expectation of quadratic form with random matrix

Question

Suppose $y \sim N(\mu,\sigma^2I)$ be normally distributed random vector. Now $z = y^TAy$ is a quadratic form. Is there any method to calculate the expectation of $z$ when $A$ is projection matrix, but function of $y$, i.e., $A= f(y)$?

Answer 1

I'm not sure that you should still consider $y^T A y$ to be a quadratic form if $A = f(y)$. Consider a scalar case: if $f(y) = \frac{g(y)}{y^2}$ then $yf(y)y = g(y)$ need not behave like a quadratic (and there are plenty of other even simpler ways to ruin its quadratic-ness, it seems).

Semantics aside, we can still find the expectation through an arbitrary transformation using the Law of the Unconscious Statistician,

$$ \text{E}(g(y)) = \int_{\Omega} g(y) \rho_y(y) dy $$

where $z = g(y)$ and $\rho_y(y) = \mathcal{N}(\mu, \sigma^2I)$ is the probability density function of $y$, and $\Omega $ is the sample space (likely $\mathbb{R}^n$).

I am not sure if there is a more specific simplification that can be provided for your problem, but there might be considering I am ignoring the restriction that $A(y)$ is a projection matrix. Anyway, hope this helps!