Sign of error in the central limit theorem

Question

Let $X_n$ and $Y_n$ be independent copies of two random variables $X$ and $Y$ with domain $\{-1,0,1\}$ for $n\in \mathbb{N}$. For a given $k\in \mathbb{N}$, I would like to find conditions on $X$ and $Y$ that guarantee that there is a $N\in \mathbb{N}$ such that for all $n>N$: $$P\left(\sum_{i=0}^{kn} X_i>0\right)>P\left(\sum_{i=0}^n Y_i>0\right). \tag{1}\label{eq1}$$

If the central limit theorem were exact in the sense that for large enough $n$, $F^X_n(x)=\Phi(x)=F^Y_n(x)$ where $F^X_n(x)$ and $F^Y_n(x)$ are the CDFs of the normalized sums of $X$ or $Y$ and $\Phi(x)$ is the CDF of the standard normal, condition \ref{eq1} would follow from $$\sqrt{k}\frac{\mathbb{E}( X_i)}{\sigma( X_i)}=\frac{\mathbb{E}(\sum_{i=0}^{k} X_i)}{\sigma(\sum_{i=0}^{k} X_i)}>\frac{\mathbb{E}( Y_i)}{\sigma(Y_i)}\tag{2}\label{eq2}.$$However, $P(\sum_{i=0}^{n} Y_i\leq 0)$ and $P(\sum_{i=0}^{kn} X_i\leq 0)$ seem to be of the order $e^{-n}$, while the Berry-Essen theorem only guarantees $|F^X_n(x)-\Phi(x)|\in O(n^{-0.5})$. Correspondingly, the normal approximation appears to be insufficient to draw conclusions about \ref{eq1}.

That said, condition \ref{eq2} appears to be remarkably accurate at predicting whether \ref{eq2} holds for large $n$ in a bunch of simulations I did. These simulations also suggested that in a large range of cases, $$F^{(\cdot)}_n(x)-\Phi(x)>0 \tag{3}\label{eq3}$$ at the relevant value of $x$ for both $X$ and $Y$. If that was true, one would only need to show that the error is eventually smaller for $X$ than for $Y$. I tried to prove \ref{eq3} using Edgeworth series, but the part of the residual for which I can control the sign seems to be dominated by the remaining error.

Are there any known conditions on when the CLT approximation error $F^{(\cdot)}_n(x)-\Phi(x)$ is positive for a given $x$ (either in the general case I presented or any subcase)?

Answer 1

There are several very different cases here.

The first case is when $\mathbb{E}(X_i)=\mathbb{E}(Y_i)=0$, which in your case means that the variables are symmetric. In this case, $P(\sum_{i=0}^{kn} X_i>0)>P(\sum_{i=0}^{n} Y_i>0)$ if and only if $P(\sum_{i=0}^{kn} X_i=0)<P(\sum_{i=0}^{n} Y_i=0).$ These probabilities are governed by the local CLT: $$ P\left(\sum_{i=0}^{N} X_i=0\right)\sim\begin{cases}\frac{1}{\sqrt{2\pi N}\sigma(X)},&\mathbb{P}(X=0)>0\\ \frac{2}{\sqrt{2\pi N}\sigma(X)}\mathbf{1}_{N\text{ is even}}, &\mathbb{P}(X=0)=0 \end{cases} $$ and similarly for $Y_i$. from which you can read off the required condition: e.g., if $\mathbb{P}(X_1=0)>0$, it reads $\sqrt{k}\sigma(X)>\sigma(Y).$

If at least one of the expectations is non-zero, then the question has nothing to do with CLT, since the corresponding probability tends to zero or to one exponentially fast, depending on whether the exectation is positive or negative. The exponential rate of convergence is given by the large deviation principle (see e.g. Durrett, Probability: theory and examples, section 2.7): e.g., if $\mu_X=\mathbb{E}X<0$, then $$ -\frac{1}{N}\log P\left(\sum^N_{i=1}X_i>0\right)\to\sup_{\theta\geq 0}(-\log \mathbb{E}e^{\theta X}), $$ and similarly for $Y$. So, if $\mathbb{E}X<0$ and $\mathbb{E}Y<0$, then the required condition is $$ k\sup_{\theta\geq 0}(-\log \mathbb{E}e^{\theta X})<\sup_{\theta\geq 0}(-\log \mathbb{E}e^{\theta Y}). $$ If $\mathbb{E}X>0$ and $\mathbb{E}Y>0$, apply the same reasoning to $-X_i$ and $-Y_i$. If the expectations are of different signs (or only one of them is zero), then the two sides of your inequality have different limits.