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I would like to know what does a Standard Deviation Multiplier represent?

For example: What exactly would a standard deviation of one (multiplier of 1) from a simple moving average of 5 represent? Does it represent a SMA 10?
Does a deviation multiplier of two from a SMA of 5 represent a SMA of 15?

I hope I'm expressing myself clearly, if not please ask me to try again.

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This post has been selected as an answer to the original posters question

The standard deviation is a statistical measure for the dispersion of data points around an arithmetic mean. Let us take your example of 5 data points, such as 5 consecutive closes as shown on the chart. I prefer a real example, so let us look at the last five closes of yesterday's regular session for YM 09-13. The close values were

14820, 14812, 14824, 14814, 14830

The arithmetic mean of those data points - identical to a SMA(5) - is obtained by adding up the five values and dividing by 5.

The arithmetic mean of 14820 does not tell us how far the single data points are away from the mean value. For example, if you had the values 13820, 12812, 15824, 13814, 17830, you would obtain the same mean value, but quite obviously the data points are further away from the mean.

The standard deviation is calculated as the square root of the variance. The variance is the sum of the squares of the distances of each data point from the mean. For your example above we calculate

Now this is the calculation method used for the Standard Deviation and Bollinger Band indicators. The method is not really accurate, because it systematically underestimates the expected standard deviation for a small sample size. Using a modified formula for the population variance would result int values of 54.00 for the variance and 7.35 for the standard deviation.

Meaning of 1 Standard Deviation

At least for normal (Gaussian, bell-shaped) distributions, you may expect that 68.2% of all data points can be found within the distance of 1 standard deviation from the mean. About 95.4% of all data points can be found with the distance of 2 standard deviations from the mean.

Applied to our example this means that we may expect about 3.4 out of 5 data point to lie within the interval [14820 - 7.35, 14820 + 7.35] or [14813, 14827]. In fact 3 data points can be found within this interval, while two data points are located outside the interval.

If we take the interval based on two standard deviations [14806, 14834], we would expect 95% of all data points to lie within this interval. For our example, all values can be found within that interval.

Although these estimations only apply to normal distributions, they are used for other distributions as well, not because they are correct, but because these estimations are easy to calculate.

Last edited by Fat Tails; September 4th, 2013 at 05:08 PM.

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@Fat Tails, could you elaborate a bit more especially about the deviation for the small sample size ?
In particular, is this related to the large standard error of the small sample size, which causes the inaccuracy ?
Much appreciated.

If you have a small sample that stands for a larger population, then the variance of the sample is typically smaller than the variance of the population.

For example, if your sample only contains one data point, then the sample variance is zero, because the data point is identical with its mean. However, the variance of the entire population represented by that single data point would not be zero unless the population consists of identical members.

The reason that the population variance is larger than the sample variance is easy to understand:

The sample mean does not represent the exact population mean, but is just an estimator for the population mean. Mathematically the mean can be derived as the value that minimizes the quadratic function of the deviations (the value that produces the "least squares"). Also see section 5 of the attached document. Therefore the sample mean minimizes the sample variance. If the variance is calculated from the data points and the population mean, this will always produce a higher value for the variance, compared to the value obtained, when calculated from the sample mean.

The standard estimator for the population variance is obtained by multiplying the sample variance with N/(N-1), where N is the size of the sample.