P & Q are two probability measures characterised by their mean and variance. I can transform between them with the Radon-Nikodyn derivative f(Z), say, s.t. f(Z) = dP/dQ. My question is, do P, & Q have to come form the same type of distribution, say both Normal, or can I transform between two different types of distribution - say Normal to Gamma.Thanks,TJ.

What do you mean by "characterised by their mean and variance" ? I assume you're talking about probability distributions over the real line R1, that are absolutely continuous i.e. have a density (like a Gamma distribution, unlike a Poisson distribution). If f is pdf (probability density function), let's call the set of points where f is nonzero the support of f : supp(f). Then if you have two absolutely continuous probability measures P1, P2 with respective pdf f1,f2, you can state : "the two probability measures are equivalent iff supp(f1)=supp(f2) ; in that case the Radon-Nikodym derivatves are given by dP1/dP2=f1/f2 and dP2/dP1=f2/f1".Thus any measure whose pdf is supported on the whole line is equivalent to a normal distrbution, and any measure whose pdf is supported on the whole half-line R+ is equivalent to a Gamma distribution. The normal and Gamma distributions, N and G, aren't equivalent. But you have one side of the relationship : G << N ; this means that dG/dN exists, so you can "transform" from normal to gamma. If N is normal(mu,sigma^2) and G is Gamma(a,lambda) then : dG/dN (x) = sqrt(2*pi) lambda^a 1_(x>0) x^(a-1) exp( (x-mu)^2/2sigma^2 - lambda x) / Gamma(a) ; this function is 0 for x <=0, so we're forgetting about the positive probabilities on the negative axis. On the other hand dN/dG doesn't exist because you cannot recover these positive probabilities once you have "annihilated" them.

Thanks moltabile. Do you have a reference for this result?If N is normal(mu,sigma^2) and G is Gamma(a,lambda) then : dG/dN (x) = sqrt(2*pi) lambda^a 1_(x>0) x^(a-1) exp( (x-mu)^2/2sigma^2 - lambda x) / Gamma(a).Many thanks,TJ.

Glad I could help. The formula I wrote is just the quotient of the Gamma density by the gaussian density, no reference for this particular case but for the general fact that dP1/dP2 = f1/f2 when P1 has pdf f1, P2 has cdf f2, you can find probably find it in your favorite measure-theoretic probability textbook ; for example in Billingsley or Durrett's book. Or you may as well prove it yourself : in order to establish that f1/f2 is this Radon-Nikodym derivative, you have to check the equality int h(x) dP1(x) = int h(x) f1(x)/f2(x) dP2(x) say for any bounded measurable h ; now if you remember that P1 and P2 have pdf's you can express both integrals in terms of Lebesgue measure, and indeed they are equal.

Cool.