How to prove that covariance equals $ cov[Y'AY, Y'BY]=2 \sigma^{4} tr (AB)$

Question

if $Y_{1},...,Y_{n}$ are independently distributed as $N~(0,\sigma^{2})$ and $A$ and $B$ are any $n \times n$ matrices, how to prove that $ cov[Y'AY, Y'BY]=2 \sigma^{4} tr(AB)$

Answer 1

Hint: to prove $\operatorname{cov}(A_{ij}Y_iY_j,\,B_{kl}Y_kY_l)=2\sigma^4A_{ij}B_{ji}$ for symmetric $A,\,B$, show $\operatorname{cov}(Y_iY_jY_kY_l)=\sigma^4(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})$ with this theorem.

Answer 2

Use the following fact :

Let $\textbf{x} \sim N(0,\sigma^2 \textbf{I})$, for a symmetric Matrix $\textbf{T}$,

$Var(\textbf{x}’\textbf{T}\textbf{x}) = 2\sigma^4trace(\textbf{T}^2)$

Now $Cov(\textbf{x}’\textbf{Ax},\textbf{x}’\textbf{Bx}) = \frac{1}{2}*$ $(Var(\textbf{y}’\textbf{(A+B)}\textbf{y}) - Var(\textbf{y}’\textbf{A}\textbf{y}) - Var(\textbf{y}’\textbf{B}\textbf{y}))$

When $\textbf{A}$ and $\textbf{B}$ are symmetric.