# Chapter 3 - Stochastic dominance -alternative explanation

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

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1. ### Bill SDKeen 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 HalesActEd TutorStaff 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

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