Cramer's Theorem question

Let $\left \{ X_i \right \} $ be IID discrete random variables with the following distribution:
$$P(X=-1)=P(X=1)=\frac{1}{2}$$ Let $\overline{X_n}$ be the sample mean defined as:
$$\overline{X_n}=\frac{1}{n}\sum_{i=1}^n X_i$$ The theory of Large Deviations can be used to show that

$$\lim_{n\rightarrow \infty} \frac{1}{n} P\left (\overline{X_n}\gt\frac{1}{2} \right ) =a \ln 2-b\ln 3$$ where $a$ and $b$ are positive numbers. Use Cramer's Theorem to compute the numbers $a$ and $b$.


As you suggest this is a straightforward application of Cramer's theorem. Recall that Cramer's theorem tells us that $$\lim_{n \to \infty} \frac{1}{n} P(\overline{X_n} > x) = -\Lambda^*(x)$$ where the $X_i$ are iid random variables with logarithmic moment generating function $\Lambda(t)$, and $\Lambda^*(x) = \text{Sup}_{t \in \mathbb{R}}(tx - \Lambda(t))$ is the Legendre transform. Sometimes it is very difficult to compute this, but when $\Lambda(t)$ is a well known smooth function of $t$ the Legendre transform is easy to compute. In this case the generating function $E[e^{tX}] = \frac{1}{2}(e^t + e^{-t})$, whence the logarithmic generating function $\Lambda(t) = Log(\frac{1}{2}(e^t + e^{-t}))$, we use calculus to take the Legendre transform when $x = \frac{1}{2}$: $$ \partial_t\left(\frac{t}{2} -  Log\left(\frac{1}{2}(e^t + e^{-t})\right)\right) = 0$$ if and only if $$ \frac{1}{2} = \frac{e^t - e^{-t}}{e^t + e^{-t}}$$ solving for $t$ we have $$\frac{3}{2} e^{-t} = \frac{1}{2}e^t$$ or $$e^{2t} = 3$$ so finally we see that $t = \frac{ln(3)}{2}$. Now plugging in we see that $$-\Lambda^*(\frac{1}{2}) = -\left(\frac{ln(3)}{4} - ln\left(\frac{\sqrt{3} + \frac{1}{\sqrt{3}}}{2}\right) \right)$$ or, after some manipulation using the properties of the logarithm $$ = ln(2) - \frac{3}{4} ln(3).$$

The answer is accepted.
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