# Convergence in mean of a series of random variables

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.

Thanks

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.
• 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 $$\frac{1}{n^2}*[\sum_{i=1}^{n}V(Xi) + 2\sum_{i \ne j}^{n} Cov(Xi, Xj)]$$ – JohnSnowTheDeveloper Feb 18 at 18:47
• @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,$$2\sum_{\color{red}{i>j}} \text{Cov}(X_i,X_j)=\sum_{\color{red}{i\ne j}} \text{Cov}(X_i,X_j)$$ – Song Feb 18 at 18:49