I am trying to analyse some data using the coefficient of variation (stdev/mean), but I have two problems:
- the data is not 100% positive and:
- the mean is very close to zero (and as Wikipedia says: "When the mean value is near zero, the coefficient of variation is sensitive to change in the standard deviation, limiting its usefulness")
Therefore the coefficient of variation looses it meaning.
So I have thought of adding a constant to each value of my data so to make all values positive and get the mean further away from zero.
Is this a good approach? Would the coefficient be meaningful then? If so, can someone post a proof for this?
- I need it in order to compare the dispersions of more series of data, not to analyse the dispersion of one particular series (in this case probably it wouldn't be meaningful)
I really need to use this coefficient because I need to compare more sets of data and I can't compare their standard deviations since it is an absolute value.
Thanks,
M
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I am writing a project for my university and I need it to compare the risk to return ratio for a series of stocks and sort them by this ratio (having the daily return ratios for a year). Here is a link about this: http://www.investopedia.co
Does 0 mean the investment breaks even? Of so, shouldn't you treat that as a return ratio of 1?
Or does negative mean you lose more than all of your initial investment?
Even for futures contracts, a mean close to 0 would seem a pretty bad investment.
http://en.wikipedia.org/wi
Hi Ozo,
Well, actually in one my cases the average daily change is 0.00000087 :), but because the standard deviation is much higher (0.01843) and the distribution is skewed after about 500 days the return is 207.92%.
I have uploaded an excel file with the data, and highlighted some of the columns that i've calculated (blue ones - I'm using the "Change" one as return ratio). And the red cell highlights the slight skew in the distribution.
The file is located at: www.b4life.com/edu/TLV_exa
Thanks,
M
Hi Ozo,
I have found some interesting stuff in one of the books from library about using the coefficient of variation, this is exactly why I am trying to use it.
So, usually standard deviation is considered a measure of risk for returns, but it can be misleading when comparing two investments because it's an absolute value and not a relative one.
In their example they show the following situation:
Investment A:
Expected return: 0.07
Standard deviation: 0.05
Investment B:
Expected return:0.12
Standard deviation: 0.07
At first sight, since investment B has a higher standard deviation some people might say that it is more risky. But if we calculate the coefficients of variation for both, we get:
CvA=0.714
CvB=0.583
So, actually B would be a better investment. This is why I want to use it in comparing 10 stocks and to build a chart so that people would be able to see which have a better risk to return ratio and how much this difference is between them.
But, since as I've said in some cases it gets very close to zero, for the difference looks huge between some of them. That's why I thought of adding a constant, so that this chart would look more realistical, but I would need to prove that this constant adding thing is correct in my project.
I am unsure of what you mean by
expected return
mean
average daily change
average daily change ratio
They seem to be used in roughly the same way.
Now for your question
"So I have thought of adding a constant to each value of my data so to make all values positive and get the mean further away from zero.
Is this a good approach? Would the coefficient be meaningful then? If so, can someone post a proof for this?"
I thnk the answer is no. If the constant you are adding is zero you still have the zero denominator difficulty. If the constant you are adding is very large (say 10,000) you swamp out any variation in the mean and you have a statistic without much meaning. The optimum place is obviouly in the middle someplace but there is no means of telling where that optimun place is.
> Is this a good approach?
No... If you add a constant, it will reflect directly on the mean.
Mean(xi + C) = Mean(xi) + C
Instead of computing stdev/mean, you would compute stdev/(mean+C), which would in fact "dilute" the effect of very low means. Anyway, a return of 0 is possible, so you have the wrong approach altogether. You could use instead stdev/performance, with performance = 1+return. That would make some sense, although you should be working in log-space.
The normal calculation in finance is the opposite: return/volatility... and is called the 'Sharpe ratio'. Also, ozo is rightfully questioning your math; serious analysts don't use the stdev but stdev(ln(return+1)) as measure of the volatility.
> ... daily change is 0.00000087 ... after about 500 days the return is 207.92%.
This is not correct. The first figure is the average Close/Open ratio, not the average return. The average return of your data is 0.23% between trading days, or 0.13% daily. The standard deviation of Close/Open ratio isn't very interesting either. But the volatility of 2.01% is.
To be at all readable, you need to 'annualize' both, and you would find the numbers: annualized return and volatility of 62.3% and 30.9% in any good financial report.
Compared to a 'risk free' investment at 5%, that yields a Sharpe ratio of 1.85.
I have uploaded your Excel file back, so you can see the calculations involved:
http://www.ee-stuff.com/Ex
Also read:
http://en.wikipedia.org/wi
http://en.wikipedia.org/wi
http://en.wikipedia.org/wi
http://en.wikipedia.org/wi
Have fun!
(°v°)
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by: ozoPosted on 2007-06-20 at 15:42:36ID: 19329036
why do you want to use the coefficient of variation?