Under certain conditions, such as having a high charge/mass ratio (not enough neutron constituents) the nucleus will cast off some charge, but doesn't have enough energy to eject a whole proton. Effectively, in the field of a large mass, a positron and neutrino are ejected, the atomic number (Z) of the daughter nucleus is reduced by 1 compared to the parent, and the mass number (A) is unchanged. The daughter nucleus becomes more tightly bound (it has less mass energy than before). This reaction does not appear to take place without the presence of other mass.
The proton by itself is mass restricted from decaying into a neutron plus positron ($Q = -1.804$ MeV) or even electron capture ($Q = -782$ keV). But the proton-proton $\to e^+ + \nu$ has a $Q= +420$ keV, so there is enough mass-energy present in the center of mass for the deuteron and positron to form.
The neutrino appears because in addition to conserving energy, momentum, charge, and baryon number, something called lepton number is conserved in the reaction.