I want to study the correlation between 2 parameters, this is done by fitting a straight line. I have uncertainties on both parameters.

I want to solve my problem using the Bayesian approach, i.e. I construct my likelihood, assume some prior information on the slope and $y$-intercept, and then I will get my posterior distribution. Then I would sample from my posterior distribution using MCMC.

The first step would be to construct my likelihood:
Assume the data come from a line of equation $y = mx + b$. Then the likelihood would be: $$P(D|\theta, I) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp(-\frac{(y - mx - b)^2}{2\sigma^2}) $$

However, in my case the uncertainties are asymmetric, how can I construct my likelihood? For example, my data look something like this: $x^{\sigma +}_{\sigma -}$ and $y^{\sigma +}_{\sigma -}$

One way to solve this out is to calculate the average of the lower and upper uncertainties, so that: $$ \sigma = \frac{\sigma^+ + \sigma^-}{2} $$ Now I have symmetric uncertainties, and then I will be able to construct my likelihood.

Is there another method to solve this out?

  • $\begingroup$ You need to choose a distribution that models your measurement errors. Gaussian obviously can't capture the asymmetry. Can you put your measurement error data plotted so that we can recommend a likelihood? $\endgroup$
    – Memming
    Commented Apr 12, 2015 at 14:59
  • $\begingroup$ @Memming, my data do NOT come from repetition of measurements. My data resembles properties of exoplanets. So, for example let's assume I have 10 different planets, each data point refers to the mass of the exoplanets. Example: mass of planet 1 = $3^{+1}_{-0.25}$. I am assuming that the errors are drawn from a normal distribution with mean 0 and variance $\sigma^+$ and $\sigma^-$. $\endgroup$
    – aloha
    Commented Apr 12, 2015 at 15:07
  • $\begingroup$ I see. You could cut two Gaussians in half, and stitch together to make a frankenstein distribution in that case. $\endgroup$
    – Memming
    Commented Apr 12, 2015 at 15:20
  • $\begingroup$ @Memming Can you give me please more details or references, this is the first time I come across cutting 2 gaussians in half! Thanks a lot :) $\endgroup$
    – aloha
    Commented Apr 12, 2015 at 15:23

1 Answer 1


You could cut two Gaussians in half, and stitch together to make a split normal distribution of your noise $\epsilon$: $$ P(\epsilon) = N(\epsilon; 0, \sigma^+)I(\epsilon > 0) + N(\epsilon; 0, \sigma^-)I(\epsilon < 0) $$ where $I$ is the indicator function. Below is an example probability density function.

enter image description here

As you can see, positive noise is distributed as a guassian with a larger standard deviation than the negative noise.

  • $\begingroup$ what are your thoughts? $\endgroup$
    – Memming
    Commented Apr 13, 2015 at 13:11
  • $\begingroup$ weird enough, even google does not know what a frankenstein distribution is! I am not sure if I should ask you for more details or if I should figure this on my own. $\endgroup$
    – aloha
    Commented Apr 13, 2015 at 21:38
  • $\begingroup$ Can you please provide me with references to your suggested answer? $\endgroup$
    – aloha
    Commented Apr 14, 2015 at 13:33
  • 1
    $\begingroup$ There's no reference, because I came up with it. It's a logical answer to your specific problem. $\endgroup$
    – Memming
    Commented Apr 14, 2015 at 14:29
  • 1
    $\begingroup$ This distribution is known as the split-normal or two-piece normal distribution, see en.wikipedia.org/wiki/Split_normal_distribution. If the name Frankenstein distribution was made up for this answer, this should perhaps be edited to use the 'real' name. $\endgroup$ Commented Jul 5, 2015 at 11:23

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