$X=(X_1,X_2,X_3)^T\sim N_3(\mu,\Sigma).$ Suppose $X_1,...,X_{20}$ are i.i.d. observations from $X$. The sample mean vector and the covariance matrix are then defined by

$$ \bar{x} = (1,0,2)^T,\quad S=\pmatrix{3,2,1\\2,3,1\\1,1,4}$$

What is the conditional covariance matrix of $(X_2,X_3)^T$ given $X_1$?

The conditional distribution is given by

$$X_2,X_3 \mid X_1\sim N(\mu^*,\Sigma^*)$$ where $$ \mu^*= \bar{x}_1+S_{12}S_{22}^{-1}(x_2-\mu_2) = \pmatrix{0\\2} + \pmatrix{2\\1}(3)^{-1}x_1= \pmatrix{2/3x_1\\2+1/3x_1} $$ and $$\Sigma^* = S_{11}-S_{12}S_{22}^{-1}S_{21}^{-1} = \pmatrix{3,1\\1,4}-\pmatrix{2\\1}(3)^{-1}\pmatrix{2,1} = \pmatrix{5/3,1/3\\1/3,11/3}$$


The iiid realisations of $X$ should use another notation than $X_i$ (e.g. something like $X^i)$ because it gets mixed up with $X_1,X_2$ and $X_3$ inside the vector $X$.

If you had known the true $\mu,\Sigma$, you could use the conditional distribution formulas given in wikipedia page to calculate it exactly. But, you have approximations for them, which are calculated from your random sample of size $20$, and you can calculate the resulting approximate distribution.

  • $\begingroup$ Thanks, I had a go using the formula but didn't understand how to implement it with this example. Did I do it correctly? $\endgroup$ – CCZ23 Sep 6 '20 at 18:24
  • $\begingroup$ In the formula, the second portion of the vector is given. So, you should first rearrange your mean and covariance matrix to reflect the ordering $(X_2,X_3,X_1)$, then let $\mathbf x_1=[X_2,X_3]^T$ and $\mathbf x_2=[X_1]$, and apply the remaining formulas. $\endgroup$ – gunes Sep 6 '20 at 19:27
  • $\begingroup$ how can i rearrange the covariance matrix in that order? Like this $\pmatrix{3,1,2\\1,4,1\\2,1,3}$ ? I think it works but just want to check $\endgroup$ – CCZ23 Sep 7 '20 at 6:30
  • $\begingroup$ Yes, your reordering is correct $\endgroup$ – gunes Sep 7 '20 at 6:36

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