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Suppose we have a multivariate normal random variable $X = [X_1, X_2, X_3, X_4]^⊤$. And here $X_1$ and $X_4$ are independent (not correlated). Also $X_2$ and $X_4$ are independent. But $X_1$ and $X_2$ are not independent. Assume that $Y = [Y_1, Y_2]^⊤$ is defined by $$Y_1 = X_1 + X_4,~~ Y_2 = X_2 − X_4.$$ If I know the covariance matrix of $X$, what would be the covariance matrix of $Y$?

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  • $\begingroup$ By using properties of variance and covariance, you can write $\text{Var}(Y_1)$, $\text{Var}(Y_2)$, and $\text{Cov}(Y_1, Y_2)$ in terms of variances/covariances involving the $X_i$. $\endgroup$
    – angryavian
    Nov 21, 2021 at 23:28

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You can assume w.l.o.g. that $E[X]=0$. Then $E[Y]=0$ (variances/covariances are not dependent on means).

You need to compute $Var(Y_1),Var(Y_2), E[Y_1 Y_2]$ since the covariance matrix of $Y$ is comprised of these three elements.

Since $X_1,X_4$ are independent, $Var(Y_1)=Var(X_1)+Var(X_4)$ which you should konw from the covariance matrix of $X$.

Similarly for $Var(Y_2)$.

Finally you can compute $E[Y_1 Y_2]= E[ X_1 X_2 - X_4^2] = Cov(X_1,X_2) -Var(X_4)$ which you should know from the covariance matrix of $X$

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  • $\begingroup$ How does var Y1, var y2 and E [y1y2] lead to Cov(Y)? $\endgroup$ Nov 21, 2021 at 23:33
  • $\begingroup$ The covariance matrix of $Y$ is comprised of the variances in the diagonal and covariances in the off-diagonal. If you assume that $E[X]=0$ then $E[Y]=0$ which means that $Cov(Y_1,Y_2)=E[Y_1,Y_2]$ $\endgroup$
    – Cris
    Nov 21, 2021 at 23:35
  • $\begingroup$ I see. But how will this be expressed in matrix form. $\endgroup$ Nov 21, 2021 at 23:39
  • $\begingroup$ The X has a covariance matrix 4x4 so there are many off diagonal elements. I know that we can find the variance Ford y1, y2 by looking at the diagonal elements. I’m not so sure what we are doing with the expectation [y1, y2] And if this covariance will be a 3 x 3 because X3 is not part of equations in y1, y2 $\endgroup$ Nov 21, 2021 at 23:41
  • $\begingroup$ You need the covariance matrix of $Y$ which is $2x2$, not that of $X$. See the definition of the covariance matrix here: en.wikipedia.org/wiki/Covariance_matrix $\endgroup$
    – Cris
    Nov 21, 2021 at 23:43

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