Assume as in the picture above, we have two points $A$ and $B$. Each has a location with a normal distribution of error, i.e. $A$ is believed to be at $x=0$ with an error of $σ=1$, and $B$ at $x=35$ with an error of $σ=1$.

Point $A$ now knows that it is only $33$ meters (exactly) away from point $B$.

  1. How would this affect the normal distribution of error of point $A$?
  2. What would also happen to the distribution if the $33$ meters has a margin of error?
  • $\begingroup$ What is your attempt? $\endgroup$
    – user144410
    Commented Feb 22, 2018 at 11:12
  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ Commented Feb 22, 2018 at 11:13
  • $\begingroup$ I am a bit naive in statistics i tried to come up with new distribution with a mean = previous mean(0 ) + the difference(35-33). I failed to understand the way I should combine the two distributions in order to come up with the new one. I believe that the distribution should be skewed and not normal $\endgroup$ Commented Feb 22, 2018 at 11:30

1 Answer 1


we can represent the joint probability distribution as in this picture above. Having a relation d=x-y can be represented on the joint distribution by straight line with the more probable position presented as the projection of the center on the line.as represented here The error of the distance can be represented by making the line thick.


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