0
$\begingroup$

The following solution to this problem was found by @Wolfies:

(E^(-((-2 z + \[Mu]1 + \[Mu]2)^2/(2 (\[Sigma]1^2 + \[Sigma]2^2)))) Sqrt[2/\[Pi]] (-1 + Erf[((z - \[Mu]2) \[Sigma]1^2 + (-z + \[Mu]1) \[Sigma]2^2)/(Sqrt[2] \[Sigma]1 \[Sigma]2 Sqrt[\[Sigma]1^2 + [Sigma]2^2])]))/(-1 + Erf[(\[Mu]1 - \[Mu]2)/(Sqrt[2] Sqrt[\[Sigma]1^2 + \[Sigma]2^2])] Sqrt[\[Sigma]1^2 +[Sigma]2^2])

Note that $Z = \frac{X + Y}{2}$ where $X$ and $Y$ are independent random variables that both follow the Normal distribution, with possibly different means and variances (denoted by subscripts $1$ for $X$ and $2$ for $Y$).

I am now trying to plot the two normal distributions together with the random variable $Z$ described by Wolfies' solution, on the same axis. What gets me is that $Z$ does not have the size I would expect relative the two Normal distributions; the distribution for $Z$ is too large.

Below is a plot of $Z$ and $X$ on two different axis, and the size problem is clear:

enter image description here

$\endgroup$
  • 1
    $\begingroup$ You missed a set of parentheses in translating @wolfies answer. You should be using: (E^(-((-2 z + \[Mu]1 + \[Mu]2)^2/(2 (\[Sigma]1^2 + \[Sigma]2^2)))) \ Sqrt[2/\[Pi]] (-1 + Erf[((z - \[Mu]2) \[Sigma]1^2 + (-z + \[Mu]1) \[Sigma]2^2)/(Sqrt[ 2] \[Sigma]1 \[Sigma]2 Sqrt[\[Sigma]1^2 + \ \[Sigma]2^2])]))/((-1 + Erf[(\[Mu]1 - \[Mu]2)/(Sqrt[ 2] Sqrt[\[Sigma]1^2 + \[Sigma]2^2])]) Sqrt[\[Sigma]1^2 + \ \[Sigma]2^2]). $\endgroup$ – JimB Apr 22 at 21:59
  • $\begingroup$ Oh! You are right. Now I get something that makes much more sense. $\endgroup$ – user120911 Apr 22 at 22:25
3
$\begingroup$

After correcting the density you should get something like the following:

Manipulate[Plot[{PDF[NormalDistribution[μ1, σ1], z],
   PDF[NormalDistribution[μ2, σ2], z],
   (E^(-((-2 z + μ1 + μ2)^2/(2 (σ1^2 + σ2^2)))) Sqrt[2/π]*
      (-1 + Erf[((z - μ2) σ1^2 + (-z + μ1) σ2^2)/(Sqrt[2] σ1 σ2 Sqrt[σ1^2 + σ2^2])]))
    ((-1 + Erf[(μ1 - μ2)/(Sqrt[2] Sqrt[σ1^2 + σ2^2])]) Sqrt[σ1^2 + σ2^2])}, {z, -10, 10},
  PlotLegends -> {"\!\(\*SubscriptBox[\(X\), \(1\)]\)", 
    "\!\(\*SubscriptBox[\(X\), \(2\)]\)", 
    "(\!\(\*SubscriptBox[\(X\), \(1\)]\)+\!\(\*SubscriptBox[\(X\), \
\(2\)]\))/2 given \!\(\*SubscriptBox[\(X\), \(1\)]\) < \
\!\(\*SubscriptBox[\(X\), \(2\)]\)"},
  PlotRange -> All],
 {{σ1, 1, "\!\(\*SubscriptBox[\(σ\), \(1\)]\)"}, 0.1, 5, Appearance -> "Labeled"},
 {{σ2, 1, "\!\(\*SubscriptBox[\(σ\), \(2\)]\)"}, 0.1, 5, Appearance -> "Labeled"},
 {{μ1, 0, "\!\(\*SubscriptBox[\(μ\), \(1\)]\)"}, 0, 5, Appearance -> "Labeled"},
 {{μ2, 0, "\!\(\*SubscriptBox[\(μ\), \(2\)]\)"}, 0, 5, Appearance -> "Labeled"},
 TrackedSymbols :> {μ1, μ2, σ1, σ2}]

Density curves

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.