A question on the applicability Chebyshev inequality for sequence of random quantities

Question

Let $(X_n)_n$ and $(Y_n)_n$ be two mutually independent sequences of random tensors (i.e scalars, vectors, matrices, etc.) defined on the same probability space, and let $f$ be a measurable function.

Question 1. If we establish that $\mathbb E[f(X_n,Y_n) \mid X_n] = \alpha+o_{n,\mathbb P}(1)$ (for some $\alpha \in \mathbb R$), and $var(f(X_n,Y_n) \mid X_n) = o_{n,\mathbb P}(1)$, can one conclude that $f(X_n,Y_n) = \alpha + o_{n,\mathbb P}(1)$ without further assumptions ?

Notation. $o_{n,\mathbb P}(1)$ stands for a quantity which goes to zero in probability.

Naively, I'd say Yes, by Chebyshev's inequality. But I worry that something strange might be going on in general, to require a bit more care.

Question 2. In case Question 1 does not answer in the affirmative, is there a "slight" modification thereof which does ?

Answer 1

The answer to your first question is positive.

For brevity, I will use the notation $a\pm b$ to indicate the interval $(a-b,a+b)$.

Let $U_n:=f(X_n,Y_n)$. By the assumption, for every $\varepsilon>0$,

\begin{align} \mathbb{P}\big(\mathbb{E}[U_n\,|\,X_n]\notin\alpha\pm\varepsilon\big)&\to 0 \qquad \text{as $n\to\infty$,} \tag{A1} \\ \mathbb{P}\big(\mathbb{V}\text{ar}[U_n\,|\,X_n]\geq\varepsilon\big)&\to 0 \qquad \text{as $n\to\infty$.} \tag{A2} \end{align} We want to show that for every $\varepsilon>0$, $\mathbb{P}(U_n\notin\alpha\pm\varepsilon)\to 0$ as $n\to\infty$.

To see this, let $\varepsilon>0$ be fixed. Note that \begin{align} &\hspace{-1em}\mathbb{P}(U_n\notin \alpha\pm \varepsilon) \\ &= \mathbb{E}\big[\mathbb{P}(U_n\notin \alpha\pm \varepsilon\,|\,X_n)\big] \\ &\leq \underbrace{\mathbb{P}\big(\mathbb{E}[U_n\,|\,X_n]\notin\alpha\pm\varepsilon/2\big)}_{Q_n} + \underbrace{\mathbb{E}\Big[\mathbb{P}\big(U_n\notin\mathbb{E}[U_n\,|\,X_n]\pm 3\varepsilon/2\,\big|\, X_n\big)\Big]}_{R_n} \end{align}

From $\text{(A1)}$, we know that $Q_n\to 0$ as $n\to\infty$. To bound $R_n$, note that by Chebyshev's inequality, for every $\delta>0$, \begin{align} R_n &\leq \underbrace{\mathbb{P}\big(\mathbb{V}\text{ar}[U_n\,|\,X_n]\geq\delta\big)}_{S_n} + \frac{4\delta}{9\varepsilon^3} \end{align} From $\text{(A2)}$, we know that $S_n\to 0$ as $n\to\infty$. It follows that for every $\delta>0$, $\limsup_{n\to\infty} R_n\leq (4\delta)/(9\varepsilon^2)$. Since $\delta>0$ is arbitrary, this implies $R_n\to 0$ as $n\to\infty$.

Altogether, we conclude that $\mathbb{P}(U_n\notin\alpha\pm\varepsilon)\to 0$ as $n\to\infty$, as claimed.