  The Coefficient of Variation Dear Students, People who study the stock market use several different methods to assess any given stock's "volatility." One of these is our old friend, the standard deviation! If we were to record the closing price of a stock over several trading days, we could compute the standard deviation of those numbers. The SD would be equal to 0 if the stock's price never changed, it would be equal to a small number if the stock fluctuated just a little, and it would be equal to a large number of the stock's price jumped around wildly over the days we've studied. As proof that the SD is sometimes used as a measure of stock volatility, consider these 2 statements that I've just now pulled off some financially-oriented websites: "Volatility: This describes the fluctuations in the price of a stock or other type of security. If the price of a stock is capable of large swings, the stock has a high volatility." "Volatility may be gauged by several measures, one of which involves calculating a security's standard deviation." Now, it seems to me the coefficient of variation does a better job of assessing volatility than does the standard deviation. (As you may recall, the coefficient of variation is equal to the SD divided by the mean.) To defend my reasoning, let's say there are two investors (A and B) who each have \$1,000 to invest. Assume that Investor A buys 100 shares of a stock that's selling for \$10 a share, while Investor B buys 20 shares of a different stock that's selling for \$50 a share. Over 5 days, suppose the price of A's stock moves like this: Day1=\$10, Day2=\$9, Day3=\$13, Day4=\$7, Day5=\$11. Over this same period, suppose investor B's stock has this kind of fluctuation: Day1=\$50, Day2=\$49, Day3=\$53, Day4=\$47, Day5=\$51. As I hope you noticed, the two stocks under consideration fluctuated the same absolute amount. In each case, the SD of the 5 prices is equal to 2. However, this measure of volatility misrepresents the investors' "ups and downs" over the 5-day period we're considering. Investor A's holdings fluctuated from \$700 to \$1,300 while Investor B's nestegg fluctuated between \$940 and \$1,060. If we calculate the coefficient of variation (CV) instead of the SD, look what happens. For investor A, the CV = 2/10 = .20; for Investor B, the CV = 2/50 = .04. Comparing these CV indices, we see that Investor A's stock is 5 times as volatile as Investor B's stock. And doesn't that conform to the fact that the overall value of Investor A's stock (with a range from \$1,300 to \$700) changes far more than does the total value of Investor B's stock (where the range extends only from \$1,060 to \$940). I hope this little example shows how the coefficient of variation can be useful when trying to assess the degree of "spread" within a set of numbers. As illustrated by the performance of the two hypothetical stocks, the SD disregards the level of the mean when it assess variability. The SD computes how variable the scores are around the mean, but the size of the mean is not taken into consideration. In contrast, the coefficient of variation looks at the spread of scores (around the mean), adjusted for the size of the mean. Sky Huck