### 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:

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.

**(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.

## 3 Comments:

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.

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.

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

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