How to express the distribution of measurements ( X m </msub> &#x223C;<!-- ∼ -->

Oakey1w

Oakey1w

Answered question

2022-06-23

How to express the distribution of measurements ( X m ∼ D u n k n o w n ) given that the underlying physical process being measured follows a Gaussian distribution ( X p ∼ N ( μ p , σ p )) and the measurement uncertainty follows a Gaussian distribution ( N ( μ u , σ u )) too?
Note: μ u here is variable and takes the value of the specific instance/sample X p .
Update:
Can one start in terms of PDFs as follows?
Given the PDF of the physical process:
f X p ( x ) = 1 σ p 2 π e − 1 2 ( x − μ p σ p ) 2
In order to express the PDF of the measurements affected by measurement error ( N ( X p , σ u )), the idea is to use an specific sample t of the physical process as the mean of the measurement uncertainty distribution. Since the specific sample t follows a PDF f X p ( t ), therefore integration is used over the whole range of t, and weighted by the PDF:
f X m ( x ) = ∫ − ∞ ∞ 1 σ u 2 π e − 1 2 ( x − t σ u ) 2 f X p ( t ) d t
Is the above formulation reasonable?
Update after David K's answer: This question can also serve to get a more intuitive understanding of the convolution of two normal distributions.

Answer & Explanation

Dustin Durham

Dustin Durham

Beginner2022-06-24Added 31 answers

In order to better appreciate the mathematics, let's step back and concentrate more on the theory.
Suppose we have a random variable X 1 ∼ N ( μ 1 , σ 1 ) and another random variable X 2 ∼ N ( 0 , σ 2 ). There are actually two universal Gaussian distributions, although the second one has a mean of 0.
The density function of X 1 is
f X 1 ( x ) = 1 σ 1 2 π e − ( ( x − μ 1 ) / σ 1 ) 2 / 2 .
The density function of X 2 is
f X 2 ( x ) = 1 σ 2 2 π e − ( x / σ 2 ) 2 / 2 .
The density function of the sum X 1 + X 2 is the convolution of the two individual density functions, which can be computed in several equivalent ways, one of which is
f X 1 + X 2 ( x )  = ∫ − ∞ ∞ f X 2 ( x − t ) f X 1 ( t )  d t   = ∫ − ∞ ∞ 1 σ 2 2 π e − ( ( x − t ) / σ 2 ) 2 / 2 f X 1 ( t )  d t . 
This equation should be recognizable as the integral you developed for the right-hand side of the last line is f X m ( x ), provided that X 1 = X p and σ 2 = σ u .
In other words, the X m you are trying to describe is simply X p + X u , where X u ∼ N ( 0 , σ u ).
The variable X u is what I would consider the error, which has mean zero. The discrepancy exists between the measurement and the true value.

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