2
$\begingroup$

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 ?

$\endgroup$
2
  • $\begingroup$ I don't understand your notation. Does $\mathbb{E}_{Y_n}$ refer to conditional expectation? What does $o_{n,\mathbb{P}}(1)$ mean? $\endgroup$
    – Algernon
    Sep 25, 2021 at 14:36
  • $\begingroup$ I meant taking expectations w.r.t $Y_n$ alone. I've rewritten this more explicitly. I've also added a definition of $o_{n,\mathbb P}(1)$. $\endgroup$
    – dohmatob
    Sep 25, 2021 at 16:49

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks for the input (upvoted). I'll try to read your post line by line to make sure I understand everything before accepting. $\endgroup$
    – dohmatob
    Sep 28, 2021 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.