Rate of convergence in law of large numbers Let $X_1, X_2, ...$ be i.i.d random variables  in $\mathbb{L}^p$ for some real number $p\geq 1$ and with $\mathbb{E}[X]=0$.


*

*Find $\alpha>0$ (as large as possible) such that $n^{\alpha}\frac{S_n}{n}\rightarrow 0$ in $\mathbb{P}$.

*Find $\alpha>0$ (as large as possible) such that $n^{\alpha}\frac{S_n}{n}\rightarrow 0$ a.s.


My Approach: For the second one, using Borel Contelli lemma we may take advantage of the first one to come up with an $\alpha$. For the first one I have tried to use truncation, by defining ${X}_i^R = -R\vee X_i \wedge R$ to get the following:
$$
\mathbb{P}(\frac{\mid S_n \mid}{n^{1-\alpha}}>\epsilon)\leq \mathbb{P}(\frac{\mid S_n-S_n^R \mid}{n^{1-\alpha}}>\frac{\epsilon}{2}) + \mathbb{P}(\frac{\mid S_n^R \mid}{n^{1-\alpha}}>\frac{\epsilon}{2})
$$ 
Now I want to send $n\to \infty$ first and then $R \to \infty$. So far I have not used independency of random variables. It seems that I need to use chebyshev inequality at this stage and use independency of random variables to get to a good point but I got stuck here. Any help? How can I proceed? Or maybe I should take another route?
 A: We will consider the case where $p \geq 2$. The $p<2$ case is discussed as a footnote.
Claim: Suppose that $p \geq 2$. Then there exists $C_p>0$ (not depending on the sequence $(X_i)$) such that for every $n \in \Bbb N$, we have that $\|S_n\|_p \leq n^{1/2} C_p \|X_1\|_p$.
We will prove the claim later (that is where independence will be used), but for now we return to the question at hand. Recall the general form of Chebyshev's inequality, which says that for any $Y\in L^p$, $$P(|Y|>\epsilon) \leq \epsilon^{-p}\|Y\|_p^p$$
Using this as well as the above claim tells you that $$P\bigg( \frac{|S_n|}{n^{1-\alpha}}>\epsilon \bigg) \leq \frac{\|S_n\|_p^p}{n^{p(1-\alpha)}\epsilon^p} \leq \frac{C_p^p n^{p/2}\|X_1\|_p^p}{n^{p(1-\alpha)}\epsilon^p} = \frac{C_p^p \|X_1\|_p^p}{\epsilon^p} n^{p(\alpha-1/2)} = C_p' \cdot n^{p(\alpha-1/2)} $$
As $n \to \infty$, this tends to zero whenever $p(\alpha-1/2)<0$, i.e, whenever $\alpha<\frac{1}{2}$. Therefore $\frac{S_n}{n^{1-\alpha}} \to 0$ in probability whenever $\alpha<\frac{1}{2}$. (Note that by CLT, this is the best possible bound.)
Similarly, $n^{p(\alpha-1/2)}$ is summable whenever $p(\alpha-1/2)<-1$, i.e, whenever $\alpha< \frac{1}{2}-\frac{1}{p}$. Therefore Borel-Cantelli tells you that $\frac{S_n}{n^{1-\alpha}} \to 0$ almost surely whenever $\alpha< \frac{1}{2}-\frac{1}{p}$.
Proof of claim: Suppose that the $X_i$ are defined on the probability space $(\Omega, \mathcal F, P)$.  

From now on fix $n \in \Bbb N$. Let us define $\Phi$ to be the measure space $\Omega \times \{1,...,n\}$ with measure $P \times c$, where $c$ is just counting measure. Each $f \in L^p(\Phi)$ can be denoted $f=(f_1,...,f_n)$ where $f_i \in L^p(\Omega)$ is just the map $\omega \mapsto f(\omega,i)$. These $f_i$ are the "slices" of $f$ on $\Omega \times \{i\}$, and completely determine $f$ (thus the $n$-tuple notation is unambiguous). We will also think of $\Omega^n$ as a probability space with product measure $P^n$.   

