I have a problem in my textbook:
Show that $\text{Cov}(X,Y)=(\text{Cov}(Y,X))^T$
My approach:
We have that $\text{Cov}(X,Y) = \mathbb{E}\biggl[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])^T\biggr]=\mathbb{E}\biggl[(Y-\mathbb{E}[Y])(X-\mathbb{E}[X])^T\biggr]^T=(\text{Cov}(Y,X))^T$
Would this be correct?
This problem can be solved using several approaches and your solution is also correct.
Method 1:
Since $AB^T = (BA^T)^T$, take $A = (X - E[X])$ and $B = (Y-E[Y])$ and use the fact $E[X]^T = E[X^T]$
Method 2:
Just expand everything
$Cov(X,Y) = E[(X-\mu_X)(Y-\mu_Y)^T] = E[(X-\mu_X)(Y^T-\mu_Y^T)] = E[XY^T] - \mu_X\mu_Y^T = E[(YX^T)^T] - (\mu_Y\mu_X^T)^T = (E[YX^T] - \mu_Y\mu_X^T)^T = Cov(Y,X)^T$.
Method 3:
Let $K = Cov(X,Y)$ and $K' = Cov(Y,X)$, then $K_{i,j} = E[(X_i - E[X_i])(Y_j - E[Y_j])] = K'_{j,i}$ and therefore, $K^T = K'$