Suppose random variable $X_1$ is a mixture of two Normal distributions with means of $\mu_A$ and $\mu_B$ respectively, standard deviations of $\sigma_A$ and $\sigma_B$ respectively, and weights given by $w_{1_A}$ and $w_{1_B}$. Suppose further that another random variable exists, $X_2$, which is independent of $X_1$, but is also a mixture of two Normal distributions. Furthermore, $X_2$ has the same means and standard deviations as $X_1$, but has its own weights given by $w_{2_A}$ and $w_{2_B}$. Can I use Mathematica to find $X_1$ + $X_1$?

I have tried:

\[ScriptCapitalD] = MixtureDistribution[{w1A, w1B}, {NormalDistribution[muA, sdA], NormalDistribution[muB, sdB]}]  + MixtureDistribution[{w2A, w2B}, {NormalDistribution[muA, sdA], NormalDistribution[muB, sdB]}] 

but apparently the problem is beyond Mathematica. Any suggestions as to how I can, nevertheless, use Mathematica to work this out?

  • 1
    $\begingroup$ Adding 2 distributions directly is meaningless in MMA. You need to use TransformedDistribution to add the r.v.'s and even then what you get is usually a symbolic distribution indicating your intentions. Use PDF etc. to derive concrete results. $\endgroup$ – asterix314 Dec 6 '19 at 2:56

To get the distribution of the sum of random variables use TransformedDistribution


\[ScriptCapitalD]1 = 
  MixtureDistribution[{w1A, w1B}, {NormalDistribution[muA, sdA], 
    NormalDistribution[muB, sdB]}];

\[ScriptCapitalD]2 = 
  MixtureDistribution[{w2A, w2B}, {NormalDistribution[muA, sdA], 
    NormalDistribution[muB, sdB]}];

\[ScriptCapitalD] = TransformedDistribution[x1 + x2,
   {x1 \[Distributed] \[ScriptCapitalD]1, 
    x2 \[Distributed] \[ScriptCapitalD]2}];

PDF[\[ScriptCapitalD], x]

(* w1B ((E^(-((muA + muB - x)^2/(2 (sdA^2 + sdB^2)))) sdB w2A)/(
    Sqrt[sdA^2 + sdB^2] (w2A + w2B)) + (E^(-((-2 muB + x)^2/(4 sdB^2))) w2B)/(
    Sqrt[2] (w2A + w2B))))/(Sqrt[2 π] sdB (w1A + w1B)) + (
 w1A ((E^(-((-2 muA + x)^2/(4 sdA^2))) w2A)/(Sqrt[2] (w2A + w2B)) + (
    E^(-((muA + muB - x)^2/(2 (sdA^2 + sdB^2)))) sdA w2B)/(
    Sqrt[sdA^2 + sdB^2] (w2A + w2B))))/(Sqrt[2 π] sdA (w1A + w1B)) *)

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