Choosing Risk Measures for Discrete Two-Outcome Returns
Summary
The document examines whether standard deviation remains a useful risk proxy when returns take only two values, such as zero or a positive payoff. Its central recommendation is to begin with the decision the measure is meant to support. If the practical question is the chance of exceeding a threshold, calculate that probability directly from the discrete distribution rather than interpreting a Sharpe ratio through a Gaussian assumption. For the example, the probability is determined by whether the threshold lies below, above, or between the two possible outcomes.
A second answer shows why standard deviation can be misleading when larger dispersion comes only from a larger favorable payoff: comparing two binary lotteries, the higher-payoff case has both a greater expected return and a greater standard deviation, without necessarily representing more downside risk. Value at Risk is suggested as another measure. The discussion is illustrative rather than a general ranking of risk measures; VaR has its own limitations, and the appropriate measure depends on the loss or decision criterion. Treating the distribution as normal with matching mean and variance can obscure the actual probability structure.
Key ideas
- Choose a risk measure based on the decision or loss criterion it is meant to represent.
- For discrete outcomes, threshold probabilities can be computed directly from the outcome probabilities.
- Standard deviation may rise with favorable payoff size without indicating greater downside risk.
- A Gaussian interpretation of a Sharpe ratio does not automatically apply to binary returns.
- Value at Risk is an alternative to consider, but the document does not establish it as universally superior.
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# Proxy for risk in portfolio theory when return can take only two values
# Proxy for risk in portfolio theory when return can take only two values
I'm trying to adapt tools from portfolio theory for another use, and I have a question about how I might do so.
Suppose that instead of having normally distributed returns, the return $R_i$ is either, say, 3 or 0. So, $E(R_i)=3P(R_i=3)$.
On the wiki page for Modern Portfolio, it says
> Note that the theory uses standard deviation of return as a proxy for risk, which is valid if asset returns are jointly normally distributed or otherwise elliptically distributed.
I'm mainly interested in computing expected returns and risk a la wiki. I'd like to use those as metrics in comparing a relatively small number of possible portfolios. But obviously in the case I'm interested in, the returns aren't normally distributed. What are the consequences of still using standard deviation of returns as a "proxy for risk"? Is there an alternative measure that makes more sense?
How bad would it be to pretend that the returns are normally distributed, centered at the mean of the bernoulli return, with the same variance?
The stakes aren't particularly high and currently the only tool for what I'm trying to do is human judgment and experience (the application isn't in finance).
## Answer by lehalle (score 3)
https://quant.stackexchange.com/a/3708
First you need to define what you need a risk measure for. It is usually to take a decision, so you have an operational criterion that defines your risk. You should go back at this point and see what is the impact of a change of distribution on it.
Just say for instance that you need a risk measure to take decisions according to a Sharpe ratio and define it like:
$${\cal S}(R) = \frac{R-R_0}{\sigma(R)} $$
In such a case the Sharpe ratio is useful because it can be read as a straightforward proxy of the probability that your returns are greater than $R_0$ assuming that $R$ follows a Gaussian process (here we are), because if you define $\Phi$ the repartition function of a Gaussian (i.e. $\mathbb{P}(G>g)=\Phi(g)$) you have:
$$\mathbb{P}(R>R_0)=\Phi( {\cal S}(R)) = \Phi\left( \frac{R-R_0}{\sigma(R)} \right)$$
So now "what if $\tilde R$ is no more Gaussian?". For the same operational criterion (i.e. probability to be greater than a base $R_0$), you can find the answer very easily:
- if $R_0$ is greater than 3: $$\mathbb{P}({\tilde R}>R_0)=0$$
- if $R_0$ is lower than 0: $$\mathbb{P}({\tilde R}>R_0)=1$$
- else $$\mathbb{P}({\tilde R}>R_0)=1-\mathbb{P}({\tilde R}=3)$$
So you have the answer (it can be reproduced for any other distribution that your toy example):
- go back to your operational criterion
- write its meaning for your distribution
- you have the answer
## Answer by Akavall (score 0)
https://quant.stackexchange.com/a/3704
Here is an example where using standard deviation as a measure of risk does not make sense:
Let's say that Case 1: Probability of getting 3 is 0.5, and probability of getting 0 is 0.5. Your expected return is `1.5`, and you expected standard deviation is `1.5` (I hope my calculation is right).
Case 2: Probability of getting 5 is 0.5, and probability of getting 0 is 0.5. Your expected return is `2.5`, and you expected standard deviation is `2.5` (I hope my calculation is right).
But does Case 2 really have higher risk? If so how? Would anyone ever prefer option in Case 1 because it is less riskier? In the above case increasing standard deviation when going from Case 1 to Case 2 has only positive effect and no negative effect, and option in Case 2 is clearly a better choice. Something like this will not happen in we have normal distribution.
Value at Risk is an alternative risk measure you could look at.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.