Uncertainty in the standard deviation

[Shukie]

Homework Statement

A health physicist is testing a new detector and places it near a weak radioactive sample. In five separate 10-second intervals, the detector counts the following numbers of radioactive emissions:

16, 21, 13, 12, 15.

a) Find the mean and standard deviation (SD) of these five numbers.

b) Compare the standard deviation with its expected value, the square root of the average number.

c) Naturally, the two numbers in part b) do not agree exactly, and we would like to have some way to assess their disagreement. This problem is, in fact, one of error propagation. We have measured the number v. The expected standard deviation in this number is just $$\sqrt{v}$$, a simple function of v. Thus, the uncertainty in the standard deviation can be found by error propagation. Show, in this way, that the uncertainty in the SD is 0.5. Do the numbers in part b) agree within this uncertainty?

The Attempt at a Solution

a) Mean = 15.4
Standard deviation = 3.5

b) Expected SD = $$\sqrt{15.4} = 3.9$$

c) I know that the following is true: $${\sigma}(\sqrt{v})= {\sigma}_v*\frac{d(\sqrt{v})}{dv} = 0.5$$ where $${\sigma}_v = \sqrt{v}$$. How do I show by error propagation that the uncertainty in the SD is 0.5?

 

[mfb]

I don't see why this should be 1/2, neither from a scaling argument, nor from a simulation, nor from calculation.

* If we replace the 5 data points by a million then certainly our sample variance will be extremely close to the expected variance, and our uncertainty won't be 1/2. The question doesn't use the N=5 information.
* If we replace the 5 data points by 5 data points with an average of millions then clearly the sample variance will have a much larger variance. The 1/2 doesn't reflect that either.

I produced hundreds of sets of 5 numbers with a mean of 15.4 and a standard deviation of sqrt(15.4). I used a normal distribution because I didn't find a nice way to get Poisson-distributed numbers but that shouldn't matter. The standard deviations had a standard deviation of about 1.3, clearly not 1/2.The standard deviation is $$\sigma_{obs} = \sqrt{\frac{1}{N}\sum_i x_i^2 - \frac{1}{N^2}\left(\sum_i x_i\right)^2}$$
Looking at the first data point only:
$$\frac{d\sigma_{obs}}{dx_1} = \frac{1}{2 \sigma_{obs}} \left( \frac{1}{N} 2 x_1 - \frac{2}{N^2} \sum_i x_i\right) = \frac{1}{N \sigma_{obs}} ( x_i - \bar{x})$$
This intuitively makes sense: If a datapoint is below the average then raising it reduces the standard deviation, otherwise it increases it.
It's uncertainty is its own square root, so $x_1$'s contribution to the overall uncertainty is $\displaystyle \frac{1}{N \sigma_{obs}} | x_1 - \bar{x}| \sqrt{x_1}$. Adding the contributions in quadrature we get
$$(\Delta \sigma_{obs})^2 = \frac{1}{N^2 \sigma_{obs}^2} \sum_i (x_i-\bar{x})^2x_i$$

Not sure how to calculate the expectation value of that now. It seems to have the right scaling, it's larger for larger numbers and smaller for larger N.

 

[haruspex]

We have measured the number v.

Umm.. what is v?

the uncertainty in the SD

I assume this means the uncertainty in the expected SD of the counts given only their mean.