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 $X_p$.


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.

  • $\begingroup$ 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? $\endgroup$
    – David K
    Feb 17, 2020 at 18:22
  • $\begingroup$ 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? $\endgroup$
    – Omer
    Feb 18, 2020 at 10:19
  • $\begingroup$ 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.) $\endgroup$
    – David K
    Feb 18, 2020 at 11:44
  • $\begingroup$ 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 $\endgroup$
    – David K
    Feb 18, 2020 at 12:28
  • $\begingroup$ If you have a source for your definitions I think you should cite it in the question. $\endgroup$
    – David K
    Feb 18, 2020 at 12:28

1 Answer 1


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.

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

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