2
$\begingroup$

What is the second moment of the conditional distribution of one multivariate normal variable given another multivariate normal variable? I know the formula for the second central moment (the variance), but the second moment is hard to find online.

$$\int_{x=-\infty}^{x=\infty} xx'f(x\vert y)dx$$

In my case, the marginal and conditional means and variances are known.

$\endgroup$
7
  • 1
    $\begingroup$ You know the conditional mean and variance, right? $\endgroup$
    – Dave
    Apr 7 '21 at 12:08
  • $\begingroup$ Yes indeed, and also the marginal mean and variance $\endgroup$
    – MD94
    Apr 7 '21 at 12:14
  • $\begingroup$ There is an equation relating mean, variance, and second moment. To jog your memory, you’ve probably seen it using expected value notation. $\endgroup$
    – Dave
    Apr 7 '21 at 12:20
  • $\begingroup$ You say it's a multivariate normal, but then integrate on the real line. Could you clarify? $\endgroup$
    – Firebug
    Apr 7 '21 at 12:34
  • 1
    $\begingroup$ Yes, those formula's are indeed implied $\endgroup$
    – MD94
    Apr 7 '21 at 12:57
1
$\begingroup$

For two multivariate normal vectors $\mathbf{x}\sim\mathcal N(\mathbf{\mu_x},\Sigma_{\mathbf{x,x}})$ and $\mathbf{Y}\sim\mathcal N(\mathbf{\mu_Y},\Sigma_{\mathbf{Y,Y}})$ we have that, given their covariance $\Sigma_{\mathbf{x,Y}}$:

$$E[\mathbf x|\mathbf Y]=\mathbf{\mu_x}+\Sigma_{\mathbf{x,Y}}\Sigma_{\mathbf{Y,Y}}^{-1}(\mathbf{Y-\mu_Y})$$

$$\text{Var}[\mathbf x|\mathbf Y]=\Sigma_{\mathbf{x,x}}-\Sigma_{\mathbf{x,Y}}\Sigma_{\mathbf{Y,Y}}^{-1}\Sigma_{\mathbf{Y,x}}$$

But the covariance is given in terms of the second moment:

$$\text{Var}[\mathbf x|\mathbf Y]=E[\mathbf {xx}^T|\mathbf Y]-E[\mathbf x|\mathbf Y]E[\mathbf {x}^T|\mathbf Y]$$

So

$$E[\mathbf {xx}^T|\mathbf Y] = \text{Var}[\mathbf x|\mathbf Y] + E[\mathbf x|\mathbf Y]E[\mathbf {x}^T|\mathbf Y]\\ =\Sigma_{\mathbf{x,x}}-\Sigma_{\mathbf{x,Y}}\Sigma_{\mathbf{Y,Y}}^{-1}\Sigma_{\mathbf{Y,x}}+ (\mathbf{\mu_x}+\Sigma_{\mathbf{x,Y}}\Sigma_{\mathbf{Y,Y}}^{-1}(\mathbf{Y-\mu_Y})) (\mathbf{\mu_x}+\Sigma_{\mathbf{x,Y}}\Sigma_{\mathbf{Y,Y}}^{-1}(\mathbf{Y-\mu_Y}))^T$$


I haven't checked, but it seems this all stems from the simple fact that:

$$\mathbf x=\mathbf{\mu_x}+\Sigma_{\mathbf{x,Y}}\Sigma_{\mathbf{Y,Y}}^{-1}(\mathbf{Y-\mu_Y})+\mathbf e\\ E[\mathbf {xx}^T|\mathbf Y]=(\mathbf{\mu_x}+\Sigma_{\mathbf{x,Y}}\Sigma_{\mathbf{Y,Y}}^{-1}(\mathbf{Y-\mu_Y}))(\mathbf{\mu_x}+\Sigma_{\mathbf{x,Y}}\Sigma_{\mathbf{Y,Y}}^{-1}(\mathbf{Y-\mu_Y}))^T+E[\mathbf {ee}^T|\mathbf Y] $$

So we could've started directly here, and then $E[\mathbf {ee}^T|\mathbf Y] = \text{Var}[\mathbf x|\mathbf Y]$.

$\endgroup$

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.