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Why a Payoff Proportional to the Stochastic Discount Factor Is on the Frontier

Article Quant Q&A · Author: PDUNG31

Summary

The document asks why a payoff proportional to the stochastic discount factor belongs to the mean–variance frontier. The answer’s intuition is that the payoff and the stochastic discount factor move perfectly together: scaling one by a positive constant does not change their correlation. Such a payoff therefore has a direct linear relationship with the pricing kernel, which is the core point offered for its frontier status.

The exchange is very brief and supplies no derivation of the mean–variance result, discussion of assumptions, or clarification of how the payoff’s normalization affects the argument. It refers to the payoff as proportional to the discount factor, while its correlation expression uses a normalization by the expected value. Readers seeking a formal proof would need additional asset-pricing context; the answer alone is an intuition rather than a complete demonstration.

Key ideas

  • A positive scaling of a random variable preserves its correlation with that variable.
  • The answer’s rationale is that the proposed payoff is perfectly correlated with the stochastic discount factor.
  • That perfect comovement is offered as the reason the payoff lies on the mean–variance frontier.
  • The short answer does not provide a proof or spell out the required assumptions and normalization.

Tags

Full text
# Question about the mean-variance frontier asset with payoff m/E(m^2)


# Question about the mean-variance frontier asset with payoff m/E(m^2)












In the book Asset Pricing by Cochrane (2005), page 18, point 3, the author says that an asset with pay off m/E(m^2) is on the mean-variance frontier. However he didn't provide any explanation for why that asset is on the mean-variance frontier. Can someone explain why for me? Thank you so much

## Answer by mehman (score 2)

https://quant.stackexchange.com/a/82211

It is because it covaries perfectly with the stochastic discount factor $m$, i.e. $Corr[\frac{m}{E[m]};m]=1$.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.