- #1

- 6

- 0

Cov(x,y)=[X-E(x))(Y-E(y)]

and this would equal 0 under the normal OLS assumption.

I know how to calculate the covariance, but X here is a matrix. So I don't understand the logic of this formula...

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Rabolisk
- Start date

- #1

- 6

- 0

Cov(x,y)=[X-E(x))(Y-E(y)]

and this would equal 0 under the normal OLS assumption.

I know how to calculate the covariance, but X here is a matrix. So I don't understand the logic of this formula...

- #2

I like Serena

Homework Helper

- 6,579

- 178

Did you already look at the wiki page?

http://en.wikipedia.org/wiki/Covariance

Your formula does not seem to be quite right.

X and Y could be random variables, but in that case an extra E should be present in your formula (see wiki).

If your X and Y are matrices then your formula should have an extra 1/N in it (see wiki).

- #3

Stephen Tashi

Science Advisor

- 7,739

- 1,525

For vectors X and Y, we have

[itex] | X + Y |^2 = |X|^2 + |Y|^2 - 2|X||Y| \cos \theta [/itex] where [itex] \theta [/itex] is the angle between the vectors.

For random variables X and Y we have

variance(X+Y) = variance(X) + variance(Y) + 2 Covariance(X,Y)

The analogy is even better if we express it in standard deviations:

[itex] ( std. dev(X+Y))^2 = (std. dev(X))^2 + (std. dev(Y))^2 + 2 Covariance(X,Y) [/itex]

The covariance is roughly analagous to the cosine term. The quantity that would be analagous to [itex] cos(\theta) [/itex] is [itex] \frac{- Covariance(X,Y) }{(std. dev(X)) (std. dev(Y))} [/itex].

Share: