Cramer's theorem gives the asymptotic decrease rate of the probability that the average of $n$ i.i.d. random variables $S_n=\frac{1}{n}\sum_i Z_i$ is larger than some number, i.e.,
$$\lim_{n\rightarrow\infty}\frac{1}{n} \ln P(S_n>z)=-I(z)$$
where $I$ is the rate function found by the Legendre transform of the log generating function i.e.,
$$I(z)=\max_{u}(zu-\ln G(u))$$
Here mean and variance of $Z_1$ are assumed to be finite and $E[Z_1]>z$ must be fulfilled.
My question is on the connection of stochastically ordered $S_n$s to the rate function. For example if all $Z_i$ are bernoulli random variables with success probability $p>z$, then Cramer's theorem will function and give us
$$P(S_n>z)\approx e^{-nI(z)}$$
Fix $z$ to be a number for example $1/2$. Then how can $I$ be parameterized in terms of $p>1/2$? If $I_1$ is for some $p>1/2$ and $I_2$ is for $p+\epsilon$ then how does $I_2/I_1$ look like? is there something general in this regard?
Thank you very much-
Let us calculate the Legendre transform explicitly. First of all, recall that the moment generating function of a Bernoulli distributed random variable with success probability $p>0$ equals
$$G(u) = q+p \cdot e^u$$
where $q := 1-p$. If we define
$$f(u) := f_z(u) := u \cdot z - \ln G(u)$$
then
$$f'(u) = z - \frac{p e^u}{\varphi(u)} = (u-1) + \frac{q}{\varphi(u)}.$$
Performing some standard calculations yields
$$f'(u) = 0 \Leftrightarrow u = \ln \frac{z \cdot q}{(1-z) \cdot p} \tag{1}$$
for any $z \in (0,1)$. For $z \in (0,1)$, it follows from $(1)$ that
$$I(z) = z \cdot \ln \frac{z \cdot q}{(1-z) \cdot p}- \ln \left( q \frac{1-2z}{1-z} \right) = z \ln \frac{z}{p} + (1-z) \ln \frac{1-z}{1-p}.$$
For $z \notin (0,1)$, we have $I(z) \in \{\infty,-\infty\}$. In particular, we find that
$$\frac{d}{dp} I(z) = - \frac{z}{p} + \frac{(1-z)}{1-p} = \frac{p-z}{p(1-p)},$$
i.e. $p \mapsto I(z)$ is increasing whenever $p>z$ and decreasing for $p < z$.