mathbb E\left[X_n\mid\mathcal G\right]-\mathbb E\left[X\mid\mathcal G\right]\right\rvert\leqslant \left\lvert \mathbb E\left[\Phi_{2^k}(X_n)\mid\mathcal G\right]-\mathbb E\left[\Phi_{2^k}(X)\mid\mathcal G\right]\right\rvert+\left\lvert \mathbb E\left[X_n-\Phi_{2^k}(X_n)\mid\mathcal G\right]\right\rvert+\left\lvert \mathbb E\left[X-\Phi_{2^k}(X)\mid\mathcal G\right]\right\rvert$$ and for a random variable $$Z$$, $$\lvert Z-\Phi_{2^k}\rvert\leqslant \lvert Z \rvert\mathbf{1}_{\{\lvert Z \rvert>2^k\}}$$, it follows that $$\left\lvert \mathbb E\left[X_n\mid\mathcal G\right]-\mathbb E\left[X\mid\mathcal G\right]\right\rvert\leqslant \left\lvert \mathbb E\left[\Phi_{2^k}(X_n)\mid\mathcal G\right]-\mathbb E\left[\Phi_{2^k}(X)\mid\mathcal G\right]\right\rvert+\left\lvert \mathbb E\left[\lvert X_n\rvert\mathbf{1}_{\{\lvert X_n \rvert>2^k\}}\mid\mathcal G\right]\right\rvert+\left\lvert \mathbb E\left[\lvert X \rvert\mathbf{1}_{\{\lvert X \rvert>2^k\}}\mid\mathcal G\right]\right\rvert$$ hence $$(C2)$$ is sufficient.

Actually, the proof shows that (C5) is sufficient. We have $$(C1)\Rightarrow (C5)\Rightarrow (C2)$$.

• It seems that you used implicitly the following formula: $E[Y|\mathcal{G}]=\lim_{n}\frac{1}{2^n}\sum_{k \in \mathbb{N}}P(Y>\frac{k}{2^n}|\mathcal{G})$
– user
Oct 11, 2021 at 2:03
• The result, does it remain true if $(C_2)$ holds for $\lambda$-almost every $x \in \mathbb{R_+}$ ?
– user
Oct 11, 2021 at 2:57
• Maybe I am missing something. Wouldn't it be enough just to consider a regular conditional distribution of $(\{X_n\},X,Y)$ w.r.t. $\mathcal G$ and invoke the relevant "unconditional" facts? E.g. (C5) is nothing else but UI for the regular conditional distribution. Oct 11, 2021 at 6:41
• I mean, to prove sufficiency of (C2) or (C5). Oct 11, 2021 at 6:51
• @user I think so, since we actually need (C2) to hold for at least one $x_k$ in the interval $[2^k,2^{k+1})$ and if $(C2)$ holds almost surely, then we can find such an $x_k$ for every $k$. Oct 11, 2021 at 8:54