Lets say you get to choose any 2 stocks.

Is it possible that a portfolio can be built from the two stocks (long or short either, any weighting) that has a lower volatility (standard deviation of movements), and simultaneously higher percent return than either of the two underlying securities alone.

I want to say yes, but I think the answer is no. In the event of a correlation of -1 between A and B:

A: a stock with high stddev, and small positive return

B: a stock with same stddev, and small negative return

Long A short B = much higher returns, much higher stddev

Short A Long B = 0 return, 0 stddev

I currently have an infinite loop randomly building portfolios to test this at home. Bonus: Is it possible with a portfolio with more than 2 stocks in the portfolio?


1 Answer 1


$X$ and $Y$ perfectly correlated with same vol. So $X-Y$ has no volatility at all and for any $n$, $n(X-Y)$ has still no volatility.

If $r_X > r_Y$ are returns of $X$ and $Y$, then $n(X-Y)$ has return $n(r_X-r_Y)$ which can be as big as you want with $n$ big enough.

  • $\begingroup$ Something like this can happen accidentally in a large problem if your input data is crappy, and that is one reason why people sometimes put in a long only constraint before running the optimization. $\endgroup$
    – nbbo2
    Aug 31, 2016 at 14:48
  • $\begingroup$ Sorry, I do not fully understand your solution. Are you making assumptions about the securities? n appears to be a multiplier here, which is not relevant to the problem. if it is a weight it should be (n)(x)-(1-n)(y). $\endgroup$ Aug 31, 2016 at 15:34
  • $\begingroup$ Let A and B have same variance, correlation 0.9999. Return of A 10% a year, B 5% a year. Then for example let w1=1001 and w2=−1000. Then w1+w2=1 and you make approx 1000 times the difference in return between the two securities i.e. 50% a year. $\endgroup$
    – nbbo2
    Aug 31, 2016 at 15:43
  • $\begingroup$ Understood, so the return can easily be higher, and as long as (A-B) has lower volatility then either A or B then the second condition is fulfilled, implying that it is possible. Thanks! $\endgroup$ Aug 31, 2016 at 16:10

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