I'm stuck on the following proof:

Let $(X_1,.. X_n)$ random variables so that $E(Xi) = \mu$, $V(Xi) = \sigma^2$ final for all $1 \leq i \leq n$.
It is also given that for all $i \neq j$, $Cov(Xi, Xj) < 0$

Denoting: $$\bar{Xn} = \dfrac 1 n \sum_{n=1}^{n} Xi$$

Proof that $$\bar{Xn} \xrightarrow{\text{L2}} \mu $$

When $\xrightarrow{\text{L2}}$ is Convergence in mean. That means that I need to prove that $$E((\bar{Xn} - \mu)^2) \xrightarrow{n\to\infty} 0 $$

I've tried: $$E((\bar{Xn} - \mu)^2) = E(\bar{Xn}^2 - 2\bar{Xn}*\mu + \mu^2) =$$

$$= E(\bar{Xn}^2) - 2E(\bar{Xn}*\mu) + E(\mu^2) = E(\bar{Xn}^2) - 2\mu^2 + \mu^2 =E(\bar{Xn}^2) - \mu^2$$

But don't know how to continue with $E(\bar{Xn}^2)$ and don't know how to use the covariance fact.



Note that $$ \Bbb E[|\overline X_n-\mu|^2]=\frac1{n^2}\Bbb E\left[\sum_{i,j=1}^n (X_i-\mu)(X_j-\mu)\right]=\frac1 {n^2}\sum_{i,j=1}^n \text{Cov}(X_i,X_j). $$ For $i\ne j$, $\text{Cov}(X_i,X_j)<0$ by the assumption, hence $$ \Bbb E[|\overline X_n-\mu|^2]\le \frac1{n^2}\sum_{i=1}^n \text{Var}(X_i)= \frac{\sigma^2}n\xrightarrow{n\to\infty} 0. $$ So, $\overline{X}_n\xrightarrow{n\to\infty} \mu$ in $L^2$ as desired.

  • $\begingroup$ Understood everything except this: Why $$ \Bbb E[|\overline X_n-\mu|^2]=\frac1 {n^2}\sum_{i,j=1}^n \text{Cov}(X_i,X_j). $$ I got to $$ \frac{1}{n^2}*[\sum_{i=1}^{n}V(Xi) + 2\sum_{i \ne j}^{n} Cov(Xi, Xj)] $$ $\endgroup$ – JohnSnowTheDeveloper Feb 18 at 18:47
  • 1
    $\begingroup$ @JohnSnowTheDeveloper We can see that $$ \text{Var}(X_i) =\text{Cov}(X_i,X_i),$$ so diagonal terms are equal. You should sum $2\sum_{i>j} \text{Cov}(X_i,X_j)$ over $i>j$, not $i\ne j$, that is,$$2\sum_{\color{red}{i>j}} \text{Cov}(X_i,X_j)=\sum_{\color{red}{i\ne j}} \text{Cov}(X_i,X_j)$$ $\endgroup$ – Song Feb 18 at 18:49

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