Positron decay direction from muon I read that in the rest frame of a positive muon, decay positrons are preferentially emitted in the direction of the muon spin. Why is that the case?
The decay is $\mu^+\to e^+\nu_e\bar{\nu_\mu}$. Assuming that the positron is emitted at almost the speed of light, it will be a right-handed particle. The 2  neutrinos are right-handed and left-handed. So we have a 3-body decay, with the 3 particles having their spin aligned or anti-aligned with their momentum. But I am not sure how can I get from here that the positron is more likely to be along the muon spin.
 A: This is the cleanest application of chirality in the weak interactions, and your instructor should have drilled it in your SM course.
The positron must be right-handed, as an antiparticle coupling to the charged weak current. The neutrino/antineutrino pair are left/right handed, respectively. All products are fast enough that to lowest order
chiralities amount to the dominant helicities.
Go to the antimuon rest frame, so take its spin pointing up. Consider the limiting case where the two neutrinos are collinear, so they are both emitted back to back to the positron emitted at maximum possible momentum (hence energy).  The net spin projection of the neutrino pair is then zero, so the spin projection sum will be the spin of the right-handed positron pointing in its direction of motion, so, then, its momentum will be preferentially in the direction of the antimuon's spin. (This will be a maximum of the corresponding spin 1/2 d-rotation matrix. All such helicity arguments pick a maximum along a spin axis, and rotations off it introduce diminution of the superposition component involved!)

Might consider my old notes of my teaching muon decay in the past. The incomparable booklet of L Okun, Leptons & Quarks, which, needless to say, you should be reading regularly, does all this in Sec. 3.3, Figures 3.2 and 3.3.

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*(Geeky) Discussion of angular distribution.

From my notes or Okun you may glean the full angular distribution

so $(3-2\varepsilon -\cos\theta (1-2\varepsilon ))/2$ for $\varepsilon\equiv 2E_e/m_\mu$ the energy of the positron as a fraction of its maximum. You then see that, near this maximum, $\varepsilon \to 1$, the angular distribution is but $(1+\cos\theta)/2= |d^{1/2}_{1/2,1/2}(\theta)|^2$, "preferring" small angles and "abhorring" π ! 
