To address the questions regarding bounding the asymptotic conditional expectations, we analyze each part as follows:

1. Bound for ( \big|\mathbb{E}\big[S^d_n \big] - \mathbb{E}\big[Y^d\big]\big| )

Since ( {X_i^d} ) are zero-mean i.i.d. random variables, the sum ( \sum_{i=1}^n X_i^d ) has mean zero. Normalizing by ( \sqrt{n} ), the mean of ( S_n^d ) remains zero: ( \mathbb{E}[S_...

Short answer

1. It is exactly 0. Because E[Xi^d]=0, we have E[Sn^d]=(1/√n)∑i E[Xi^d]=0. You also assumed Y^d has mean 0. Hence |E[Sn^d]−E[Y^d]|=0 for all n,d.

2. As written, there is no general useful upper bound. The two conditional expectations are random variables defined on different probability spaces; if you interpret them as conditional expectation functions mS(x)=E[Sn(i)|Sn(1:i−1)=x] and...

Bounding Asymptotic Conditional Expectations

This question deals with bounding differences between expectations of a normalized sum of i.i.d. random variables and its normal approximation, particularly when the dimension grows with the sample size.

Question 1:

For the first question, we actually don't need a complicated bound because:

$\mathbb{E}[S^...

I'll analyze this problem about bounding asymptotic conditional expectations when dimension grows with .

Analysis of Question 1

Your proposed method has merit but needs refinement. The key issue is the last inequality where you try to bound the distance between densities using Bentkus's result.

Issues with the Current Approach

1. Density vs. Probability Bound: Bentkus's...

Let us write the –dimensional random variables in block form

 

where etc. Throughout we assume

 • (hence ),
 • is non-singular,
 • .

QUESTION 1
|E[S^d_n]-E[Y^d]| ? ...

Here is a sketch of how one can control both

(1)  |\E [S⁽ᵈ⁾ₙ] − \E [Y⁽ᵈ⁾]|,
(2)  sup_{z∈ℝⁱ⁻¹} | \E [S⁽ᵈ⁾ₙ(i) | S⁽ᵈ⁾ₙ(1:i−1)=z ] − \E [Y⁽ᵈ⁾(i) | Y⁽ᵈ⁾(1:i−1)=z ] |,

when d grows with n. We will rely on Bentkus’ inequality in total‐variation (TV) distance, together with a few simple Cauchy–Schwarz or Lipschitz‐test‐function arguments.

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1. Th...

<think> We are given a sequence of i.i.d. d-dimensional random vectors {X_i^d} with mean zero and covariance matrix Σ^d. We define S_n^d as the normalized sum: S_n^d = (sum_{i=1}^n X_i^d) / sqrt(n). Let Y^d be a d-dimensional normal random vector with mean zero and covariance Σ^d.

The paper by Bentkus provides a bound on the supremum over a class A of sets of the absolute difference between the p...

This is an excellent and insightful question that touches on some deep results in probability theory. Let's break down each part.

Analysis of Question 1:

This part has a surprisingly simple answer. The bound is exactly zero.

1. Expectation of : The random variables are i.i.d. with mean zero, i.e., $\mat...

Below we denote by
Sₙ = Sₙᵈ : = n^{-1/2}∑_{k=1}^{n}X_k  and Y = Yᵈ∼N(0,Σ), 
β : = E|X¹|³ (the same β is used in Bentkus’ inequality) and we keep the assumption E X=0, Σ = EXXᵗ non–singular and β<∞.

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1 The difference of the unconditional means

Because eve...