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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 $$ \pi^D_{\text{prior}}(m) \propto \exp \left( - \frac{1}{2 \gamma^2} || L_D m ||^2 \right) $$

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    $\begingroup$ Why do you need to integrate it? $\endgroup$
    – Glen_b
    Commented Nov 5, 2016 at 11:02
  • $\begingroup$ I want to estimate the variance as shown in Fig.1 $\endgroup$
    – MBM
    Commented Nov 5, 2016 at 11:31
  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Commented Nov 5, 2016 at 11:57
  • $\begingroup$ Thanks @Glen_b. I know that it is done. I am asking about the middle steps between?!! $\endgroup$
    – MBM
    Commented Nov 5, 2016 at 12:23
  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Commented Nov 6, 2016 at 0:25

1 Answer 1

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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

$$\pi^D_{\mathrm{prior}}(m) \propto \exp\left( -\frac{1}{2} m^\top \left[ \left( \frac{\left[ L_d^\top L_D \right]}{\gamma^{2}} \right)^{-1} \right]^{-1} m \right).$$

Just compare this to the density given on the Wikipedia page of the multivariate normal.

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