# Integration of exponential having ||Lx|| inside?

I am stuck with performing this integration. Can anyone help please?

$$\mathbb{V}\mathrm{ar}[M_j] \stackrel{\text{def}}{=} \mathbb{E}[M_j^2] = e_j^T \left( \int_{\mathbb{R}^{n+1}} m^2 \pi^D_{\text{prior}}dm \right) e_j \stackrel{\text{def}}{=} \gamma^2 e_j^T \left(L_D^T L_D\right)^{-1} e_j$$

The $e_j$ is called: canonical basis $$\pi^D_{\text{prior}}(m) \propto \exp \left( - \frac{1}{2 \gamma^2} || L_D m ||^2 \right)$$

• Why do you need to integrate it? – Glen_b -Reinstate Monica Nov 5 '16 at 11:02
• I want to estimate the variance as shown in Fig.1 – MBM Nov 5 '16 at 11:31
• but then see that the term with the integral in it is followed by an equality (with $\text{def}$ above it, presumably indicating a definition). So it seems there's nothing to integrate, you just use the formula at the end. Someone integrated it for you. – Glen_b -Reinstate Monica Nov 5 '16 at 11:57
• Thanks @Glen_b. I know that it is done. I am asking about the middle steps between?!! – MBM Nov 5 '16 at 12:23
• So it's not that you want to estimate the variance (see your previous response to me) -- for which you could just use the formula -- but that you want to derive the result. Clarifying that issue (what you ultimately wanted to achieve) is why I asked. – Glen_b -Reinstate Monica Nov 6 '16 at 0:25

Presumably by $m^2$ you mean $mm^\top$ since $m$ is a vector. The integral computes the second moment matrix of a zero-mean multivariate normal distribution whose covariance is $\gamma^2 (L^\top_D L_D)^{-1}$. Since it is zero-mean, the second moment matrix is the covariance matrix. Write
$$\pi^D_{\mathrm{prior}}(m) \propto \exp\left( -\frac{1}{2} m^\top \left[ \left( \frac{\left[ L_d^\top L_D \right]}{\gamma^{2}} \right)^{-1} \right]^{-1} m \right).$$