 Solution

ID: past-exam-of-the-random-matrix-theory-course-of-the-university-of-oxford/2025/1/b/i/solution

k

th powers of the eigenvalues with the trace of the

k

m_{k} (n) = n^{- 1 - k /2} tr (M^{k}) = n^{- 1 - k /2} i_{1}, \dots, i_{k} = 1 \sum n M_{i_{1} i_{2}} M_{i_{2} i_{3}} \dots M_{i_{k} i_{1}} .

For odd

k

, every summand has zero expectation by part (a)(ii), so

E [m_{k} (n)] = 0

for every

n

For the second moment, use

M_{ji} = M_{ij}

. There are

n

diagonal entries with variance two and

n (n - 1)

ordered off-diagonal entries with variance one. Their means vanish, giving

E [m_{2} (n)] = n^{- 2} \sum_{i, j} E [M_{ij}^{2}] = \frac{2 n + n ( n - 1 )}{n ^{2}} = 1 + \frac{1}{n} .

Hence

n \to \infty lim E [m_{2} (n)] = 1

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