Show that $\text{Cov}(\mu_X+X,\mu_Y+Y)=\text{Cov}(X,Y)$

Asked May 11 '21 at 12:42

Active May 11 '21 at 12:42

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I have a problem in my textbook:

Show that $\text{Cov}(\mu_X+X,\mu_Y+Y)=\text{Cov}(X,Y)$ for all deterministic $\mu_X,\mu_Y$ of the appropriate size.

My approach:

$$\text{Cov}(\mu_X+X,\mu_Y+Y)=\mathbb{E}\biggl[(\mu_X+X-\mathbb{E}[\mu_X+X])(\mu_Y+Y-\mathbb{E}[\mu_Y+Y])^T\biggr] = \mathbb{E}\biggl[(\mu_X+X-\mathbb{E}[X]- \mu_X)(\mu_Y+Y-\mathbb{E}[Y]-\mu_Y)^T\biggr]=\mathbb{E}\biggl[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])^T\biggr]=\text{Cov}(X,Y)$$

Would this approach be correct?

asked May 11 '21 at 12:42

3

Yes, it is correct. – drhab May 11 '21 at 12:51
@drhab thank you very much for reviewing my approach!:) – May 11 '21 at 13:05

Show that $\text{Cov}(\mu_X+X,\mu_Y+Y)=\text{Cov}(X,Y)$

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