A Gaussian distribution can be uniquely defined by mean and variance (or std deviation). This means, I can use the equation of Gaussian to plug in different values of input and use the mean, std deviation values to determine the output of the function. If I change any of the two parameters, my bell curve would look different.

Let’s assume that there is another distribution $X$, which is again represented by mean and std deviation, but with an equation different from Gaussian.

  • 1) Does such a distribution already exist?

  • 2) If yes, can it be used in situations in which Gaussian is applicable. For example, unknown natural process are assumed to follow a Gaussian distribution. Quantities from physical world that are a sum of different quantities/ variables are expected to follow Gaussian. Can the above mentioned $X$ be used in place of Gaussian to represent distribution of such quantities?

Edit: I am trying to understand why Kalman Filter has Gaussian assumption. One thing evident from the steps of Kalman Filter is that it uses mean and variance to represent the distribution. However, that raises the question, why just Gaussian, why not any other distribution which has similar representation? I am not able to relate central limit theorem with Kalman Filter applications such as SLAM or sensor fusion.

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    $\begingroup$ Many two-parameter distributions can be determined by their mean and variance: an obvious example is the binomial distribution. But the Gaussian distribution is special because of the central limit theorem $\endgroup$ – Henry Aug 19 '18 at 0:04
  • $\begingroup$ So, the distributions that can be represented by mean and variance alone cannot be used to replace Gaussian in the application areas in which it is used? $\endgroup$ – skr_robo Aug 19 '18 at 0:05
  • $\begingroup$ That depends on where it is used, and whether it is appropriate. There may be cases where other distributions (and indeed other methods of fitting them) are more appropriate $\endgroup$ – Henry Aug 19 '18 at 0:08
  • $\begingroup$ @Henry I think I should provide better context. I am adding an edit. $\endgroup$ – skr_robo Aug 19 '18 at 0:22
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    $\begingroup$ Then you might want to look at stats.stackexchange.com/questions/260996/… $\endgroup$ – Henry Aug 19 '18 at 1:00

Yes of course. Lots of distribution with two parameters, like the $\text{Beta}$ law :

$$ X \sim \text{Beta}(\alpha, \beta) $$

with $\alpha$ and $\beta$ determined by $X$ mean and std deviation. But it will be used differently than a gaussian law ! It depends a lot on the context.

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  • $\begingroup$ Thank you for the answer. Please see the edit. $\endgroup$ – skr_robo Aug 19 '18 at 0:29

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