linear algebra and variance

Question

Please help me understand the following concept. Let A be a matrix of constants and Y is a random vector. Var(AY) = AVar(Y)A', where A' is A transpose. I understand that AA' = A^2. What I don't understand is why A' goes in the end, instead of AA'Var(Y). Is there some kind of property in linear algebra that would help me understand the correct order? Thank you!

Answer 1

By definition, ${\rm Var}(Y) = \mathbb{E}[YY^\top] - \mathbb{E}[Y]\mathbb{E}[Y]^\top$. Recall that $\mathbb{E}$ is a linear map. Let's compute ${\rm Var}(AY)$ directly: $$\begin{align} {\rm Var}(AY) &= \mathbb{E}[AY(AY)^\top] - \mathbb{E}[AY]\mathbb{E}[AY]^\top \\ &= \mathbb{E}[AYY^\top A^\top] - (A\mathbb{E}[Y])(A\mathbb{E}[Y])^\top \\ &= A\mathbb{E}[YY^\top]A^\top - A\mathbb{E}[Y] \mathbb{E}[Y]^\top A^\top \\ &= A(\mathbb{E}[YY^\top] - \mathbb{E}[Y]\mathbb{E}[Y]^\top)A^\top \\ &= A{\rm Var}(Y)A^\top,\end{align}$$as required. The key property you seem to have missed is that the transpose of a product of matrices is the product of the transposes, in the reverse order: $(AB)^\top = B^\top A^\top$.