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I face a problem in which I have to evaluate a integral of the form:

$$ \int_{-\infty}^{\infty} \exp(-\lambda_t^\prime C \lambda_t + \lambda_t^\prime b)d\lambda_t, $$ where $\lambda_t$ is a $n$-dimensional column vector, $C$ is a symmetric positive definite $n$ by $n$ matrix and $b$ is $n$ by 1 column vector. I denote the transpose of a vector with "$\prime$". I know the answer in the one-dimensional case is given by $$ \frac{\sqrt{\pi}\exp(\frac{b^2}{4C})}{\sqrt{C}}. $$

Can anyone help me in the $n$-dimensional case?

simbod
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user369157
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1 Answers1

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Hint. As $C$ is symmetric positive, there is a positive diagonal matrix $M = {\rm diag}(\mu_1, \ldots, \mu_n)$, and an orthogonal $P$ such that $C = PMP^t$. Now, \begin{align*} -\lambda^t C \lambda + \lambda^t b &= -(P\lambda)^t PCP^t P\lambda + (P\lambda)^t Pb\\ &= \sum_{i=1}^n -\mu_i (P\lambda)^2_i + (P\lambda)_i(Pb)_i \end{align*} We have for the integral \begin{align*} \int_{\mathbf R^n} \exp\bigl(-\lambda^t C \lambda + \lambda^t b\bigr)\, d\lambda &= \int_{\mathbf R^n} \exp \left(\sum_{i=1}^n -\mu_i (P\lambda)^2_i + (P\lambda)_i(Pb)_i\right)\,d\lambda\\ &= \int_{\mathbf R^n} \prod_{i=1}^n \exp\bigl(-\mu_i(P\lambda)^2_i+ (P\lambda)_i(Pb)_i\bigr) \,d\lambda\\ &= \int_{\mathbf R^n} \prod_{i=1}^n \exp\bigl(-\mu_i x_i^2 + x_i (Pb)_i \bigr) \, dx\\ &= \prod_{i=1}^n \int_{\mathbf R} \exp\bigl(-\mu_i x_i^2 + x_i (Pb)_i \bigr) \, dx_i \end{align*} Now use what you know for $n=1$.

martini
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