- #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...

- 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

- 6,577

- 176

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

- 7,570

- 1,465

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].

- Replies
- 3

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 3K

- Replies
- 1

- Views
- 1K

- Replies
- 3

- Views
- 4K

- Last Post

- Replies
- 2

- Views
- 3K

- Replies
- 7

- Views
- 739

- Last Post

- Replies
- 4

- Views
- 21K

- Replies
- 1

- Views
- 3K

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 18

- Views
- 5K