1
$\begingroup$

I am wondering what the relationship is between skewness, kurtosis and mean variance efficiency is.

Is it correct that particular investors are willing to give up mean variance efficiency in return for greater probability of positive returns. Does this mean that a investor trades a lower average return for bearing more exposure towards positively skewed return distributions and a higher kurtosis? Similar to a lottery type pay-off function? Or do they accept a lower risk adjusted return instead of average return? Holding other factors constant such as risk aversion.

$\endgroup$

2 Answers 2

1
$\begingroup$

Mean-variance efficiency generally only considers the first two moments. Skew and kurtosis are outside of it, and Sharpe is commonly criticized for not accounting for skew and kurtosis (leading to metrics like Omega as a replacement).

Regarding your specific question, it depends entirely on what risks a given investor/trader wants to take. One person may prefer a return stream with smaller average returns but a small chance of a large payoff as a result of skew and/or kurtosis. Another may prefer small steady profits with a small risk of large losses (eg, selling deep OTM puts). All depends on which risks you're willing to take on.

$\endgroup$
6
  • $\begingroup$ Thanks for your answer, but the question was whether one leads to other. Not whether the sharpe ratio measures skewness or kurtosis, and again holding other factors constant. A higher sharpe ratio does imply a greater mean-variance efficiency am I correct? $\endgroup$
    – incognito
    Oct 9, 2019 at 16:34
  • 1
    $\begingroup$ Well, your headline question was different than the question posed in the first line. Neither Sharpe nor MV says anything about skew or kurtosis. A MV-efficient portfolio is equivalent to one constructed to maximize Sharpe ratio (assuming a single portfolio gives a maximal Sharpe ratio). They're two sides of the same coin. $\endgroup$
    – Chris
    Oct 9, 2019 at 17:30
  • $\begingroup$ Thank you. My bad indeed. My question stems from the fact that I have optimised two portfolios, the first yielding highly positive returns on average while being characterized by high positive skewness & kurtosis relative to another portfolio earning lower average returns, have lower skewness & kurtosis but higher Sharpe ratios (mean variance efficiency). I would think that the sharpe ratio would provide biased results. Due to the non-normality of the variance, am I correct? $\endgroup$
    – incognito
    Oct 9, 2019 at 17:45
  • $\begingroup$ That's one of the drawbacks of using Sharpe ratio as a, or the, decision criterion, as it doesn't consider the third or fourth moments. You could theoretically get two portfolios with the same Sharpe, but with different skew and excess kurtosis resulting in fundamentally different risk profiles. People get around this by considering other performance metrics (I've used something like an omega ratio for this purpose) and also by considering several metrics when making investment decisions. $\endgroup$
    – Chris
    Oct 9, 2019 at 19:37
  • $\begingroup$ Even something like Sharpe can be calculated over different return intervals (eg, daily v monthly) giving different results. There's not really a silver bullet or 'right' way to make decisions like these, ideally you just get as much information as possible to pull the trigger. Please consider accepting/upvoting my answer if it answers your question. $\endgroup$
    – Chris
    Oct 9, 2019 at 19:38
0
$\begingroup$

Preferences vary by person.

Some people have loss aversion. Their satisfaction from gaining \$1 is significantly smaller than their pain from losing \$1. People with this condition may give up potential gains to avoid small potential losses.

$\endgroup$
1
  • $\begingroup$ I understand the assymetric loss aversion of investors, however given that an investor is willing to accept more skewness will this result in a lower average return? or a lower risk adjusted return (Sharpe) $\endgroup$
    – incognito
    Oct 8, 2019 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.