Show that this sequence is positive with probability approaching one

Question

Let $(X_n)$ be a sequence of square integrable real random variables on a probability space $(\Omega,\mathcal F,P)$. Suppose that

$$E[X_n\mid \mathcal A]\to 1_A \quad P\text{-a.s.}$$

$$V[X_n\mid \mathcal A]\to 0 \quad P\text{-a.s.}$$

as $n\to \infty$, where $\mathcal A\subset \mathcal F$ is a sub $\sigma$-algebra, and $A\in\mathcal A$ has positive $P$-measure. Can I show then that

$$P[X_n\geq 0]\to 1 \quad \text{as } n\to\infty \quad?$$

Any help is very appreciated.

Answer 1

No, I don't think we can conclude $P(X_n \ge 0) \rightarrow 1$. Let $\Omega = [0,1]$ with $P$ the Lebesgue measure and $\mathcal A$ the sigma-algebra on $[0,1]$ generated by the Borel subsets of $[0,\frac 12]$. Let $A = [0,\frac 12]$ and \begin{align*} X_n(x) := \begin{cases} 1 & x \in A \\ \frac 1n & x \in (\frac 12, \frac 34] \\ -\frac 1n & x \in (\frac 34,1] \end{cases}. \end{align*} Then $\mathbb{E}[X_n | \mathcal A] = 1_A$ for all $n$ so clearly $\mathbb{E}[X_n | \mathcal A] \rightarrow 1_A$, and $V(X_n | \mathcal A) = \frac{1}{n^2}1_{(\frac 12, 1]} \rightarrow 0$ almost surely. However, we have $P(X_n < 0) = \frac 14$ for all $n$, so $P(X_n \ge 0) \not \rightarrow 1$.

Answer 2

In general no: let $X_n=\mathbf{1}_A+Y_n$, where $Y_n$ is independent of $\mathcal A$. The first assumption translates as $\mathbb E\left[Y_n\right]\to 0$ and the second one as $\operatorname{Var}(Y_n)\to 0$.

Moreover, $$ \mathbb P(X_n\geqslant 0)=\mathbb P(A\cap \{1+Y_n\geqslant 0)+\mathbb P(A^c\cap \{Y_n\geqslant 0) $$ and by independence, $$ \mathbb P(X_n\geqslant 0)=\mathbb P(A)\mathbb P(1+Y_n\geqslant 0)+\mathbb P(A^c)\mathbb P(Y_n\geqslant 0). $$ Let $(Y_n)_{n\geqslant 1}$ be a sequence of independent random variables such that for each $n$, $Y_n$ taked the values $1/n$ and $-1/n$ with probability $1/2$. Letting $\mathcal A=\sigma(Y_1)$ and $A=\{Y_1=1\}$, the assumptions on fulfills the assumptions and for $n\geqslant 2$, $$ \mathbb P(X_n\geqslant 0)=\mathbb P(A\cap \{1+X_n\geqslant 0\})+\mathbb P(A^c\cap \{X_n\geqslant 0\})= \mathbb P(A)+\mathbb P(A^c)/2<1, $$ as $\mathbb P(1+Y_n\geqslant 0)=1$ and $\mathbb P(Y_n\geqslant 0)=1/2$.