0
$\begingroup$

Lemma Let $Q(\theta)=\theta'A\theta+b'\theta+c$, where $\theta,b\in\mathbb{R}^n$, c is a scalar and $A$ is a symmetric positive definite $n\times n$ matrix. Then

$\int_{\mathbb{R}^n} e^{-Q(\theta)}\mathrm{d}\theta=e^{-Q_0}\frac{\pi^{n/2}}{\sqrt{det\;A}}$ where $Q_0=min_{\theta\in\mathbb{R}^n}Q(\theta)$

To prove the above lemma we let $\theta_0\in argmin\; Q$. Also say that $\epsilon=\theta-\theta_0$ and $\hat{Q}(\epsilon)=Q(\epsilon+\theta_0)$. Why do we get $\hat{Q}(\epsilon)=\epsilon'A\epsilon+Q_0$ and why we are left with $\int_{\mathbb{R}^n} e^{-\epsilon'A\epsilon}\mathrm{d}\theta=\frac{\pi^{n/2}}{\sqrt{det\;A}}$ ?

$\endgroup$
0
$\begingroup$

Note that $\hat{Q}(\varepsilon) = \varepsilon^t A \varepsilon + v^t \varepsilon + d$ for some vector $v \in \mathbb{R}^n$ and scalar $d \in \mathbb{R}$. Since $\hat{Q}$ attains its minimum at $\varepsilon = 0$, we have $v = \nabla \hat{Q}(0) = 0$ and because $A$ is positive definite we have $d = \hat{Q}(0) = \inf \limits_{x \in \mathbb{R}^n}Q(x)$.

The remaining integral is a standard integral, a proof can be found here.

$\endgroup$
  • $\begingroup$ Why scalar c is replaced by vector d? $\endgroup$ – Waqas Nov 10 '16 at 13:15
  • $\begingroup$ If you expand $Q(\varepsilon + \theta_0)$, you will see that $d = c + b^t\theta_0 + \theta_0^t A \theta_0$. $\endgroup$ – Dominik Nov 10 '16 at 13:16
  • $\begingroup$ @A.G. thanks for the clarification, this was actually a mistake that I've corrected now. $\endgroup$ – Dominik Nov 10 '16 at 14:42
  • $\begingroup$ Can you please give the expanded term for $v$, and explain more explicitly, why $v=0$ and yes now I understand $1\times n$, $n\times n$, $n \times 1$ gives $1\times 1$, which makes $d$ a scalar. $\endgroup$ – Waqas Nov 10 '16 at 14:45
  • $\begingroup$ It doesn't matter what $v$ is exactly, you get $v = 0$ from the fact that the gradient of a differentiable function at an extreme point is zero. $\endgroup$ – Dominik Nov 10 '16 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.