If the probability distribution function of two sequences of random variables have the same weak limit and one of the sequences satisfies a Large deviation principle, then does it imply that the other one also satisfies a LDP with the same rate function? Here is a more precise version of my question: 


Let 
$$ X\_n: (\Omega\_n, P\_n) \rightarrow \mathbb{R} $$ and 
$$ X\_n^{\prime}: (\Omega\_n^{\prime}, P^{\prime}\_n) \rightarrow \mathbb{R}$$be a sequence of random variables. We define the probability distribution function as 
$$ \mu\_n(A) := P\_n(X\_n^{-1}(A)), \qquad \mu\_n^{\prime}(A) := P\_n^{\prime}(X\_n^{\prime^{-1}}(A)) $$
for every set $A \subset \mathbb{R}$. We are given that for 
every bounded continuous function $\phi : \mathbb{R} \rightarrow \mathbb{R}$ (or equivalently for every continuous function with compact support) 
$$ \lim\_{n\rightarrow \infty} \int \phi d \mu\_n = \lim\_{n\rightarrow \infty} \int \phi d \mu\_n^\prime = \int \phi d \mu $$
Furthermore, we also know that the random variable $X\_n$ satisfies the 
Large deviation principal, ie for a given number $x \in \mathbb{R}$
the following limit exists 
$$ \lim\_{n \rightarrow \infty} \frac{-1}{n}\log(\mu\_n(t\in \mathbb{R}: t > x )) = I(x) $$


Does it follow that the other random variable also satisfies a large deviation principle with the same rate function, ie 
$$ \lim\_{n \rightarrow \infty} \frac{-1}{n}\log(\mu\_n^{\prime}(t\in \mathbb{R}: t > x )) = I(x) $$


Note that I am asking two questions: First of all does the limit exist? 
Secondly, is it the same limit. 



If two probability distributions have the same weak limit and one of them satisfies Large Deviation Principle, what can we say about the other? 
No and no. To see why the limit need not exist, suppose $\mu = \delta\_0$ and $X^\prime\_n = \frac{(-1)^n}{n}$ a.s. Then $\liminf \frac{-1}{n}\log \mu^\prime\_n(0,\infty) = 0$ while $\limsup \frac{-1}{n}\log \mu^\prime\_n(0,\infty) = \infty$. Even if the limit exists, it need not be the same limit. Suppose $\mu = \delta\_0$. Suppose $X\_n = 1/n$ and $X^\prime\_n = -1/n$ a.s. Then $\lim\frac{-1}{n}\log \mu^\prime\_n(0,\infty)=\infty$ while $\lim\frac{-1}{n}\log \mu\_n(0,\infty) = 0$.


