# Why is 0 $\nu \beta\beta$ decay often written with electron emission and not positron?

According to http://www.cobra-experiment.org/double_beta_decay/ I can see that double $\beta$+ decay is possible, but I often find neutrinoless double beta sources with the double $\beta$- decay equation. Can someone explain to me why one is more prevalent than the other? Thanks!

Both single beta-decay and double beta-decay may occur with $e^+$ as well as $e^-$.
However, in both cases, the emission of $e^-$ is predicted to appear (and in the single beta case, is also observer to appear) in a larger number of processes essentially because it's energetically easier for neutrons to decay to protons plus electrons; than it is for protons to decay to neutrons and positrons.
It's because the neutrons are heavier than the protons which is why the decay with the neutron(s) in the initial state and proton(s) and electron(s) in the final state are more likely to have enough energy. Note that the proton is stable or very, very long-lived, so the elementary reaction $$p\to n+ e^+ + \nu$$ is impossible by energy conservation. Up to some "chaotic" binding energies, this process is a part of any $\beta^+$ decay, so they're suppressed, too. The electron mass (and the positron mass) is equal to 0.5 MeV which is about 2.6 times smaller than the "neutrino minus proton" mass difference; neutrino masses are negligible. The total rest mass of the initial state has to exceed the total rest mass of the final state.
All these comments are about the "average types of nuclei". Proton-rich nuclei are more likely to decay via $\beta^+$. All these "general" rules-of-thumb are valid both for single beta-decay and double beta-decay.