Let $Y$ be a $p$-dimensional random vector $\mathbb{E}|Y|^2\lt \infty$, and $A$ be a $q\times p$ deterministic matrix. Show that
$$\text{Cov}(AY)=A\text{Cov}(Y)A^T$$
My solution:
$$\text{Cov}(AY)\mathbb{E}\biggr[(AY-\mathbb{E}[AY])(AY-\mathbb{E}[AY)^T\biggr]=\mathbb{E}\biggl[(AY-A\mathbb{E}[Y])(AY-A\mathbb{E}[Y])^T\biggl]=\mathbb{E}\biggl[A(Y-\mathbb{E}[Y])(Y-\mathbb{E}[Y])^TA^T\biggr]=A\mathbb{E}\biggl[(Y-\mathbb{E}[Y])(Y-\mathbb{E}[Y])^T\biggr]A^T=A\text{Cov}(Y)A^T$$
Would this solution be correct?
Also, the answer to this question states the following:
Write $(AY)_i=\sum_jA_{ji}Y_j$ and argue from the definition of covariance.
First of all, isn't it supposed to be $(AY)_i=\sum_j(A)_{ij}Y_j$ (indices i and j were not in order)?
I attempted to solve the problem vie their method also:
We have that $(AY)_i=\sum^p_{j=1}(A)_{ij}Y_j = a_i^TY$, where $a_i=(A_{i1},A_{i2},...,A_{ip})^T$ i.e., the rows of $A$.
So, similarly as in my first method:
$$\text{Cov}(a_i^TY)=a_i^T\text{Cov}(Y)a_i$$
But I found that $$(A\text{Cov}A^T)_{ij}=\sum^p_{r=1}a_{ir}\Sigma_{ir}a_{jr}$$
Well now I'm confused now.
I don't know if I overcomplicated it, or I didn't start doing it correctly from the beginning. If you could tell me if my first method is correct and how I went wrong with the second method, I would appreciate that a lot.