Mathematical tools for natural sciences

Suppose we sample n data points \(\small{x_1,x_2,x_3,....,x_n }\) from a population that has Gaussian distribution of mean mean \(\small{\mu}\) and standard deviation \(\small{\sigma }\).

The expressions for \(\small{\overline{x} }\), the most probable estimates of \(\small{\mu}\) and the uncertainity \(\small{\sigma_{\overline{x}} }\) are given in terms of the observed data points by the expressions,

\(~~~~~~~~~~\sigma_\overline{x}^2 ~=~\small{ \dfrac{\sigma }{n} }\)

\(~~~~~~~~~~\sigma_\overline{x} ~=~\small{ \dfrac{\sigma }{\sqrt{n} } }\)

The above expression was derived using * the method of maximum likelihood estimation *. The derivation was skipped here.

In the above result it was assumed that the all the data points come from same distribution and hence have same standard deviation \(\small{\sigma}\). However, there are occasions when we have to find the mean of data points (\small{x_i }\) each with a different uncertainity \(\small{\sigma_i}\).

Again, we can derive the combined mean and uncertainity of data points with different standard deviations using the method of maximum likelihood. **Each data point is weighted with the inverse of its own standard deviation so that a point with smaller uncertainity are given more weight than a point with larger uncertainity. **.

Skipping the derivation, we present the expressions for the weighted mean and its uncertainity in terms of individual values and their uncertainities as,

\(~~~~~~~~~~\sigma_\overline{x}^2 ~=~\small{ \dfrac{1}{\sum\limits_{i=1}^n 1/\sigma_i^2 } }\)

\(~~~~~~~~~~\sigma_\overline{x} ~=~\small{\sqrt{ \dfrac{1}{\sum\limits_{i=1}^n 1/\sigma_i^2 }} }\)

Compute the weighted average of these 4 measurements and the uncertainity on the weighted average.

We assign,

\(\small{x_1 = 11.2,~~x_2=10.1,~~x_3=11.9,~~x_4=12.3 }\)

\(\small{\sigma_1 = 1.9,~~\sigma_2~=~2.4,~~\sigma_3 = 2.1,~~\sigma_4 = 2.5 }\)

Using the above formula for the weighted mean and uncertainity on it, we get,

\(~~~~~~~~~~\small{\overline{x} ~=~ \dfrac{\sum\limits_{i=1}^n x_i/\sigma_i^2}{\sum\limits_{i=1}^n 1/\sigma_i^2} ~~=~~\dfrac{ \left(\dfrac{11.2}{1.9^2} + \dfrac{10.1}{2.4^2} + \dfrac{11.9}{2.1^2} + \dfrac{12.3}{2.5^2} + \right)}{\left(\dfrac{1}{1.9^2} + \dfrac{1}{2.4^2} + \dfrac{1}{2.1^2} + \dfrac{1}{2.5^2} \right) } ~~=~~ 11.4 }\)

\(~~~~~~~~~~\small{\sigma_\overline{x}^2 ~=~\dfrac{1}{\sum\limits_{i=1}^n 1/\sigma_i^2 }~~=~~ \dfrac{1}{\left(\dfrac{1}{1.9^2} + \dfrac{1}{2.4^2} + \dfrac{1}{2.1^2} + \dfrac{1}{2.5^2} \right)} ~~=~~ 1.2 }\)

The weighted mean of the 4 results is \(\small{11.4 \pm 1.2}\)