Multivariate Normal Distribution

**statmajor** · October 6th 2009, 03:58 PM

I know that $f(x_1, x_2, x_3) = \frac{1}{(2 \pi)^{3/2}|\Sigma|^{1/2}}exp(-\frac{1}{2}x \Sigma^{-1} x)$ since n = 3 and mu = 0.
I've never used the multivariate normal distribution. My prof just derived it, but never taught us how to use it.

Any help would be appreciated.

**matheagle** · October 6th 2009, 09:55 PM

You may think this is a multivariate problem, but it isn't
Just find the distribution of $W=X_1-2X_2+X_3\sim N(0,\sigma^2)$

where $\sigma^2=(1,-2,1)\Sigma(1,-2,1)^t$

**statmajor** · October 7th 2009, 02:21 PM

W is a chi-squared distribution with 3 degrees of freedom.

If that's true, what do I do with?

**matheagle** · October 7th 2009, 03:28 PM

why don't you get the square root of 15.36 and do it directly?
REMEMBER x^2>4 is x>2 OR x<-2.
I'm not sure about this chi-square 3 thingy
It's not X_1^2 +X_2^2....

**statmajor** · October 7th 2009, 05:15 PM

So I did what you said:

$P(- \sqrt{15.36}<X_1 -2X_2 + X_3<\sqrt{15.36}$

Can I do the following:

$P(- \sqrt{15.36}<X_1<\sqrt{15.36})$ + $P(- \sqrt{15.36}<-2X_2<\sqrt{15.36})$ + $P(- \sqrt{15.36}<X_3<\sqrt{15.36})$

and $<br /> \sigma^2=(1,-2,1)\Sigma(1,-2,1)^t = 5<br />$

**matheagle** · October 7th 2009, 08:24 PM

Originally Posted by statmajor

So I did what you said:

$P(- \sqrt{15.36}<X_1 -2X_2 + X_3<\sqrt{15.36}$

Can I do the following:

$P(- \sqrt{15.36}<X_1<\sqrt{15.36})$ + $P(- \sqrt{15.36}<-2X_2<\sqrt{15.36})$ + $P(- \sqrt{15.36}<X_3<\sqrt{15.36})$

and $<br /> \sigma^2=(1,-2,1)\Sigma(1,-2,1)^t = 5<br />$

no, those events are not mutually exclusive.
JUST let

$W=X_1 -2X_2 + X_3$

Then figure out what the mean (0) and the variance of W is.
THEN standardize

$Z={W-E(W)\over \sqrt{V(W)}}$

and use the st normal table

**statmajor** · October 16th 2009, 09:01 AM

Sorry to bother you again (and sound like an idiot), but I still can't figure out what what W is.

It's probably something simple, but just can't get it.

Sorry.

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