Let $X_1,X_2,\dots$ be a sequence of random variables such that the distribution of their sum $S_n = \sum_{i=1}^n X_i$ satisfies a large deviation principle in the sense that there exists a lower semicontinuous function $I:\mathbb {R}\to [0,\infty ]$ such that for $a>0$
$$\liminf_{n\to\infty} \frac{1}{n}\log P(\vert S_n\vert > an) \geq -\inf_{z\in \lbrack-a,a\rbrack^C} I(z)$$
and
$$\limsup_{n\to\infty} \frac{1}{n}\log P(\vert S_n\vert \geq an) \leq -\inf_{z\in (-a,a)^C} I(z).$$
Now, let $Y_1,Y_2,\dots$ be another sequence of random variables and write $R_n$ for their sum. Assume that for all natural numbers $n$ it holds that almost surely
$$\vert S_n- R_n\vert \leq \log n.$$
Does the same large deviation principle or a similar one hold for the distribution of $R_n$?
One can write $P(\vert R_n\vert > an) = P\big(\vert S_n\vert > an+\mathcal{O}(\log n)\big) = P\big(\vert S_n\vert > (a+\mathcal{O}(\log n)/n)n\big)$ with $\frac{\mathcal{O}(\log n)}{n}\to 0$. But does this yield an LDP somehow with these conditions?
Best Answer
The sequences $S_n/n$ and $R_n/n$ are superexponentially close, in the sense that, for any $\varepsilon>0$, $$ \lim_{n\rightarrow\infty}\frac{1}{n}\log P(n^{-1}|S_n - R_n| > \varepsilon) = -\infty,$$ where it is agreed that $\log 0 = -\infty$. It follows that $S_n/n$ satisfies an LDP if and only if $R_n/n$ does, and with the same rate function. See results on superexponential approximation, e.g. Proposition 1.19 in http://staff.utia.cas.cz/swart/lecture_notes/LDP2.pdf.