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 $\mathbb E_{Y_n}[f(X_n,Y_n)] = \alpha+o_{n,\mathbb P}(1)$ (for some $\alpha \in \mathbb R$), and $var_{Y_n}(f(X_n,Y_n)) = o_{n,\mathbb P}(1)$, can one conclude that $f(X_n,Y_n) = \alpha + o_{n,\mathbb P}(1)$ without further assumptions ?

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 ?

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 ?

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 $\mathbb E_{Y_n}[f(X_n,Y_n)] = a(X_n)+o_{n,\mathbb P}(1)$ with $a(X_n) \to \alpha$ (a.s), and $var_{Y_n}(f(X_n,Y_n)) = o_{n,\mathbb P}(1)$, can one conclude that $f(X_n,Y_n) = \alpha + o_{n,\mathbb P}(1)$ without further assumptions ?

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 modification of the question which does ?

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 $\mathbb E_{Y_n}[f(X_n,Y_n)] = \alpha+o_{n,\mathbb P}(1)$ (for some $\alpha \in \mathbb R$), and $var_{Y_n}(f(X_n,Y_n)) = o_{n,\mathbb P}(1)$, can one conclude that $f(X_n,Y_n) = \alpha + o_{n,\mathbb P}(1)$ without further assumptions ?

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 ?

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 functtion.

Question 1. If we establish that $\mathbb E_{Y_n}[f(X_n,Y_n)] = a(X_n)+o_{n,\mathbb P}(1)$ with $a(X_n) \to \alpha$ (a.s), and $var_{Y_n}(f(X_n,Y_n)) = o_{n,\mathbb P}(1)$, can one conclude that $f(X_n,Y_n) = \alpha + o_{n,\mathbb P}(1)$ without further assumptions ?

Naively, I'd say Yes, by Chebychev'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 modification of the question which does ?

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 $\mathbb E_{Y_n}[f(X_n,Y_n)] = a(X_n)+o_{n,\mathbb P}(1)$ with $a(X_n) \to \alpha$ (a.s), and $var_{Y_n}(f(X_n,Y_n)) = o_{n,\mathbb P}(1)$, can one conclude that $f(X_n,Y_n) = \alpha + o_{n,\mathbb P}(1)$ without further assumptions ?

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 modification of the question which does ?