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I have three standard normal variables, $X_1$, $X_2$, and $X_3 \sim N(0,1)$.

Let $Y_1 = X_1 - X_2$, $Y_2 = X_3 - \bar X_2$, $Y_3 = \bar X_3$

Where $\bar X_2 = (X_1 + X_2)/2$ and $\bar X_3$ is found in the same way.

Show that $Y_1$, $Y_2$, $Y_3$ have a multivariate normal distribution, find $\mu$ and $\Sigma$.

Can someone please guide me on this or point me to a textbook which illustrates these types of transformations?

Thank you

Edit: Thanks for editing my math, Michael.

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  • $\begingroup$ You have not told us the joint distribution of $X_1,X_2,X_3$. Did you intend them to be independent? $\endgroup$ – Michael Hardy Sep 19 '15 at 18:34
  • $\begingroup$ Yes, sorry about that. $\endgroup$ – Stephen Sep 19 '15 at 18:35
  • $\begingroup$ It's not clear what you mean by $\bar X_2$ and $\bar X_3$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 19 '15 at 18:37
  • $\begingroup$ Apologies, $\bar X_2 = (X_1 + X_2)/2$ and $\bar X_3$ should follow in the same way. $\endgroup$ – Stephen Sep 19 '15 at 18:44
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$\newcommand{\cov}{\operatorname{cov}}\newcommand{\var}{\operatorname{var}}$ $$ \var(X_1-X_2) = \var(X_1)+\var(-X_2) = \var(X_1) + (-1)^2\var(X_2). $$ \begin{align} \cov(Y_1,Y_2) & = \cov(X_1-X_2,X_3 - \bar X_2) \\[6pt] & = \cov(X_1,X_3-\bar X_2) - \cov(X_2,X_3-\bar X_2) \\[6pt] & = \cov(X_1,X_3) - \cov(X_1,\bar X_2) - \cov(X_2,X_3) + \cov(X_2,\bar X_2). \end{align}

And $$ \cov(X_1,\bar X_2) = \cov\left( X_1, \frac{X_1+X_2} 2 \right) = \frac 1 2 \left( \cov(X_1,X_1)+\cov(X_1,X_2) \right), $$ etc.

Summary: You can get all of the variances and covariances by using $\cov(U,V+W)$ $= \cov(U,V)+\cov(U,W)$ and $\cov(U,V) = \cov(V,U)$ and $\cov(cU, V) = c\cov(U,V)$.

You can also use matrix methods, but I might prefer to hear more from you before going into that.

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  • $\begingroup$ Thanks. So we can find all of the entries of $\Sigma$ by doing these calculations and the expected values are easy enough to find. Then we have each vector that we need and can use the known pdf of a multivariate normal. It is that straightforward, right? $\endgroup$ – Stephen Sep 19 '15 at 19:35
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    $\begingroup$ Yes. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 19 '15 at 20:31

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