At first I had thought that one could apply Ellis' theorem, [E:1984], which states that for a sequence, $U_n$, of random vectors in $\mathbb R^d$,

**if**

$\lim_{n\rightarrow \infty} \frac{1}{n} \log \mathbb E[\exp(n \alpha X_n)] = \Lambda(\alpha)$ exists and is convex for $\alpha$ in an open ball about $0$

**then**

[1] $\quad\liminf_{n\rightarrow \infty} \frac{1}{n}\log\mathbb P(X_n \in G) \geq -\inf_{x \in G}\Lambda^{\star}(x)$ for $G$ open

[2] $\quad\limsup_{n\rightarrow \infty} \frac{1}{n}\log\mathbb P(X_n \in F) \leq -\inf_{x \in F}\Lambda^{\star}(x)$ for $F$ closed

**and** for convex sets, $C$, the limit exists

[3] $\quad\lim_{n\rightarrow \infty} \frac{1}{n}\log\mathbb P(X_n \in C) = -\inf_{x \in C}\Lambda^{\star}(x)$ for $C$ convex

Here, $\Lambda^{\star}$ is the Legendre-Frenchel transform of $\Lambda$ :

[4] $\quad\Lambda^{\star}(x) = x \,\alpha_x - \Lambda(\alpha_x)$, where $\alpha_x$ is the solution in $\alpha$ to

[5] $\quad\frac{d}{d\alpha}\Lambda(\alpha) = x, \quad \alpha_x = (\Lambda')^{-1}(x)$

In one dimension, any interval on the line (half line) is a convex set and therefore the limit exists. The infemum is given by the "dominating point" which is the point closest to the mean. In this case, which covers the inequalities requested,

[6] $\quad\lim_{n\rightarrow \infty} \frac{1}{n}\log\mathbb P(X_n > \mu + \epsilon) = -\Lambda^{\star}(\mu + \epsilon)$

[7] $\quad\lim_{n\rightarrow \infty} \frac{1}{n}\log\mathbb P(X_n < \mu - \epsilon) = -\Lambda^{\star}(\mu - \epsilon)$

Expanding $\Lambda^{\star}(\bar \xi + \epsilon)$ about $\epsilon=0$, using the inverse function theorem in [5] to calculate derivatives in [4], we have to second order for the general case,

$\Lambda^{\star}(\mu + \epsilon) = \Lambda^{\star}(\mu - \epsilon) = \frac{1}{2\,\Lambda''(0)} \,\epsilon^2$

**Now** in the particular case at hand, consider $\mathcal I = 2^{\mathbb N}$, and allowing abuse of notation whereby a subset of naturals is identified with its corresponding increasing sequence,

let $\mathcal J_n = \{ \;\vec j \in \mathcal I\,: | \mathbb N \setminus \vec j | \leq n \}$

Intuitively, $U_n$ takes values

$U_n = \sum_{\ell=1}^{\infty} p_{j_{\ell}}\quad$ with probability $\quad(1 - \sum_{\ell=1}^{\infty} p_{j_{\ell}})^n\qquad$ for $\;\vec j \in \mathcal J_n$

but since the cardinality of $\mathcal I$ and hence, also of $\mathcal J_n$, is $\aleph_1$ then intuition must be excercised with care. If there is a limiting random variable

concentration inequalities which are known for $U_n$. – Did Oct 7 '12 at 10:54