Density of the product of two complex gaussian random variables

Matteo Frigo

February 9, 2002

Lemma 1 Let a and b be independent complex normal random variables with mean 0 and variance 2. The probability density function of |ab| is f(c) = cK_{0}(c), where K_{0} is the modified Bessel function of second kind and index 0.

Proof: Let F (c) = Pr[|ab| < c] be the probability distribution function of |ab|, so that f(c) = \frac{d}{dc}F (c). We have:

1 − F (c) = Pr[|ab| ≥ c]
= \frac{1}{(2π)^{2}} ∫_{ℂ} da ∫_{|b|≥c/|a|} e^{−(|a|^{2}+|b|^{2})/2} db
= \frac{1}{2π} ∫_{ℂ} e^{−|a|^{2}/2}da ∫_{0}^{2π} \frac{dθ}{2π} ∫_{c/|a|}^{∞}re^{−r^{2}/2} dr
= \frac{1}{2π} ∫_{ℂ} e^{− \frac{1}{2} (|a|^{2}+ \frac{c^{2}}{|a|^{2}} )} da
= ∫_{0}^{∞} re^{− \frac{c}{2} ( \frac{r^{2}}{c} + \frac{c}{r^{2}} )}dr .

The substitution e^{z} = r^{2}/c yields

1 − F (c) = ∫_{−∞}^{∞} \frac{c}{2}e^{z} e^{−c cosh z} dz

The Bessel function K_{1} obeys the following identity [1, 9.6.24]:

K_{1}(c) = ∫_{0}^{∞} e^{−c cosh z} cosh z dz
= ∫_{−∞}^{∞} \frac{1}{2}e^{z}e^{−c cosh z} dz .

Therefore, 1 − F (c) = cK_{1}(c) holds. From [1, 9.6.28], we have \frac{d}{dc}[−cK_{1}(c)] = cK_{0}(c), and therefore the equality f(c) = \frac{d}{dc}F (c) = \frac{d}{dc} [1 − cK_{1}(c)] = cK_{0}(c) holds.

References

[1] M. Abramowitz and I. A. Stegun. Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables. Dover, New York, 1965.