Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 25 / 1 / b

Table of contents
- Solution b

Solution

For any

r_{1}, \dots, r_{n} \in R

, the linearity from part (a) gives

\sum_{j = 1}^{n} r_{j} X (h_{j}) = X (\sum_{j = 1}^{n} r_{j} h_{j})

almost surely, and the right side has a normal distribution. This is the defining linear-combination criterion for a multivariate normal distribution; singular covariance matrices are allowed.

Passing to the limit in the inner products of the partial sums gives

Cov (X (g), X (h)) = E [X (g) X (h)] = j \sum ⟨ g, e_{j} ⟩ ⟨ h, e_{j} ⟩ = ⟨ g, h ⟩ .

The passage to the limit is justified by Cauchy-Schwarz inequality and

L^{2}

convergence. Equivalently, the vector's characteristic function is

E exp (i \sum_{j} r_{j} X (h_{j})) = exp (- \frac{1}{2} \sum_{j, k} r_{j} r_{k} ⟨ h_{j}, h_{k} ⟩) .

Thus both joint normality and the complete covariance matrix follow from the Hilbert-space inner product.

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