Large deviations Exercise (Durret 2.7.6)

Question

I'm studying probability theory and doing the exercises in Durret v5.

Let $X_1,\ldots, X_n$ iid with $EX_1 = 0$. Show that if $\epsilon,a>0$, then then $$\liminf_{n\to\infty }\frac{P(S_n \geq na)}{nP(X_1 \geq n(a+\epsilon))}\geq 1$$

Hint: Let $F_n =\{ X_i \geq n(a+\epsilon) > \text{for exactly one } i\leq n\}$.

Edit: I tried using the inequality $P(S_n \geq na) \geq P(S_{n-1}\geq -n\epsilon)P(X_n \geq n(a+\epsilon))$, but i don't know what to do next.

Answer 1

Here's my answer after working on it for some time. A cleaner or shorter proof is welcome:

Let $F_n =\{X_i \geq n(a+\epsilon)\textrm{ for exactly one } i\leq n\}$ and note that $F_n = \bigsqcup_{i=1}^{n} G_i$, where $G_i$ is the event that the large element has index $i$. This gives us that: \begin{align*} \frac{P(S_n \geq na)}{nP(X_1\geq na)} &\geq \frac{P(F_n \cap \{S_n \geq na\})}{nP(X_1\geq n(a+\epsilon))} \\[0.5em] &=\frac{\sum_{i=1}^{n}P(G_i \cap \{S_n \geq na\})}{nP(X_1\geq n(a+\epsilon))} \\[0.5em] &=\frac{nP(G_1 \cap \{S_n \geq na\})}{nP(X_1\geq n(a+\epsilon))} \\[0.5em] &=\frac{P(X_1\geq n(a+\epsilon))P\left(\left\{\max\limits_{i<n}X_i < n(a+\epsilon)\right\} \cap \{ S_{n-1}\geq -n\epsilon\} \right)}{P(X_1\geq n(a+\epsilon))} \\[0.5em] &= P\left(\left\{\max\limits_{i<n}X_i < n(a+\epsilon)\right\} \cap \{ S_{n-1}\geq -n\epsilon\} \right) \end{align*} Since $EX_1 = 0$, then $E|X_1| <\infty$ and $nP(X_1 \geq n(a+\epsilon)) \xrightarrow{n\to\infty}0$. This gives us that: $$P\left(\max\limits_{i<n}X_i \geq n(a+\epsilon)\right) \leq nP\left(X_1 > n(a+\epsilon)\right) \xrightarrow{n\to\infty}0. \ (*)$$ We have that $\liminf P\left(\left\{\max\limits_{i<n}X_i < n(a+\epsilon)\right\} \cap \{ S_{n-1}\geq -n\epsilon\} \right)$ is $1$ by the weak law of large numbers and $(*)$. This gives us the result.

Answer 2

We have $$ \frac{P(S_n\geq na)}{nP(X_1\geq n(a+\epsilon))}\geq \frac{P(S_{n-1}\geq -n\epsilon \vert X_j<n(a+\epsilon) \text{ for all }j<n)P(F_n)}{nP(X_1\geq n(a+\epsilon))}. \ \ (1) $$

Note that the conditional probability should be used due to lack of independence of "All $X_j <n(a+\epsilon)$" and "$S_{n-1}>-n\epsilon$".

Let $\alpha = P(X_1\geq n(a+\epsilon))$. Then $P(F_n)=n\alpha(1-\alpha)^{n-1}$, which gives (1) is at least $$ P(S_{n-1}\geq -n\epsilon\vert X_j<n(a+\epsilon) \text{ for all }j<n)(1-\alpha)^{n-1}. $$ By Markov inequality, we have $$ \alpha=P(X_1\geq n(a+\epsilon))\leq \frac{E(|X_1| I_{X_1\geq n(a+\epsilon)})}{n(a+\epsilon)}, \ \mathrm{and} $$ $$ 1-\alpha\geq 1-\frac{E(|X_1| I_{X_1\geq n(a+\epsilon)})}{n(a+\epsilon)}. $$ Thus, $$ 1\geq (1-\alpha)^{n-1} \geq \left(1-\frac{E(|X_1| I_{X_1\geq n(a+\epsilon)})}{n(a+\epsilon)}\right)^{n-1} \rightarrow 1 \ \mathrm{as} \ n\rightarrow\infty. $$ By Weak Law of Large Numbers (Textbook Theorems 2.2.12 and 2.2.14), $P(S_{n-1}\geq -n\epsilon\vert X_j<n(a+\epsilon) \text{ for all }j<n)\rightarrow 1$ as $n\rightarrow\infty$. The result follows.