复高斯向量$\boldsymbol{z}$的$\boldsymbol{z}\boldsymbol{z}^H$倒数的数学期望求解

Alright, let's work through this problem clearly. First, a quick notation correction: $\boldsymbol{z}\boldsymbol{z}^H$ is an $M \times M$ rank-1 matrix, so a scalar reciprocal isn't defined here. I'm confident you're asking for $\mathbb{E}\left{ \frac{1}{\boldsymbol{z}^H \boldsymbol{z}} \right}$ — the expectation of the reciprocal of the squared Euclidean norm of the complex Gaussian vector $\boldsymbol{z}$. Let's dive in.

Key Setup and Observations #

We have $\boldsymbol{z} \sim \mathcal{CN}(\boldsymbol{0}_M, \Theta)$, where $\Theta$ is a non-diagonal, non-identity Hermitian matrix (so it's positive definite, with distinct positive eigenvalues $\lambda_1, \lambda_2, ..., \lambda_M$).

First, use the eigen decomposition of $\Theta$ (since it's Hermitian, this is always possible):
$$\Theta = U \Lambda U^H$$
where $U$ is a unitary matrix, and $\Lambda = \text{diag}(\lambda_1, \lambda_2, ..., \lambda_M)$ is the diagonal matrix of eigenvalues.

Define $\boldsymbol{w} = U^H \boldsymbol{z}$. Since unitary transformations preserve Gaussianity, $\boldsymbol{w} \sim \mathcal{CN}(\boldsymbol{0}_M, \Lambda)$ — meaning the components $w_1, w_2, ..., w_M$ are independent complex Gaussian variables, with $w_k \sim \mathcal{CN}(0, \lambda_k)$.

Crucially, the squared norm is invariant under unitary transformations:
$$\boldsymbol{z}^H \boldsymbol{z} = \boldsymbol{w}^H U^H U \boldsymbol{w} = \boldsymbol{w}^H \boldsymbol{w} = \sum_{k=1}^M |w_k|^2$$
So we just need to compute $\mathbb{E}\left{ \frac{1}{\sum_{k=1}^M |w_k|^2} \right}$.

Case 1: $M = 1$ (Scalar Case) #

If $\boldsymbol{z}$ is a single complex scalar, $z \sim \mathcal{CN}(0, \lambda_1)$. The reciprocal $\frac{1}{|z|^2}$ has no finite expectation: the integral defining the expectation diverges near $|z|=0$, since the probability density behaves like $\frac{1}{\lambda_1} e^{-|z|2/\lambda_1}$, and $\int_0^\infty \frac{1}{x} \cdot \frac{1}{\lambda_1} e^{-x/\lambda_1} dx$ does not converge.

Case 2: $M \geq 2$ (Vector Case) #

For $M \geq 2$, the expectation exists. We use an integral representation for the reciprocal: for any positive scalar $a$,
$$\frac{1}{a} = \int_0^\infty e^{-ta} dt$$
By Fubini's theorem (valid for non-negative functions), we can swap expectation and integration:
$$\mathbb{E}\left{ \frac{1}{\sum_{k=1}^M |w_k|^2} \right} = \int_0^\infty \mathbb{E}\left{ e^{-t \sum_{k=1}^M |w_k|^2} \right} dt$$

Since the $w_k$ are independent, the expectation factors into a product:
$$\mathbb{E}\left{ e^{-t \sum_{k=1}^M |w_k|^2} \right} = \prod_{k=1}^M \mathbb{E}\left{ e^{-t |w_k|^2} \right}$$

For each $w_k \sim \mathcal{CN}(0, \lambda_k)$, $|w_k|^2$ follows an exponential distribution with rate $\frac{1}{\lambda_k}$. The moment-generating function (Laplace transform) is:
$$\mathbb{E}\left{ e^{-t |w_k|^2} \right} = \frac{1}{1 + t \lambda_k}$$

Substituting back, we get:
$$\mathbb{E}\left{ \frac{1}{\boldsymbol{z}^H \boldsymbol{z}} \right} = \int_0^\infty \prod_{k=1}^M \frac{1}{1 + t \lambda_k} dt$$

Closed-Form Solution via Partial Fractions #

To evaluate this integral, use partial fraction decomposition on the product term. For distinct $\lambda_k$, we can write:
$$\prod_{k=1}^M \frac{1}{1 + t \lambda_k} = \sum_{k=1}^M \frac{A_k}{1 + t \lambda_k}$$
where the coefficients $A_k$ are:
$$A_k = \frac{\lambda_k^{M-1}}{\prod_{j \neq k} (\lambda_k - \lambda_j)}$$

When we integrate term-by-term, individual integrals appear divergent, but the divergent logarithmic terms cancel out (a result of the coefficients summing to zero in a way that eliminates the infinite part). After simplification, we get the closed-form expression:
$$\mathbb{E}\left{ \frac{1}{\boldsymbol{z}^H \boldsymbol{z}} \right} = \sum_{k=1}^M \frac{\lambda_k^{M-2} \ln \lambda_k}{\prod_{j \neq k} (\lambda_k - \lambda_j)}$$

An equivalent, more intuitive form is:
$$\mathbb{E}\left{ \frac{1}{\boldsymbol{z}^H \boldsymbol{z}} \right} = \sum_{k=1}^M \frac{\ln \lambda_k}{\lambda_k \prod_{j \neq k} \left( 1 - \frac{\lambda_j}{\lambda_k} \right)}$$

Final Notes #

• If you did intend to ask about the expectation of the matrix inverse $(\boldsymbol{z}\boldsymbol{z}^H){-1}$, note that $\boldsymbol{z}\boldsymbol{z}^H$ is rank-1 (and thus singular) almost surely for $M \geq 2$, so this inverse does not exist with probability 1, and its expectation is undefined.

内容的提问来源于stack exchange，提问作者Felipe Augusto de Figueiredo