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.

  • 1
    $\begingroup$ two sequences can converge to the same limit but at very different speed. $\endgroup$
    – Alekk
    Oct 22, 2012 at 19:18

1 Answer 1


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 $\lim\frac{-1}{n}\log \mu_n(0,\infty) = 0$.

  • $\begingroup$ Thank you, that is a clear counter example. In which case, is there any extra ``reasonable'' hypothesis that guarantees the answer to my question will be yes? $\endgroup$
    – Ritwik
    Oct 23, 2012 at 4:57
  • $\begingroup$ Exponential tightness would help. See, for example, Theorems 3.7 and 3.8(b) of math.wisc.edu/~kurtz/feng/chap3.pdf $\endgroup$
    – Dan
    Oct 23, 2012 at 13:17

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