1. **Problem a:** Show that for a random vector $\mathbf{X}$, $\mathrm{Var}(A\mathbf{X} + \mathbf{b}) = A \mathrm{Var}(\mathbf{X}) A^T$ where $A$ is a $(k \times p)$ matrix and $\mathbf{b}$ is a $(p \times 1)$ vector of constants. 2. **Formula and explanation:** The variance of a random vector $\mathbf{Y} = A\mathbf{X} + \mathbf{b}$ is given by $$\mathrm{Var}(\mathbf{Y}) = E[(\mathbf{Y} - E[\mathbf{Y}])(\mathbf{Y} - E[\mathbf{Y}])^T].$$ Since $\mathbf{b}$ is constant, it does not affect variance. 3. **Step-by-step derivation:** - Compute $E[\mathbf{Y}] = E[A\mathbf{X} + \mathbf{b}] = A E[\mathbf{X}] + \mathbf{b}$. - Then, $$\mathrm{Var}(\mathbf{Y}) = E[(A\mathbf{X} + \mathbf{b} - A E[\mathbf{X}] - \mathbf{b})(A\mathbf{X} + \mathbf{b} - A E[\mathbf{X}] - \mathbf{b})^T]$$ which simplifies to $$E[(A(\mathbf{X} - E[\mathbf{X}]))(A(\mathbf{X} - E[\mathbf{X}]))^T] = E[A(\mathbf{X} - E[\mathbf{X}])(\mathbf{X} - E[\mathbf{X}])^T A^T].$$ - Since $A$ is constant, it factors out: $$A E[(\mathbf{X} - E[\mathbf{X}])(\mathbf{X} - E[\mathbf{X}])^T] A^T = A \mathrm{Var}(\mathbf{X}) A^T.$$ 4. **Conclusion:** Thus, we have shown $$\mathrm{Var}(A\mathbf{X} + \mathbf{b}) = A \mathrm{Var}(\mathbf{X}) A^T.$$