Hi chiro,
Thanks for the reply.
I am working with the following definition:
E[X] = Integral of X dP (P is the probability measure on the space) = Integral X(ω)P(dω)
f(X,Y) is not the density function; it is simply a function applied to random variables, giving a composition of...
Hello everyone,
I have the following question. Suppose that X and Y are independent and f(x,y) is nonnegative. Put g(x)=E[f(x,Y)] and show E[g(X)]=E[f(X,Y)]. Show more generally that Integral over X in A of g(X) dP = Integral over X in A of f(X,Y) dP. Extend to f that may be negative. I've...
Hello friends,
I'm working through my book and I'm having a lot of trouble coming to terms / believing this. Could anyone assist?
Let F be a free group and N be the subgroup generated by the set {x^n : x is in F and n is fixed} then N is normal in F.
Any ideas?
Yes, the function is assuredly constant. X in R is the set of omega such that X(w) is in R. However, X is a random variable, so it maps from the the space to R. So, I feel the connections starting to coalesce in my head... I think now, Since J is the empty set or the whole space and we've...
Yes, definitely measurable. Oh! If X is not constant, then the inverse image maps to a set strictly smaller than the space, or am I completely confused now? My apologies, for some reason I'm having particular difficulty on this one.
Hello,
I thought there might be some sort of topological argument, but the book is very analysis-oriented, so I was trying to stick to that line of thinking. σ(X) is defined as the smallest sigma field that X is measurable w.r.t., i.e. the intersection of all such fields.
Hello everyone,
I'm having a little trouble with a probability problem with three parts; I think I'm having trouble wrapping my head around just what's going on here. If anyone could give me a starting point, I'd appreciate it.
Here's the problem (Billingsley 5.1) (X a random variable)...