# Distribution of measurements given the underlying physical process distribution and the measurement uncertainty distribution

How to express the distribution of measurements ($$X_m \sim \mathcal{D}_\mathrm{unknown}$$) given that the underlying physical process being measured follows a Gaussian distribution ($$X_p \sim \mathcal{N}(\mu_p, \sigma_p)$$) and the measurement uncertainty follows a Gaussian distribution ($$\mathcal{N}(\mu_u, \sigma_u)$$) too?

Note: $$\mu_u$$ 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: $$f_{X_p}(x) = \frac{1}{\sigma_p\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu_p}{\sigma_p}\right)^2}$$

In order to express the PDF of the measurements affected by measurement error ($$\mathcal{N}(X_p, \sigma_u)$$), 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: $$f_{X_m}(x) =\int^{\infty}_{-\infty} \frac{1}{\sigma_u\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-t}{\sigma_u}\right)^2}f_{X_p}(t)dt$$

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.

• I thought measurement error was usually considered to have a distribution like $\mathcal{N}(0, \sigma_u)$, that is, it's something added to the actual value in a way that introduces uncertainty but not bias. Is there a reason to formulate it differently? Feb 17, 2020 at 18:22
• The idea here is to figure out a way to express the measurements of the noise-like physical process including the measurement uncertainty. Basically, the question is how the distribution of the noisy measurements of the noise-like physical process look like? Which I think is different than adding the two distributions, do you agree?
– Omer
Feb 18, 2020 at 10:19
• I agree up to the point where you say it is different from adding two distributions. I think it is the same, unless you have some reason to believe that larger actual values of a particular output of the process tend to result in larger (or smaller) errors of measurement of that output. Then you need to say what the dependency is. (It's still a sum of two variables but the joint distribution is more complicated then.) Feb 18, 2020 at 11:44
• I just realized you're writing "measurement uncertainty" where I would have written "measurement error." Uncertainty of measurement is a single parameter, such as a standard deviation: physics.nist.gov/cuu/Uncertainty/glossary.html Feb 18, 2020 at 12:28
• If you have a source for your definitions I think you should cite it in the question. Feb 18, 2020 at 12:28

Let's step back a little and consider the mathematics, not the application.

Suppose we have a random variable $$X_1 \sim \mathcal{N}(\mu_1, \sigma_1)$$ and another random variable $$X_2 \sim \mathcal{N}(0, \sigma_2)$$. That is, we have two general Gaussian distributions, except that the second distribution has mean zero.

The density function of $$X_1$$ is $$f_{X_1}(x) = \frac{1}{\sigma_1\sqrt{2\pi}}e^{-((x-\mu_1)/\sigma_1)^2/2}.$$

The density function of $$X_2$$ is $$f_{X_2}(x) = \frac{1}{\sigma_2\sqrt{2\pi}}e^{-(x/\sigma_2)^2/2}.$$

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

\begin{align} f_{X_1+X_2}(x) &= \int_{-\infty}^\infty f_{X_2}(x - t)f_{X_1}(t)\,dt \\ &= \int_{-\infty}^\infty \frac{1}{\sigma_2\sqrt{2\pi}}e^{-((x - t)/\sigma_2)^2/2} f_{X_1}(t)\,dt. \end{align}

This equation should look familiar, because the right-hand side on the last line is the integral you developed for $$f_{X_m}(x)$$, provided that $$X_1 = X_p$$ and $$\sigma_2 = \sigma_u.$$

In other words, the $$X_m$$ you are trying to describe is simply $$X_p + X_u,$$ where $$X_u \sim \mathcal{N}(0, \sigma_u)$$.

The variable $$X_u$$ is what I would consider the error, which has mean zero. It is the difference between the measurement and the actual value.

• Thanks for your answer. This exercise unintentionally helped me get a more intuitive understanding of the convolution of distributions. :)
– Omer
Feb 19, 2020 at 21:12