Instrumental and Statistical Uncertainties
P=68% = Probability that a measurement of a Gaussian variant will lie within 1 [math]\sigma[/math] of the mean
Example of cosmic counting experiments. Is the variation statistical?
| Date |
Time (hrs) |
Singles Counts Top (N_1) |
Singles Counts Bottom (N_2) |
Coincidence Counts |
Coinc/Hour |
Coinc/min |
[math]\sqrt{N}[/math]
|
| 11/15 to 11/16 |
21 |
359217 |
383919 |
3581 |
170.52 |
2.84 |
|
Taylor Expansion
A quantity which is calculated using quantities with known uncertainties will have an uncertainty based upon the uncertainty of the quantities used in the calculation.
To determine the uncertainty in a quantity which is a function of other quantities, you can consider the dependence of these quantities in terms of a tayler expansion
Consider a calculation of a Table's Area
[math]A= L \times W[/math]
The mean that the Area (A) is a function of the Length (L) and the Width (W) of the table.
[math]A = f(L,W)[/math]
The Taylor series expansion of a function f(x) about the point a is given as
[math]f(x) = f(a) + \left . f^{\prime}(x)\right |_{x=a} \frac{x}{1!} + \left . f^{\prime \prime}(x)\right |_{x=a} \frac{x^2}{2!} + ...[/math]
- [math]= \left . \sum_{n=0}^{\infty} f^{(n)}(x)\right |_{x=a} \frac{x^n}{n!}[/math]
For small values of x (x << 1) we can expand the function about 0 such that
[math]\sqrt{1+x} = \left . \sqrt{1-0} \frac{1}{2}(1+x)^{-1/2}\right |_{x=0} \frac{x^1}{1!}+ \left . \frac{1}{2}\frac{-1}{2}(1+x)^{-3/2} \right |_{x=0} \frac{x^2}{2!}[/math]
- [math]=1 + \frac{x}{2} - \frac{x^2}{4}[/math]
The talylor expansion of a function with two variables[math] (x_1 , x_2)[/math] about the average of the two variables[math] (\bar {x_1} , \bar{x_2} )[/math] is given by
[math]f(x_1, x_2)=f(\bar {x}_1, \bar{x}_2)+(x_1-\bar {x}_1) \frac{\partial f}{\partial x_1}\bigg |_{(x_1 = \bar {x}_1, x_2 = \bar{x}_2)} +(x_2-\bar{x}_2) \frac{\partial f}{\partial x_2}\bigg |_{(x_1 = \bar {x}_1, x_2 = \bar{x}_2)}[/math]
or
[math]f(x_1, x_2)-f(\bar {x}_1, \bar{x}_2)=(x_1-\bar {x}_1) \frac{\partial f}{\partial x_1}\bigg |_{(x_1 = \bar {x}_1, x_2 = \bar{x}_2)} +(x_2-\bar{x}_2) \frac{\partial f}{\partial x_2}\bigg |_{(x_1 = \bar {x}_1, x_2 = \bar{x}_2)}[/math]
The term
[math]f(x_1, x_2)-f(\bar {x}_1, \bar{x}_2)[/math]
represents a small fluctuation of the function from its average [math]f(\bar {x}_1, \bar{x}_2)[/math] if we ignore higher order terms in the Taylor expansion ( this means the fluctuations are small).
Based on the Definition of Variance
- [math]\sigma^2 = \frac{\sum_{i=1}^{i=N} (x_i - \bar{x})^2}{N}[/math]
We can write the variance of the area
- [math]\sigma^2_A = \frac{\sum_{i=1}^{i=N} (A_i - \bar{A})^2}{N}[/math]
- [math]= \frac{\sum_{i=1}^{i=N} \left [ (L-\bar{L}) \frac{\partial A}{\partial L} \bigg |_{\bar L \bar W} + (W-\bar W) \frac{\partial A}{\partial W} \bigg |_{\bar L \bar WW} \right] ^2}{N}[/math]
- [math]= \frac{\sum_{i=1}^{i=N} \left [ (L-\bar{L}) \frac{\partial A}{\partial L} \bigg |_{\bar L \bar W} \right ] ^2}{N} + \frac{\sum_{i=1}^{i=N} \left [ (W-\bar W) \frac{\partial A}{\partial W} \bigg |_{\bar L \bar W} \right] ^2 }{N}[/math]
- [math]+2 \frac{\sum_{i=1}^{i=N} \left [ (L-\bar{L}) (W-\bar W) \frac{\partial A}{\partial L} \bigg |_{\bar L \bar W} \frac{\partial A}{\partial W} \bigg |_{\bar L \bar W} \right]^2}{N} [/math]
- [math]= \sigma^2_L \left ( \frac{\partial A}{\partial L} \right )^2 +\sigma^2_W \left ( \frac{\partial A}{\partial W} \right )^2 + 2 \sigma^2_{LW} \frac{\partial A}{\partial L} \frac{\partial A}{\partial W} [/math]
where
[math]\sigma^2_{LW} = \frac{\sum_{i=1}^{i=N} \left [ (L-\bar{L}) (W-\bar W) \right ]^2}{N}[/math] is defined as the Covariance between [math]L[/math] and [math]W[/math].
Weighted Mean and variance
If each observable ([math]x_i[/math]) is accompanied by an estimate of the uncertainty in that observable ([math]\delta x_i[/math]) then weighted mean is defined as
- [math]\bar{x} = \frac{ \sum_{i=1}^{i=n} \frac{x_i}{\delta x_i}}{\sum_{i=1}^{i=n} \frac{1}{\delta x_i}}[/math]
The variance of the distribution is defined as
- [math]\bar{x} = \sum_{i=1}^{i=n} \frac{1}{\delta x_i}[/math]
[1] Forest_Error_Analysis_for_the_Physical_Sciences