@Attila Horvath Are you aware of Conditioned?

With it, you can compute conditional probability, conditional expectation (and through the workaround/definition also conditional variance). I tested these computations with the example and problem sets from US textbooks and they work just fine. When i looked at those texts i did notice the term "conditional distribution" but i have not read up on it; i am pre-"Introduction to Probability", hehe. It may or may not be possible to define a conditional distribution with that function, i am just wondering if you were aware of the nice function. Afaik MAPLE does not have it.

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 *)

Yes, this assumptation can be made: Assumptions -> [Mu]1 <= [Mu]2 as well as [Sigma]1==[Sigma]2.

Would you mind having a look at my notebook?

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

distCensored = CensoredDistribution[{-Infinity, 0}, distXminusY]

MeanOfdistCensored = 
 Simplify[Assuming[\[Mu]1 <= \[Mu]2, Mean[distCensored]]]

Assuming[\[Mu]1 <= \[Mu]2, PDF[distCensored, z]]

Thanks for your help, I'm really grateful!

Attila

distXminusY = TransformedDistribution[x - y,
  {x \[Distributed] NormalDistribution[\[Mu]1, \[Sigma]1], 
   y \[Distributed] NormalDistribution[\[Mu]2, \[Sigma]2]}]
distCensored = CensoredDistribution[{-Infinity, 0}, distXminusY]
Mean[distCensored] // FullSimplify
(* -((E^(-((\[Mu]1 - \[Mu]2)^2/(2 (\[Sigma]1^2 + \[Sigma]2^2))))
    Sqrt[\[Sigma]1^2 + \[Sigma]2^2])/Sqrt[2 \[Pi]]) + 
 1/2 (\[Mu]1 - \[Mu]2) Erfc[(\[Mu]1 - \[Mu]2)/(
   Sqrt[2] Sqrt[\[Sigma]1^2 + \[Sigma]2^2])] *)
Variance[distCensored]
(* a long output - maybe can be simplified by assuming \[mu]1 <= \[mu]2  *)

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]