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View Full Version : Why are units on Mason's SD formula wrong?

MarkD
03-10-2003, 10:13 PM
If you look at his formula:

- the average (U) is in \$/hr (assuming \$ is your monetary measure and hours is your time measurement)
- N is unitless
- X is in dollars
- T is hours

Therefore, simple unit analysis of the formula shows that variance has units of \$^2 / hrs, and therefore SD would have units of \$ / sqrt(hrs).

We measure our average in terms of \$/hr so logically the SD would have the same units, why is this not so? Has to be a reason.

BruceZ
03-11-2003, 04:51 AM
They aren't wrong. The units of SD are always \$/sqrt(hr). The units of variance are \$^2/hr. Playing n hours increases your variance by a factor of n and increases your standard deviation by a factor of sqrt(n). When you read something like "my sd is 10 bb/hr" that's incorrect, what we really mean is that it is 10 bb for exactly 1 hour. For n hours it is 10*sqrt(n) bb.

MarkD
03-11-2003, 10:52 AM
Ok, but why? Units don't make sense for me at the moment.

If I drive from New York to Los Angeles and my average speed is 60 MPH, I would calculate my SD and find that it had units of MPH as well. All other examples in my stats textbook also show the mean to have the same units as the SD. The poker formula is the only time I've seen it where the units were different than the mean (mean = \$/hr, SD = \$/sqrt(hr)).

BruceZ
03-11-2003, 03:52 PM
This applies not only to poker, but to any situation in which we are computing the standard deviation of a sample average. It comes from the central limit theorem for the sample mean, which is one of the most important theorems in statistics. It states that if we have n samples from any distribution with mean u and variance sigma^2, and we compute the sample mean (sum of samples divided by n), this sample mean will have a mean which is u and a variance which is sigma^2/n. The standard deviation of the sample mean is the square root of the variance or sigma/sqrt(n).

If you divide the distance of your drive by the total number of hours to get average mph, this is a sample average. If you were to just log the miles driven in each hour of your drive and compute the standard deviation of all these numbers, then the standard deviation would have the same units as the numbers, and the same units as the mean, but this not the standard deviation of the sample average. If you wanted to know the standard deviation of your average speed so that could figure out the probability of making the drive in various numbers of hours, then this would be the standard deviation you computed for your data divided by the square root of the number of hours sigma/sqrt(n) and has units of miles/sqrt(hours). If you drive for n hours, the standard deviation of your distance does not increase by a factor of n, but by a factor of sqrt(n).

MarkD
03-11-2003, 04:01 PM
Exactly the answer I was looking for, I'll have to take the time to look at it more closely and then review the central limit theorem.

This question has been bugging me for about 2 years, thanks for getting at the heart of the problem I was having.