0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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