Chapter 3 - Stochastic dominance -alternative explanation

Discussion in 'CM2' started by Bill SD, May 15, 2022.

  1. Bill SD

    Bill SD Keen member

    Hi,
    The Chap 3 notes offer alternative explanations for first and second-order stochastic dominance.
    On the first line of pg 6 it explains that "Z first-order stochastically dominates X if Z can be obtained from X by shifting probability from lower to higher outcome levels." The Acted question and answer on that page then aim to provide an illustration of this alternative explanation. But i have 3 basic questions on this alternative explanation:

    Q1: I'm missing something obvious but don't follow how this explanation makes sense with Z and X since both possible outcomes for Z and X have prob of 0.5 (the bottom of pg4) so 'shifting' them should make no difference!

    Q2: Additionally, with 2 assets (A and B) but 3 possible returns (like $1,$2,$3 in the question on pg6) how many of the original probabilities would one trial 'shifting' for both A and B to check if the new expected outcome of A (using the 'shifted' probabilities) equals the original outcome of B (and vice versa)? [The 3 possible shifts for A, B are: shifting the prob from $1 to $2, from $1 to $3 and from $2 to $3?]

    Q3: Finally, is there any intuition behind this alternative explanation or is it intended to be an incomprehensible magic trick?
     
  2. Steve Hales

    Steve Hales ActEd Tutor Staff Member

    Think of "shifting the probability" as sliding the distribution function to the right. So whilst the outcomes are still split 50/50 in terms of probabilities, the returns will have increased.
    For this shortcut to work the patterns need to be the same, ie $1,$2,$3 would need to shift linearly to something like $2.5, $3.5, $4.5.
    A "shift" is like boosting each possible return by the same fixed amount. So in the example above the boost is $1.5 - no magic I'm afraid :)
     
    Bill SD likes this.

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