Let $Z_n \in \mathcal{X}$ be a sequence of independent random variables where $\mathcal{X}$ is a topological vector space and let $\mu_n$ the probability measures associated with $Z_n$. Suppose that $Z_n$ that satisfies the conditions for the abstract Gärtner-Ellis Theorem. So the rate function $I(\cdot)$ associated with $Z_n$ is the Fenchel–Legendre transform of $M(\lambda) =\lim_{n \rightarrow \infty} \frac{1}{n} \log E[\exp \langle n\lambda, Z_n \rangle ]$.
Now consider a second random variable $Y_n$ that is a Markov chain ($Y_n$ is sampled from $Y_{n-1}$) and the probability measure associated with $Y_n$ is $\mu_n$.
My question is:
Is $I(\cdot)$ the rate function associated with $Y_n$?
If answer is yes to the above question, let me provide a counter-example of the equivalence of the rate function of $Z_n$ and $Y_n$.
Consider two random variables on $\{0,1\}$ let $g_n$ be 1 with probability $1/2$, else 0 let $h_1$ = $g_1$ and then for $n>1$, let $h_n=1-h_{n-1}$ with probability $(1/2)^n$ and else $h_n=h_{n-1}$.
Note that for all $n$, $h_n$ and $g_n$ have the same probability distribution, $Prob(g_n = 1) = 1/2$ (by definition). $$ Prob(h_n = 1) = Prob(h_{n-1}=1)\cdot(1/2)^n + Prob(h_{n-1}=0)\cdot(1-(1/2)^n) = (1/2) \cdot (1/2)^n+(1/2) \cdot (1-(1/2)^n) = (1/2)\cdot((1/2)^n + 1-(1/2)^n) = 1/2 $$ Hence $Prob(g_n=1) = Prob(h_n=1)$.
At the limit both random variables have the same distribution, but the convergence of the sequences is rather different. ($h_n$ converges, $g_n$ doesn't)
My other question is: Does this example invalidate the positive answer to first question?