 Solution

ID: 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, $E [Z] = \sum_{i} E [X_{i}] = 0$ . Expanding the square and again using linearity of expectation gives

Var (Z) = E [Z^{2}] = i \sum E [X_{i}^{2}] + 2 i < j \sum E [X_{i} X_{j}] = i \sum σ_{i}^{2} + 2 i < j \sum Cov (X_{i}, X_{j}) .

Here the zero means imply

E [X_{i}^{2}] = σ_{i}^{2}

and

E [X_{i} X_{j}] = Cov (X_{i}, X_{j})

. The covariance expansion itself does not require independence. Under the additional independence stated here,

E [X_{i} X_{j}] = E [X_{i}] E [X_{j}] = 0

, so it simplifies further to $Var (Z) = \sum_{i} σ_{i}^{2}$ .

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