 Solution

ID: past-exam-of-the-algorithmic-foundations-of-learning-course-of-the-university-of-oxford/2025/2/a/ii/solution

Let

a_{1}, \dots, a_{d} > 0

V_{n}

. Its trace satisfies

\sum_{j = 1}^{d} a_{j} = tr V_{n} = d λ + \sum_{t = 1}^{n} ∥ X_{t} ∥_{2}^{2} \leq d λ + n .

det V_{n} = \prod_{j = 1}^{d} a_{j} \leq (\frac{\sum _{j} a _{j}}{d})^{d} \leq (λ + \frac{n}{d})^{d} .

Since

det V_{0} = λ^{d}

, taking the logarithm proves

lo g \frac{det V _{n}}{det V _{0}} \leq d lo g (1 + \frac{n}{d λ}) .

The determinant is largest, for a fixed trace, when the eigenvalues are equal; this is the geometric content of the estimate.

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