For each $p$ we define $L_0^p(\Phi)$ as the closed subspace consisting of all $(f_1,...,f_n) \in L^p(\Phi)$ such that all of the $f_i$ have the same distribution, and $E[f_1]=0$. For $p \geq 1$, define a map $A: L^p_0(\Phi) \to L^p(\Omega^n)$ which sends $(f_1,...,f_n) \mapsto \sum_1^n f_i \circ \pi_i$ where $\pi_i : \Omega^n \to \Omega$ is the natural projection. From now on we will denote $\overline{f_i}:= f_i \circ \pi_i$. One can see that for $k \in \Bbb N$ and for $f = (f_1,...,f_n) \in L^{2k}_0(\Phi)$, the independence of the $\overline{f_i}$ implies that $$\| Af \|_{L^{2k}(\Omega^n)}^{2k} = E_{P^n}\bigg[ \bigg(\sum_1^n \overline{f_i} \bigg)^{2k}\bigg] = \sum_{j_1+...+j_n=2k} \binom{2k}{j_1,...,j_n}E_{P^n}[\bar{f_1}^{j_1} \cdots \bar{f_n}^{j_n}] $$$$ =\sum_{\substack{j_1+...+j_n=2k \\ j_1,...,j_n \neq 1}} \binom{2k}{j_1,...,j_n}E_{P^n}[\bar{f_1}^{j_1} \cdots \bar{f_n}^{j_n}]$$$$ \leq (2k)! \cdot \big| \big\{ (j_1,...,j_n) : j_1+...+j_n = 2k, \; j_1,...,j_n \neq 1\big\} \big| \cdot E_P[f_1^{2k}] $$$$ \leq  B_k \; n^k \; E_P[f_1^{2k}] = B_k \; n^{k-1} \|f\|_{L^{2k}(\Phi)}^{2k}$$ where $B_k$ is some integer not depending on $n$. Some justification is required here. In the first equality, we merely used the definition of the $L^{2k}$ norm on $\Omega^n$. For the second equality, we used the multinomial expansion. For the third equality, we used the fact that the $\bar{f_j}$ are independent and have mean zero in order to eliminate terms in which a first power appears. In the following inequality, we used the fact that the multinomial coefficients are bounded above by $(2k)!$, as well as the fact that the $\bar{f}_j$ are i.i.d. to deduce that $E_{P^n}[\bar{f_1}^{j_1} \cdots \bar{f_n}^{j_n}] \leq E_P[f_1^{2k}]$. In the next inequality, we used the fact that if $\;j_1+...+j_n=2k$ with $j_1,...,j_k \neq 1$, then at most $k$ of the $j_i$ can be nonzero (and hence the $O(n^k)$ bound). And in the final equality, we used that $E_P[f_1^{2k}] = n^{-1} \|f\|^{2k}_{L^{2k}(\Phi)}$, simply by definition of the $L^{2k}$ norm on $\Phi$.

 Thus by taking $(2k)^{th}$ roots, we see that $\|Af\|_{L^{2k}(\Omega^n)} \leq n^{\frac{1}{2}-\frac{1}{2k}} C_{2k} \|f\|_{L^{2k}(\Phi)}$ where $C_{2k}:= B_k^{\frac{1}{2k}}$. Since $k$ was arbitrary, the Riesz-Thorin Interpolation Theorem tells us that whenever $p \in [2k,2k+2]$ for some $k \in \Bbb N$, we have that $\|Af\|_{L^{p}(\Omega^n)} \leq n^{\frac{1}{2}-\frac{1}{p}} C_p \|f\|_{L^{p}(\Phi)} $, where $C_p = C_{2k}^t C_{2k+2}^{1-t}$ with $t \in [0,1]$ satisfying $\frac{1}{p} = \frac{1-t}{2k+2} + \frac{t}{2k}$.  

 To finish the proof of the claim, note that since the $X_i \in L^p(\Omega)$ are independent, we see that $S_n$ has the same distribution as $A(X_1,...,X_n)$, and therefore $$\|S_n\|_{L^p(\Omega)} = \| A(X_1,...,X_n) \|_{L^p(\Omega^n)} \leq n^{\frac{1}{2}-\frac{1}{p}} C_p \cdot \|(X_1,...,X_n) \|_{L^p(\Phi)} $$$$ = n^{\frac{1}{2}-\frac{1}{p}} C_p \cdot n^{\frac{1}{p}} \|X_1\|_{L^p(\Omega)} = n^{1/2} C_p \|X_1\|_{L^p(\Omega)}$$
Note: For the case when $1 \leq p \leq 2$, we can obtain the weaker bound $\|S_n\|_p \leq n^{1/p} \|X_1\|_p$ using similar interpolation techniques between $p=1$ and $p=2$. Using Chebyshev's inequality as above then gives us convergence in probability whenever $\alpha< 1-\frac{1}{p}$, but there is not enough information get bounds on a.s. convergence (however, SLLN gives zero).
