Convergence in distribution

Click For Summary
SUMMARY

The discussion focuses on proving the continuous mapping theorem, which states that if a sequence of real-valued random variables \(X_n\) converges in distribution to a random variable \(X\), then for any continuous function \(h\), the transformed variables \(h(X_n)\) also converge in distribution to \(h(X)\). The user seeks a mathematical proof without relying on Skorokhod's representation theorem. Key points include the use of bounded continuous functions and the relationship between expectations of transformed variables.

PREREQUISITES
  • Understanding of convergence in distribution for random variables
  • Familiarity with continuous functions and their properties
  • Knowledge of expectation and bounded continuous functions
  • Basic proficiency in mathematical proofs and notation
NEXT STEPS
  • Study the properties of convergence in distribution in detail
  • Learn about the implications of the continuous mapping theorem in probability theory
  • Explore alternative proofs of the continuous mapping theorem without Skorokhod's representation
  • Investigate the role of bounded continuous functions in probability and statistics
USEFUL FOR

Mathematicians, statisticians, and students of probability theory who are looking to deepen their understanding of convergence concepts and the continuous mapping theorem.

shan
Messages
56
Reaction score
0
Given the definition:
For real-valued random variables X_n, n\geq1 and X, then X_n\stackrel{D}{\rightarrow}X if for every bounded continuous function g: R \rightarrow R, E_n[g(X_n)]\rightarrow E[g(X)]

I want to prove the continuous mapping theorem:
If X_n\stackrel{D}{\rightarrow}X then h(X_n)\stackrel{D}{\rightarrow}h(X) for any continuous function h: R \rightarrow R
without using Skorokhod's representation theorem.

The theorem makes sense to me intuitively but I'm lost as to how to prove it mathematically.

Edit: apologies for the really bad latex, my browser keeps hanging on the preview/save
 
Last edited:
Physics news on Phys.org
If anyone was interested:

Say h(Y_n) = Z_n, h(Y) = Z

E(g(Z_n)) \rightarrow E(g(Z)) for every g that is bounded and continuous (from definition)

E(f(Y_n)) \rightarrow E(f(Y)) for every f that is bounded and continuous (from definition)

E(g(h(Y_n)) \rightarrow E(g(h(Y)) is true because h is continuous and g o h is also continuous, h is also bounded by g
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K