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Let $X$ be a circular symmetric complex Gaussian random variable with zero mean and unit variance. Define $Y=\frac{1}{A+x}$ for some real-valued constant A.

What is the distribution of $Y$?

When is its mean well defined?

When is the second moment well defined?

I can find almost nothing in the literature about this case

Thanks

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closed as off-topic by Ricardo Andrade, Chris Godsil, Andrey Rekalo, Vladimir Dotsenko, Stefan Kohl Mar 29 '14 at 22:04

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$$Q_X(z) = \frac{e^{-|z|^2}}{\pi} d\lambda(z)$$ $$w(z) = \frac{1}{A+z}$$ $$z(w) = \frac 1w - A$$ $$\frac{dz}{dw} = \frac 1{w^2}$$ $$\frac{d\lambda(z)}{d\lambda(w)} = \frac 1{|w|^4}$$

You can compute the density function of the new distribution by Integration by Substitution/Change of Variables: $$Q_Y(w) = \frac{e^{-|z(w)|^2}}{\pi} \cdot \frac {d\lambda(z)}{d\lambda(w)} \cdot d\lambda(w) = \frac{e^{-\left|\frac 1w - A\right|^2}}{\pi} \cdot \frac 1{|w|^4} \cdot d\lambda(w)$$ $$Q_Y(w) = \frac{e^{-\left|\frac 1w - A\right|^2}}{\pi |w|^4} d\lambda(w)$$

$Y$'s mean is always well-defined, but its 2nd moment is never finite. In fact, its $t$'th absolute moment is finite for $t < 2$ and infinite for $t \ge 2$.

More generally, if $X: \Omega \to \mathbb C$ is a 2D random variable with a density function $\mu_X = \frac {dQ_x}{d\lambda}$ such that $\liminf_{z\to A} \mu_X(z) > 0$ and $\limsup_{z\to A} \mu_X(z) < \infty$ (for example, if $\mu_X$ is continuous and nonzero in $A$), then $Y = \frac{1}{X-A}$ will have finite $t$'th absolute moments for $t < 2$ and infinite for $t \ge 2$.

Let's say that for $|z-A|<\delta$, the density function $\mu_X(z)$ satisfies $0 < \varepsilon < \mu_X(z) < K < \infty$. First, if $t < 2$: $$\mathbb E(|Y|^t) = \int |w|^t dQ_Y(w) = \int {|w|^t \cdot \frac{\mu_X(z(w))}{|w|^4} d\lambda(w)} = \int_{r=0}^\infty {\int_{\theta=0}^{2\pi} {r^t \cdot \frac{\mu_X(z(re^{i\theta}))}{r^4} r d\theta dr} } \le 1\cdot \frac 1{\delta^t} + \int_{r=\frac 1\delta}^\infty {\int_{\theta=0}^{2\pi} {r^t \cdot \frac{K}{r^4} r d\theta dr} }= \frac 1{\delta^t} + 2\pi\cdot K\cdot \int_{r=\frac 1\delta}^\infty {r^{t-3}dr} < \infty ;$$ however, when $t \ge 2$: $$\mathbb E(|Y|^t) = \dots \ge \int_{r=\frac 1\delta}^\infty {\int_{\theta=0}^{2\pi} {r^t \cdot \frac{\varepsilon}{r^4} r d\theta dr} }= 2\pi\cdot\varepsilon\cdot \int_{r=\frac 1\delta}^\infty {r^{t-3} dr} = \infty .$$

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