For example, $x$ is a vector which is complex gaussian distribution with mean value $0$ and variance $1$. $A$ and $B$ is a self-definite matrix, then what is the expectation of the Quadratic fraction： $E(x'Ax/(x'Bx))$
I want to get a closed form about the expectation, or the corresponding closed form of the upper bound.
Let $(X_1,Y_1,\ldots,X_n,Y_n)$ be independent real rv with distribution $N(0,1)$ and $Z_k=X_k+iY_k$. If $C=(c_{ij})$ is a complex matrix we denote by $C^*=(\overline{c_{ji}}) $ its complex conjugate.
Let $A$ and $B$ two Hermitian matrices of order $n$ (namely such that $A=A^*$ and $B=B^*$) and assume that $B$ is positive definite. We compute $$I=\mathbb{E}\left(\frac{Z^*AZ}{Z^*BZ}\right)$$ There exists a unitary matrix $U$ (meaning $U^*=U^{-1}$) and a matrix $D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$ such that $B=U^*DU$ with $\lambda_j>0$ for all $j=1,\ldots,n.$ Considering $\tilde{A}=UAU^*$ and $\tilde{Z}=UZ$ we have $Z^*AZ=\tilde{Z}\tilde{A}\tilde{Z}.$ Therefore $I=\mathbb{E}\left(\frac{\tilde{Z}\tilde{A}\tilde{Z}}{\tilde{Z}D\tilde{Z}}\right).$ Since $\tilde{Z}\sim Z$ without loss of generality we will assume $U=I_n.$ We have to compute $$I=\mathbb{E}\left(\frac{\sum_{jk}a_{jk}(X_j-iY_j)(X_k+iY_k)}{\sum_{k=1}^n \lambda_k(X_k^2+Y_k^2)}\right)=\sum_{jk}a_{jk}\mathbb{E}\left(\frac{(X_j-iY_j)(X_k+iY_k)}{\sum_{k=1}^n \lambda_k(X_k^2+Y_k^2)}\right).$$ If $j\neq k$ the above fraction is a random variable $F$ such that $F\sim -F$ with a null expectation. We arrive to $$I=\sum_{j=1}^na_{jj}K_j$$ where $$K_j=\mathbb{E}\left(\frac{X_j^2+Y_j^2}{\sum_{k=1}^n \lambda_k(X_k^2+Y_k^2)}\right).$$ Since $X_k^2+Y_k^2=2V_k$ has a chisquare distribution with degree 2, then $\Pr(V_k>v)=e^{-v}.$ As a consequence $$K_j=\mathbb{E}\left(\frac{ V_j}{\sum_{k=1}^n \lambda_kV_k}\right)=\mathbb{E}\left(V_j\int_0^{\infty}e^{-s\sum_{k=1}^n \lambda_kV_k}ds\right).$$ Now we use the independence of $V_1,\ldots,V_n$ and we observe that $$\mathbb{E}(V_1e^{-\lambda_1s V_1})=\frac{1}{(1+\lambda_1s)^2},\ \mathbb{E}(e^{-\lambda_ks V_k})=\frac{1}{1+\lambda_k s}$$ leading to $$K_1=\int_0^{\infty}\frac{ds}{(1+\lambda_1s)^2\prod_{k=2}^n(1+\lambda_k s)},$$ which is a computable integral.