Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 9 / b / Solution

By linearity of expectation, $\mathbb{E}[Z]={\sum }_i\mathbb{E}[X_i]=0$. Expanding the square and again using linearity of expectation givesHere the zero means imply $\mathbb{E}[X_i^2]={\sigma }_i^2$ and $\mathbb{E}[X_i{X}_j]=\mathrm{Cov}(X_i,X_j)$. The covariance expansion itself does not require independence. Under the additional independence stated here, $\mathbb{E}[X_i{X}_j]=\mathbb{E}[X_i]\mathbb{E}[X_j]=0$, so it simplifies further to $\mathrm{Var}(Z)={\sum }_i{\sigma }_i^2$.