Show two claims about expectation and covariance

Question

1. Let $X_1,\ldots,X_n$ be pairwise uncorrelated with $E(X_1)=\ldots =E(X_n)=0$. Show that then $$ E\left\{\left(\sum_{i=1}^{n}X_i\right)^2\right\}=\sum_{i=1}^{n}E(X_i^2). $$
2. Let $X_1,\ldots,X_n$ be pairwise uncorrelated and $\alpha_i,\beta_i\in\mathbb{R}$ for $i=1,\ldots,n$. Show that then $$ \text{Cov}\left(\sum_{i=1}^{n}\alpha_i X_i,\sum_{i=1}^{n}\beta_i X_i\right)=\sum_{i=1}^{n}\alpha_i\beta_i\text{Var}(X_i). $$

Good evening,

here are my proofs. It would be great to hear if they are right.

1.

\begin{align} E\left(\left(\sum_{i=1}^{n}X_i\right)^2\right)&=\text{Var}\left(\sum_{i=1}^{n}X_i\right)+\left(E\left(\sum_{i=1}^{n}X_i\right)\right)^2\\ &=\sum_{i=1}^{n}\text{Var}(X_i)+\left(\sum_{i=1}^{n}\underbrace{E(X_i)}_{=0}\right)^2\\ &=\sum_{i=1}^{n}\left(E(X_i^2)-\underbrace{\left(E(X_i)\right)^2}_{=0}\right)\\ &=\sum_{i=1}^{n}E(X_i^2). \end{align}

For the second identity it is used that the $X_i$ are pairwise uncorrelated, i.e. that $\text{Var}(\sum_{i=1}^{n} X_i)=\sum_{i=1}^{n}\text{Var}(X_i)$. $\Box$

2.

\begin{align} &\text{Cov}\left(\sum_{i=1}^{n}\alpha_i X_i,\sum_{i=1}^{n}\beta_i X_i\right)\\ &=E\left(\left(\sum_{i=1}^{n}\alpha_i X_i\right)\cdot\left(\sum_{i=1}^{n}\beta_i X_i\right)\right)-E\left(\sum_{i=1}^{n}\alpha_i X_i\right)\cdot E\left(\sum_{i=1}^{n}\beta_i X_i\right)\\ &=E\left(\sum_{i=1}^{n}\alpha_i\beta_i X_i^2+\sum_{1\leqslant i,j\leqslant n \atop i\neq j}\alpha_i\beta_j X_i X_j\right)-\sum_{i=1}^{n}\alpha_i E(X_i)\sum_{i=1}^{n}\beta_i E(X_i)\\ &=\sum_{i=1}^{n}\alpha_i\beta_i E(X_i^2)+\sum_{1\leqslant i,j\leqslant n \atop i\neq j}\alpha_i\beta_j E(X_i X_j)-\left(\sum_{i=1}^{n}\alpha_i\beta_i (E(X_i))^2+\sum_{1\leqslant i,j\leqslant n \atop i\neq j}\alpha_i\beta_j E(X_i X_j)\right) \\ &=\sum_{i=1}^{n}\alpha_i\beta_i E(X_i^2)-\sum_{i=1}^{n}\alpha_i\beta_i (E(X_i))^2\\ &=\sum_{i=1}^{n}\alpha_i\beta_i (\text{Var}(X_i)+(E(X_i))^2)-\sum_{i=1}^{n}\alpha_i\beta_i (E(X_i))^2\\ &=\sum_{i=1}^{n}\alpha_i\beta_i\text{Var}(X_i)+\sum_{i=1}^{n}\alpha_i\beta_i (E(X_i))^2-\sum_{i=1}^{n}\alpha_i\beta_i (E(X_i))^2\\ &=\sum_{i=1}^{n}\alpha_i\beta_i\text{Var}(X_i) \end{align}

For the step from the 2nd to the 3rd identity it is used that the $X_i$ are pairwise uncorrelated, i.e. that $E(X_iX_j)=E(X_i)E(X_j)$. $\Box$

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With kind regards

math12

Answer 1

Good proofs. A bit long, you may reduce part 2 using the centered RV $X_i^* = X_i - E(X_i)$