# Expectation of double quadratic form

I want to compute the following expectation $$E(\hat{Y_k}'A\hat{Y_l}\hat{Y_k}'A\hat{Y_l})$$ where $$A$$ is a symmetric non-random matrix and $$E(\hat{Y_k}) = Y_k$$, $$E(\hat{Y_l}) = Y_l$$. Additionally, $$\hat{Y_k}$$ and $$\hat{Y_l}$$ are independent.

I tried to get an answer by myself by using the trace-trick or $$E(.)= E(E(.|.))$$. I also took a look into Expectation of Quadratic Form with Two Random Vectors which seems to go into the right direction but the arrangement of the random variables is slightly different and causes problems for me. Addinitonally, the random variables have expectation of zero which makes it a bit easier than in my case.

Does anyone have an idea? Thanks alot in advance!

• What is the dimensionalities of $\hat{Y}_k$ and $\hat{Y}_l$? Also, do you have an assumption on the variances of them? Sep 6 at 11:27
• The matrix $A$ is a dxd matrix and $\hat{Y_k}$ and $\hat{Y_l}$ are dx1 vectors. I don't have any special assumptions about the variance so far. Sep 6 at 12:19

Although the solution is essentially and already contained in W. Huber's answer, here is a detailed derivation with a non-zero mean: \begin{align} \mathbb E(\overbrace{\hat{Y_k}'A\hat{Y_l}}^{\text{real}}\times\overbrace{\hat{Y_k}'A\hat{Y_l}}^{\text{real}}) &= \mathbb E(\hat{Y_k}'A\hat{Y_l}\hat{Y_l}'A'\hat{Y_k})\\ &= \mathbb E(\hat{Y_k}'A\hat{Y_l}\hat{Y_l}'A'\hat{Y_k})\\ &= \mathbb E[\hat{Y_k}'A\mathbb E(\hat{Y_l}\hat{Y_l}'|\hat{Y_k})A'\hat{Y_k})]\\ &= \mathbb E[\hat{Y_k}'\underbrace{A\{\text{Cov}(Y_l)+\mathbb E[\hat{Y_l}]\mathbb E[\hat{Y_l}]'\}A'}_{\Large \Sigma}\hat{Y_k})]\\ &=\mathbb E[\text{tr}(\hat{Y_k}'\Sigma\hat{Y_k})]\\ &=\mathbb E[\text{tr}(\hat{Y_k}\hat{Y_k}'\Sigma)]\\ &=\text{tr}(\mathbb E[\hat{Y_k}\hat{Y_k}']\Sigma)\\ &=\text{tr}(\{\text{Cov}(Y_k)+\mathbb E[\hat{Y_k}]\mathbb E[\hat{Y_k}]'\}\Sigma) \end{align}