Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 27K / a

Table of contents
- Solution a

Solution

For

d = 1

and

X \sim N (0, σ^{2})

, the characteristic function is

φ_{X} (u) = e^{- σ^{2} u^{2} /2} .

For a multivariate normal distribution with covariance matrix

σ^{2} I_{d}

, the coordinates are uncorrelated and hence independent by independence of uncorrelated jointly normal variables. Therefore

φ_{X} (u) = \prod_{j = 1}^{d} E [e^{i u_{j} X_{j}}] = \prod_{j = 1}^{d} e^{- σ^{2} u_{j}^{2} /2} = exp (- \frac{σ ^{2}}{2} j = 1 \sum d u_{j}^{2}) .

This is the characteristic function of an isotropic Gaussian vector.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version