# Is it reasonable to measure standard deviation from true value rather than mean?

I am evaluating the accuracy of GPS watches, taking many readings over a known distance. I've been calculating standard deviation using the mean reading, but because I know what the reading should be, I could use that instead of the mean. Would this be a reasonable thing to do?

• Of what are you computing the standard deviation? The coordinates, or the distance to the known location, or something else? In most cases standard deviations are not relevant for assessing accuracy: they tell you about precision. – whuber May 8 '14 at 21:13
• It depends on what you want to measure. if you want to measure how consistent the GPS readings are, you might use a standard deviation. If you want to measure how close they are to the know answer, you might use the RMSE (from the known value). If you do that, note that your denominator should be n, not n-1. – Glen_b -Reinstate Monica May 8 '14 at 21:25
• A related question is discussed at stats.stackexchange.com/questions/65640. It explains the use of the RMS mentioned by @glen_b. – whuber May 8 '14 at 21:37
• Possible duplicate of How to calculate 2D standard deviation, with 0 mean, bounded by limits – Tim Mar 21 '16 at 12:20

Re:"Would this be a reasonable thing to do?"

As Wuber pointed out, it depends on what you are trying to measure. GPS signals for civilians are purposely degraded to prevent GPS signals being used for missile guidance against US targets. So the "location" of a particular position will vary with time on purpose and due to overall error in GPS.

When measuring the difference between two positions (at two different times) there will be about a SQRT(2) increase over the error of measuring one position.

• Department of Defense in the United States stopped degradation of GPS signals in 2000. – StatsStudent Nov 4 '15 at 10:07
• As MaxW pointed "true value" and "true mean of population" might be different. Anyway, if they are the same (that is, the GPS device has no systematic error or bias), it will be very reasonable to estimate standard deviation using true value, but then you should divide by n instead of n-1. – Pere Jul 31 '16 at 18:49

Maybe I'm oversimplifying this, but it sounds like just a mean squared error. If $\theta$ is the truth, then

$$\text{E}(X_i - \theta)^2 = \text{E}(X_i - \bar{X} + \bar{X} - \theta)^2 \\ = \text{E}(X_i - \bar{X})^2 + \text{E}(\bar{X} - \theta)^2$$

So on the variance scale, what you're computing is the ordinary variance plus an estimate of the bias squared.