Well in general you would apply the delta method to derive the
variance in question:
g(a,b) = a*b
\partial g over \partial a = b, \partial g over \partial b = a,
so the variance (or rather MSE) in question is
E( g(a,b) - g(Ea,Eb) )^2 asymptotically = (Eb, Ea) Cov(a,b) (Eb, Ea)^T
It does not have the term V(a)*V(b) from Jann's and Marcello's demonstration
Var(ab) = [E(a)]^2 * Var(b) + [E(b)]^2 * Var(a) + Var(a)*Var(b)
as that term will be of order 1/n^2 asymptotically, which is an order
of magnitude smaller than Cov(a,b) matrix that is of order 1/n.
--
Stas Kolenikov
http://stas.kolenikov.name
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