Dear Community,

I would like to ask your help again. Related to the problem above, I would like to define the PDF and CDF of the following events:

X1 and X2 are random variable from normal distributions with equal sigma and mu1 <= mu2.

What is the conditional distribution of X1 when X1>X2?

As I have programming background, I've made a simulation in R:

mu1 <- 0
mu2 <- 3
r.u1 <- rnorm(n = 10000000, mean = mu1, sd = sigma)
r.u2 <- rnorm(n = 10000000, mean = mu2, sd = sigma)
RS2.u1 <- dnorm(xx, mean = mu1, sd = s1)
RS3.u2 <- dnorm(xx, mean = mu2, sd = s2)
tt <- as.data.frame(cbind(r.u1, r.u2, collision = r.u1 > r.u2))
hh <- hist(tt[tt$collision == 1,"r.u2"], freq = F, breaks = xx, xlim = c(-10, 10))

    ![enter image description here][1]

To figure out the CDF first in Mathematica, my attempt was:

distX = (1/Sqrt[2 Pi \[Sigma]1^2]) Exp[-((x - \[Mu]1)/\[Sigma]1)^2/2]
distY = (1/Sqrt[2 Pi \[Sigma]1^2]) Exp[-((y - \[Mu]1)/\[Sigma]1)^2/2]

distXminusY = 
 Assuming[\[Sigma] > 0, 
  TransformedDistribution[
   x - y, {x \[Distributed] NormalDistribution[\[Mu]1, \[Sigma]], 
    y \[Distributed] NormalDistribution[\[Mu]2, \[Sigma]]}]]

dist_condyx = 
 Assuming[Element[{\[Mu]1, \[Mu]2, \[Sigma]1}, Reals] && \[Sigma]1 > 
    0 , Integrate[
   distX[t]*(1 - CDF[distY, t])/CDF[distXminusY, 0], {t, -Infinity, 
    x}]]

It seems it can't integrate the function but maybe I define it incorrectly. To get the PDF I have to derivate this function afterwards.

Can you advise me how to proceed with this?

Many thanks for your help.

Best, Attila

Attachments:

Screen Shot 2021...png

PDF[distCensored, z]

just returns the code (i.e., Mathematica can't figure it out). But because distXminusY is just a normal distribution the pdf of distCensored is found with defining a piecewise function:

pdf[z_] := Piecewise[{{PDF[distXminusY, z]/CDF[distXminusY, 0], z <= 0}}, 0]

Many thanks for your advice. My purpose is to define the PDF of the difference of two gaussian random variables (X and Y) if X-Y<0 and see the mean and variance. At this stage I just really want to learn how to define a custom PDF and ask for the mean and variance.

pdfx[x_] = 
 ConditionalExpression[
  PDF[(1/Sqrt[2*Pi]*\[Sigma]1) *(E^-1/
       2*((x - \[Mu]1)/\[Sigma]1)^2), {x, -Infinity, 
    Infinity}], \[Sigma]1 > 0]

Mean[pdfx]

and

Variance[pdfx]

It would be helpful if you posted actual code rather than a picture. And should the y be an x?

Also, if you describe a bit more what you want to do (in words), then more of us could probably give you something explicit.

I wonder if you mean to write the following:

dist = ProbabilityDistribution[(1/Sqrt[2 Pi \[Sigma]1^2]) Exp[-((x - \[Mu]1)/\[Sigma]1)^2/2],
   {x, -Infinity, Infinity}, Assumptions -> \[Sigma]1 > 0];
PDF[dist, x]
(* E^(-((x - \[Mu]1)^2/(2 \[Sigma]1^2)))/(Sqrt[2 \[Pi]] Sqrt[\[Sigma]1^2] *)
Mean[dist]
(* \[Mu]1 *)
Variance[dist]
(* \[Sigma]1^2 *)