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.