# Understanding proof of a Lemma

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 $\int_{\mathbb{R}^n} e^{-\epsilon'A\epsilon}\mathrm{d}\theta=\frac{\pi^{n/2}}{\sqrt{det\;A}}$ ?

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 $d = \hat{Q}(0) = \inf \limits_{x \in \mathbb{R}^n}Q(x)$.
• If you expand $Q(\varepsilon + \theta_0)$, you will see that $d = c + b^t\theta_0 + \theta_0^t A \theta_0$. – Dominik Nov 10 '16 at 13:16
• 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. – Waqas Nov 10 '16 at 14:45
• 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. – Dominik Nov 10 '16 at 14:47