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1 vote

Large deviations for sequences that are not sums of iid

I believe you are missing an $n$ in your definition of $K_n(t)$, that is $K_n(t)=\log E(e^{tnS_n})$. I assume in the sequel that this is what you meant. If $S_n$ satisfies the large deviations princip …
ofer zeitouni's user avatar
1 vote
Accepted

LDP for Marchenko Pastur with k/n tending to 0

For standard Gaussians, and with the matrix $W/n$, the proof of the LDP given by Ben Arous-Guionnet adapts to the Wishart setup. However, you will have different scalings and so the non-commutative en …
ofer zeitouni's user avatar
3 votes
Accepted

Large deviation for Brownian occupation time

This is precisely the Donsker-Varadhan LDP, coupled with an application of the contraction principle. Namely, the rate function is $$I(x)=\inf\{ J(\mu): \int f d\mu =x\}$$ where $J$ is the Donsker-Var …
ofer zeitouni's user avatar
0 votes
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Joint typicality of sequences

Essentially, yes, although it is ambiguous the way you stated it (so it is not clear to me what $T_{P^n}$ really means). The general statement is that you have a LDP for the empirical process $n^{-1}\ …
ofer zeitouni's user avatar
2 votes
Accepted

Family of large deviation principles

First, to see why this is not enough, suppose all variable involved take value in some compact interval. Suppose the sequence $X_n$ satisfies the LDP, with rate function $J(x)$, and let $X_n^\epsilon …
ofer zeitouni's user avatar
2 votes
Accepted

Large deviations for integrands

It all depends on the shape of $P_2$ and on the assumptions you put on $X_i$. In what follows I'll assume that $\Lambda(\lambda)=\log E_1 e^{\lambda X_1}$ is finite for all $\lambda$. I will also ass …
ofer zeitouni's user avatar
14 votes
Accepted

What is "tilting" in the context of large deviations?

Tilting refers to a change of measure of the form $e^{\lambda \cdot x}/C(\lambda)$ where $C(\lambda)=E(e^{\lambda\cdot X})$. (The setup is that the measure for which you are proving large deviations i …
Matthew Daws's user avatar
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