Energy conservation if photon absorbed below resonance

Suppose I have some quantum system (like atom) with excitation energy $E_{exc}$ which is homogeneously broadened due to finite lifetime.

I shine light with narrow spectrum centred around energy $\langle E_{abs} \rangle < E_{exc}$. Due to broadening the system can absorb such photons.

Assume that the excited state can decay to ground state just by emission of an other photon. The average energy of emitted photons would be $\langle E_{em} \rangle = E_{exc}$.

If $N$ absorbed photon of average energy $\langle E_{abs} \rangle < E_{exc}$ produce $N$ photons with average energy $E_{exc}$ this is against energy conservation law.

So, What is the catch?

I think this have something to do with statistics (?) Maybe I need (in average) destroy $> N$ photons of average energy $\langle E_{abs} \rangle < E_{exc}$ to create $N$ excited states of resonance frequency $E_{exc}$ ?

• I don't think it's an issue of statistics; that would imply that in individual events you could observe violations of energy conservation, which would be Nobel-winning work. Rather, it probably has to do with the fact that other systems are getting "coupled" to this system and inducing this broadening anyway. So for example in a quantum dot you'll get Lorentzian broadening due to coupling to phonon baths; when you send in an electron whose energy is too low, it's absorbing a random vibration in order to get the required energy. – CR Drost Feb 20 '15 at 17:53
• And even for a single isolated atom there is broadening due to coupling to the electromagnetic field itself. Nonetheless, energy is conserved. – Rococo Mar 11 '16 at 23:48