Why $\text{E}(AX+b) = A\text{E}(X)+b$ but $\text{Cov}(A,X+b) = \text{Cov}(A,X)$?

Question

Suppose $X$ a random vector and $b$ a constant vector, with $A$ a corresponding constant matrix, then we have the expectation and the covariance of $A$ and $X+b$ satisfying

$$\mathbb{E}(AX+b) = A\mathbb{E}(X)+b$$

$$\text{Cov}(A,X+b) = \text{Cov}(A,X)$$

1. Is the first equation due to the linearity of expectation?
2. Why does the second equation miss $b$?

I see a textbook that claims these properties without proof. Please show me some calculation to derive these two equations.

Answer 1

The reason is that the operator $\mathbb{E}$ is a linear operation, for example in the continuos case

$$ \mathbb{E}[\alpha X + \beta] = \int{\rm d}x~ (\alpha x + \beta)f_X(x) = \alpha\int {\rm d}x~ x f_X(x) + \beta\int {\rm d}x~ f_X(x) = \alpha \mathbb{E}[X] + \beta $$

In order to calculated the covariance you first subtract the effect of the mean, that's the reason why the constant term does not make any contribution

$$ \mathbb{C}{\rm ov}[X, Y] = \mathbb{E}[(X - \mathbb{E}[X])(Y - \mathbb{E}[Y])] $$

So that

\begin{eqnarray} \mathbb{C}{\rm ov}[\alpha X + \beta, Y] &=& \mathbb{E}[(\alpha X + \beta - \mathbb{E}[\alpha X + \beta])(Y - \mathbb{E}[Y])] \\ &=& \mathbb{E}[(\alpha X + \beta - \alpha\mathbb{E}[X] - \beta])(Y - \mathbb{E}[Y])] \\ &=& \mathbb{E}[(\alpha X + \alpha\mathbb{E}[X])(Y - \mathbb{E}[Y])] \\ &=& \mathbb{E}[\alpha (X -\mathbb{E}[X]])(Y - \mathbb{E}[Y])] \\ &=& \alpha \mathbb{E}[(X -\mathbb{E}[X]])(Y - \mathbb{E}[Y])] \\ &=& \alpha \mathbb{C}{\rm ov}[X, Y] \end{eqnarray}

Answer 2

If $Y$ is some random vector then:$$\mathsf{Cov}(Y):=\mathsf E(Y-\mathsf EY)(Y-\mathsf EY)^T$$

Now observe that:

$$(Y+b)-\mathsf E(Y+b)=Y-\mathsf EY$$