Total Drek

Or, the thoughts of several frustrated intellectuals on Sociology, Gaming, Science, Politics, Science Fiction, Religion, and whatever the hell else strikes their fancy. There is absolutely no reason why you should read this blog. None. Seriously. Go hit your back button. It's up in the upper left-hand corner of your browser... it says "Back." Don't say we didn't warn you.

Thursday, January 20, 2005

Somebody check our work.

For reasons that aren't important, my officemate and I spent some time yesterday trying to construct a proof demonstrating that var(kx) = k^2var(x) where k equals an integer constant, and x is a vector. Here's what we've got:

(1) var(kx)= E(kx-E(kx))^2 =

(2) E(kx-kE(x))^2 =

(3) E(k(x-E(x)))^2 =

(4) k^2E(x-E(x))^2 =

(5) k^2var(x)

In the above, of course, "E" refers to the expectation of x, or the mean. So? Show of hands, how did we do?

And, on a more timely note, when the inaugural finally overwhelms your will to live, check out this website. It makes the hurting stop. Then start again. So, hey, at least the pain is fresh and novel.


Blogger Tom Bozzo said...

As long as you aren't requiring yourself to prove that the mathematical expectation is a linear operator (which is actually trivial), you have it exactly right. Cheers.

Thursday, January 20, 2005 1:23:00 PM  
Blogger Tom Bozzo said...

Check that. I didn't see "X is a vector" so "var" is really the variance-covariance (VC) matrix.

Write VC(x) = E[(x-E(x))(x-E(x))'] where the prime is a vector transposition.

The rest of the proof still works out as you had it before.

Thursday, January 20, 2005 1:32:00 PM  
Blogger Drek said...

Thanks for the feedback, Tom. I'll tell you- this blogging thing is handy sometimes.

Friday, January 21, 2005 8:35:00 AM  

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