$L^2$ and $L^p$ convergence of sample mean

Question

I have following convergence problem:

Let $X_1$, $X_2$, ... be a sequence of identically distributed random variables, they are pairwise independent. Denote $\bar{X}_n = \frac{1}{n}\sum_{i=1}^nX_i$. Suppose $E[|X_i|^p]<\infty$, for some $p\in(1,2]$. Want to show following:

a) For any $\epsilon>0$, $P(|\bar{X}_n-\mu|>\epsilon)\to 0$ as $n\to \infty$, where $E[X_i]=\mu$;

b) $E[|\bar{X}_n-\mu|]\to 0$ as $n\to \infty$;

c) If $p=2$, then $E[|\bar{X}_n-\mu|^2]\to 0$ as $n\to \infty$;

d) For $1<p<2$, $E[|\bar{X}_n-\mu|^p]\to 0$ as $n\to \infty$.

I know that if I can show b), c), d) hold, then by Markov inequality, a) holds.

c) is easy to proof using $\operatorname{Var} (X_i)$ is finite and pairwise independence.

My question is how to show b) and d).

I tried following truncation $$Y_n = (X_n - \mu)1_{(|X_n-\mu|\leq n)}$$ $$\bar{Y}_n = \sum_{i=1}^n Y_i$$ $$\mu_n = E[\bar{Y}_n]$$ for b) we have $$E[|\bar{X}_n-\mu|] \leq E[|\bar{X}_n-\mu-\bar{Y}_n|]+E[|\bar{Y}_n-\mu_n|]+E[|\mu_n|]$$

I have difficulty to show $$E[|\bar{X}_n-\mu-\bar{Y}_n|]\to 0$$ $$E[|\bar{Y}_n-\mu_n|]\to 0$$ $$E[|\mu_n|]\to 0$$

Thank you for the helps.

Answer 1

Under the basis ${e^{i\omega_{n}x}}$, inner product of two functions represented by $\vec{u}$ and $\vec{v}$ is their ordinary dot product.

$\rho(x) = [c_{-\infty}, ..., c_{-1}, c_{0}, c_{1}, ..., c_{+\infty}] = [c_m]$ (represented as row vector)

$\rho(x + s) = [ c_me^{i\omega_{m}s} ]$

For $n = 2$, the probability density for $\bar{X}_n$ is:

$\bar{\rho}(x) = \int_{a + b = 0} \rho(x + a)\rho(x + b) = [ c_me^{i\omega_{m}a} ] \cdot [ c_{m} e^{i\omega_{m}b} ] = [c_m^{2}]$

For $n = 3$,

$\bar{\rho}(x) = \int_{a + b + c = 0} \rho(x + a)\rho(x + b)\rho(x + c) = [ c_me^{i\omega_{m}a} ] \cdot [ c_{m} e^{i\omega_{m}b} ] \cdot [ c_{m} e^{i\omega_{m}c} ] = [c_m^{3}]$

For any $n$, $\bar{\rho}(x) = \sum_m c_{m}^{n}{e^{i\omega_{m} x}} $

As probability density is a real function, $c(\omega)$ is Hermitian. This means if $c(\omega) = re^{i\theta}$, then $c(-\omega) = re^{-i\theta}$

For $x = 0$, the real parts of $c(\omega)^{n}$ and $c(-\omega)^{n}$ add up and their imaginary parts cancel.

In the limit that $n \rightarrow \infty$, the integral evalualtes to $\infty$ for $x = 0$, unless all coefficients are zero. For $x \ne 0$, $\rho(x) = 0$ by normalization.

Answer 2

The following proposition would answer your question:
Proposition Let $\{X_n,n\ge 1\}$ be a sequence of pairwise independent identically distributed random variables. Denote $S_n=\sum_{i=1}^nX_i$, $\bar{X}_n=S_n/n$.
a) If \begin{align*} \lim_{n\to\infty}n\mathsf{P}(|X_1|>n)=0,\tag{1}\\ \lim_{n\to\infty}\mathsf{E}[X_11_{\{|X_1|\le n\}}]=\mu,\tag{2} \end{align*}
then \begin{equation} \underset{n\to\infty}{\text{pr-lim}}\,\bar{X}_n=\mu.\tag{3} \end{equation} b) If $\mathsf{E}[|X_1|^p]<\infty$ for $p\in[1,\infty)$, then \begin{equation*} \lim_{n\to\infty}\mathsf{E}[|\bar{X}_n-\mu|^p]=0.\tag{4} \end{equation*}

Proof $\;$ a) Let \begin{align*} Y_{ni}&=X_i1_{\{|X_i|\le n\}},\quad 1\le i\le n,n\ge 1. \\ T_n&=\sum_{i=1}^nY_{ni},\quad n\ge 1. \end{align*} Then $\{Y_{ni}, 1\le i\le n\}$ are pairwise independent identically distributed and uncorrelated. Now from (2) we have \begin{align*} \mathsf{E}[Y_{ni}]&=\mathsf{E}[X_11_{|X_1|\le n}],\\ \mathsf{E}\Bigl[\frac{T_n}n\Bigr]&=\frac1n\sum_{i=1}^n\mathsf{E}[Y_{ni}]=\mathsf{E}[X_11_{|X_1|\le n}]\to \mu,\qquad\text{as } n\to\infty. \tag{5}\\ \mathsf{Var}\Bigl[\frac{T_n}n\Bigr]&=\frac1{n^2}\sum_{i=1}^n \mathsf{Var}[Y_{ni}]\le \frac1{n^2}\sum_{i=1}^n \mathsf{E}[Y_{ni}^2] =\frac1n\mathsf{E}[X_1^21_{|X_1|\le n}]\\ &=\frac1n\int_0^ny^2dF_{|X_1|}(y)\le \frac2n\int_0^ny\mathsf{P}(|X_1|>y)dy\stackrel{(1)}\to 0,\qquad\text{as }n\to\infty.\tag{6} \end{align*} Using (5) and (6) we find that \begin{equation*} \underset{n\to\infty}{\text{pr-lim}}\,\frac{T_n}n=\mu, \tag{7} \end{equation*} Meanwhile, using (1), \begin{align*} \mathsf{P}(S_n\ne T_n)&\le\sum_{i=1}^n \mathsf{P}(X_i\ne Y_{ni})=\sum_{i=1}^n\mathsf{P}(|X_i|>n)\\ &=n\mathsf{P}(|X_1|>n)\to 0,\qquad\text{as }n\to\infty.\tag{8} \end{align*} Now (3) holds from (7) and (8).
b) If $\mathsf{E}[|X_1|^p]<\infty$ for $p\ge 1$, then $\{|X_n|^p, n\ge 1\}$ are uniformly integrable. Furthermore, $\{|\bar{X}_n-\mu|^p,n\ge1\}$ are also uniformly integrable, now (4) holds from (3) and the uniformly integrable of $\{|\bar{X}_n-\mu|^p,n\ge1\}$.