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Miller, Chapter 2 video: Probabilities


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Hi @David Harper CFA FRM

I was watching the chapter two video this morning while getting ready for work. I came across this and had a question for you.

Screen Shot 2016-08-30 at 7.01.37 AM.png

What is the probability of just an "up" situation?

Is it (0.7) x (0.8) + (0.3) x (0.3) ? = 0.65?


Well-Known Member
This is a good question - read the assigned reading on binomial trees or the notes and it should provide an answer.
Hint: There are two probabilities of an up move, the risk-neutral probability and the actual "true" probability.

David Harper CFA FRM

David Harper CFA FRM
Staff member
Hi @PortoMarco79 Yes, yours is correct for the unconditional probability of an up move. I think these diagrams are a useful in helping to sort out the terminologies:
  • The conditional probabilities include the given P(U|G) = 80% and P(U|S) = 30%
  • The joint probabilities include P(U,G) or P(UG) = P(G)*P(U|G) = 70% * 80% = 56% and P(U,S) or P(US) = P(S)*P(U|S) = 30% * 30% = 9%
  • Just as you wrote, the unconditional P(U) = P(UG) + P(US) = 70% * 80% + 30% * 30% = 65%; because this includes (adds) all of the joint probabilities that include the up (U) outcome
  • Bayes tells us P(G|U) = P(UG)/P(U) which is just a rearrangement of joint P(UG) = P(U)*P(G|U), such that P(G|U) = P(UG)/P(U) = 56%/65% = 86.2%; i.e., in Bayesian language we could say that our prior probability that the economy will grow was 70%, per given unconditional P(G) = 70%, but we received new "evidence" in the form of an up stock move, U, such that our posterior (revised) probability of growing economy is P(G|U) = 86%. I hope that's helpful!
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