# Mean vs. variance - which is dominant?

I am currently trying to gain some basic understanding of the mean-variance tradeoff. However, since I do not have an economic education background, I am struggling with some issues. Currently I am wondering which of the two is the key driver for decision-making. Let's say I have option A with a low variance and a low expected value, and option B with a moderate variance and a moderate expected value - which of the two would people choose? Are we more attracted by increasing profit or by relatively lower risks? Or is there no general principle and it depends on individual preferences?

I appreciate any input on that! Thank you in advance.

It depends on both preferences and circumstances, the latter being especially important if there is some risk of a loss.

Suppose Option A has normally distributed outcomes with expected value \$10 and standard deviation (square root of variance) \$2, while Option B again with normally distributed outcomes has expected value $100 and standard deviation also \$100. In that case the risk of loss under Option A is extremely small, while that under Option B (> 1 standard deviation from mean, 1-tail) is about 16%.

Now consider three individuals: 1 and 2 each start with \$1,000, 3 starts with \$100. Individual 1, we assume, accepts Option B despite the risk of loss, but individuals 2 and 3 both prefer Option A. In that case the difference between 1 and 2, whose circumstances are the same, can only be attributed to their preferences: 2 is more risk-averse. But the difference between 1 and 3, though it could reflect different preferences, is more plausibly explained by their different circumstances: starting with only \$100, 3 judges that he cannot afford a significant risk of a loss which he might be quite content to accept if he had started with \$1,000.

There is no general principle. It depends on preferences. Especially the degree of risk aversion, i.e. the preference toward risk.

As you mention, there is a trade-off. If there would be a dominant factor, as your title asks, then there would be no trade-off, as the best decision would always be clear.

• Hi: mathmematically, the tradeoff us viewed as $\mu - \lambda \times \sigma^2$ so it critically depends on $\lambda$ whiich as the other commenter mentioned, is denoted as the risk aversion parameter. – mark leeds Sep 11 '18 at 22:12
• @Lafayote: Just a heads up that any decent finance or investments text should have a reasonable discussion of this material. Sharpe's "Investments" text come to mind. Also, Modern Portfolio Theory" by Andrew Rudd. It's been a while so there may be better ones but those would suffice. All the best. – mark leeds Sep 13 '18 at 20:20

Both are important. Mean will tell you the historical outcome while variance will tell you the historical risk. If I had to only choose one, I would make an argument for variance. Here is an example:

Population              Results
A   B   C   D   Mean    Variance    SD
1   3   5   5   3.50    2.75       1.66
2   3   4   5   3.50    1.25       1.12
10  0   0   4   3.50    16.75      4.09


The mean tells you 3.5 is the expected result while the variance tells you the risk. we can see that the second set has less risk while the 3rd set has the most. Taking the square root of the variance, which is the standard deviation (SD) gives the volatility of the choice. I would choose line 2 because it has a mean of 3.5 but the less amount of risk associated with it.