- #1
steviekm3
- 10
- 0
Suppose we have a normal distribution and a sample of n values from the normal distribution.
To estimate the variance we can use the standard sample variance formula ( average squared distance from the mean divided by either n ( biased estimator ) or n-1 ( unbiased estimator ) ).
There is another property about the normal distribution that possibly can be used to estimate variance and that is the property that the mean absolute deviation from the mean =
sqrt(2/pi) * std deviation
What I was wondering is that is it possible to calculate the sample mean absolute deviation from the sample mean and then divide this by sqrt(2/pi) to get an estimate for the standard deviation ? If so how does it compare with the regular formulas for estimating std deviation ?
To estimate the variance we can use the standard sample variance formula ( average squared distance from the mean divided by either n ( biased estimator ) or n-1 ( unbiased estimator ) ).
There is another property about the normal distribution that possibly can be used to estimate variance and that is the property that the mean absolute deviation from the mean =
sqrt(2/pi) * std deviation
What I was wondering is that is it possible to calculate the sample mean absolute deviation from the sample mean and then divide this by sqrt(2/pi) to get an estimate for the standard deviation ? If so how does it compare with the regular formulas for estimating std deviation ?