Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The characteristic function of the $N$-variate Normal distribution is $$\forall \mathbf{t} \in \mathbb{R}^N \quad \psi(\mathbf{t}) \equiv \mathbb{E}\left( e^{i\mathbf{t}X}\right) = \exp \left( i{ }^t\boldsymbol\mu \mathbf{t} - \frac{1}{2}{ }^t\mathbf{t}\boldsymbol\Sigma\mathbf{t}\right)$$

Is there any way to compute its integral on a symmetric cube? Suppose $(q_1,\ldots,q_N)\in\mathbb{R}_+^N$, then I want to calculate

$$I(q_1,\ldots,q_N)\equiv \int_{-q_1}^{q_1}\cdots\int_{-q_N}^{q_N} \psi(\mathbf{t})$$

I am particularly interested in the bivariate case ($N=2$) with $$\mu=\begin{pmatrix}\mu_1\\\mu_2\end{pmatrix} \quad \Sigma=\begin{pmatrix}1 & \rho \\ \rho & 1 \end{pmatrix} $$

Edit: If no exact solutions exist, approximations for $min(q_i) \gg 1$ are of interest. In any case, references would be greatly appreciated.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.