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Sharpe ratio
- Thread starter btowntrader54
- Start date
I know if the distribution is highly negatively skewed, then the expected return E(r) will be high......leading to a high Sharpe ratio as you said.Quote from MTE:
Generally, you would use a year of daily data points to calculate it. And as the other poster noted, for it to be a meaningful measure the returns should be approx. normal. A highly skewed distribution can result in a very high Sharpe ratio, which is misleading.
But I don't know why it's misleading? Even Normal distribution could have a high E(r)......so what's wrong with a high E(r) anyway?
But I don't understand what role would that skew play? Every distribution has its own SD afterall.....no matter if it's skewed or not.Quote from MTE:
It doesn't have to be normally distributed. The problem though is that the Sharpe ratio can be misleading if the distribution is negatively skewed, for instance. Since standard deviation doesn't take into account that skew.
And what we only require is to maximize the excess return (i.e. E(r) - rf ) while minimizing the risk (i.e. SD).......this is how we interpret the Sharpe ratio.
So all it matters is the SD, isn't it? Skewness has no business on it......this is what I think. But am I wrong and why?
Quote from raymond008375:
I know if the distribution is highly negatively skewed, then the expected return E(r) will be high......leading to a high Sharpe ratio as you said.
But I don't know why it's misleading? Even Normal distribution could have a high E(r)......so what's wrong with a high E(r) anyway?
It's not the return that is the problem, it's the risk, i.e. standard deviation that doesn't capture this skew. So the Sharpe ratio may be high during "normal times", but when that "negative skew" event happens that high ratio doesn't mean sh*t.
Quote from raymond008375:
But I don't understand what role would that skew play? Every distribution has its own SD afterall.....no matter if it's skewed or not.
And what we only require is to maximize the excess return (i.e. E(r) - rf ) while minimizing the risk (i.e. SD).......this is how we interpret the Sharpe ratio.
So all it matters is the SD, isn't it? Skewness has no business on it......this is what I think. But am I wrong and why?
A classic example is an option premium selling strategy, which in good times can have a very high Sharpe ratio, but when that "black swan" event happens the strategy takes a huge blow, which is a real risk, but it is not reflected in the Sharpe ratio.
So......can I interpret your meaning in this way?«Quote from MTE:
A classic example is an option premium selling strategy, which in good times can have a very high Sharpe ratio, but when that "black swan" event happens the strategy takes a huge blow, which is a real risk, but it is not reflected in the Sharpe ratio.
The risks involved when we perform certain trading strategies don't be able to be represented only by SD. Just like the graph.
Both may have the same SD, but obviously their third moment (skewness) is very different. The left one actually is riskier than the right one.
Am I right?
Attachments
Type wrong....should be "The right one actually is riskier than the left one."
Looks like the best way to measure risk is to have a look at the distribution AND know what strategy is being used above any number like Sharpe,etc...
It's not easy to look at the distribution.......since we have sample estimates only, and the distribution itself will change over time...Quote from fluttrader:
Looks like the best way to measure risk is to have a look at the distribution AND know what strategy is being used above any number like Sharpe,etc...
Well, distribution could really help understanding why risky, though not when.....
Quote from raymond008375:
So......can I interpret your meaning in this way?«
The risks involved when we perform certain trading strategies don't be able to be represented only by SD. Just like the graph.
Both may have the same SD, but obviously their third moment (skewness) is very different. The left one actually is riskier than the right one.
Am I right?
Exactly, a normal distribution can be described by its mean and standard deviation. So if the distribution is not normal then those two are not enough.
By the way, Markowitz portfolio theory assumes normal distribtion, because it states that a portfolio can be described by its mean and standard deviation.