# Simple question on large deviations result

I am not understanding an example. I think I missed something really obvious. Can someone point it out to me?

Let $X_i$ be iid Bernoulli random variables, taking values 0 and 1 with probability 1/2.

then $\lim\limits_{n\rightarrow\infty} 1/n \log \mathbb{P}(S_n \geq an ) = - I(a)$ for $a\in[0,1]$, where

$I(z) = \log(2) + z\log(z) + (1-z)\log(1-z)$

In the sketch proof, it started with $I(z) = I(1-z)$

I must have missed something here or I am being really dumb, or both. As I can see $\mathbb{P}(S_n \geq an )$ is decreasing in $a$, therefore $-I(z)$ should be a decreasing function, and $I$ should be an increasing function?

PS why is Large Deviation Principles not a tag? or is there a closely related one?

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the probability is decreasing in a, so the left hand side is going to be decreasing in $a$. I cannot see the symmetry. – Lost1 Oct 29 '12 at 15:37

Roughly speaking, $-I(a)$ describes the convergence to zero of $\mathbb P(S_n\approx na)$ in the sense that $\frac1n\log\mathbb P(S_n\in nA)$ converges to $-\inf\limits_AI$.
In your case, since $I$ is decreasing on $[0,\frac12]$ and increasing on $[\frac12,1]$, the result is that, for every $a\leqslant\frac12$, $\frac1n\log\mathbb P(S_n\leqslant na)$ converges to $-I(a)$ and, for every $a\geqslant\frac12$, $\frac1n\log\mathbb P(S_n\geqslant na)$ converges to $-I(a)$.