
Applied Math Applied Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 24th, 2017, 08:21 PM  #1 
Newbie Joined: Aug 2013 Posts: 14 Thanks: 0  (Brownian Motion) Proving that a random variable X is normal?
I have attached the question below! So given that W(t) is a Brownian motion, how do you prove X=W(a)+W(b) is normal? I know it must satisfy this condition: that W(t)W(s)~N(0,ts). My initial thought process: X(t)X(s)=(W(t)+W(t))(W(s)+W(s))=2(W(t)W(s)), but since the interval is 0<a<b, I'm pretty sure the variable substitutions I've done is incorrect. Is my approach to this problem incorrect? Are the W(a)+W(b) just constants? But that would entail that (W(a)+W(b))(W(a)+W(b))=0? Sorry, I don't think I am grasping the concepts very well. I would appreciate any pointers on how to solve this problem! Last edited by facebook; February 24th, 2017 at 08:39 PM. 
February 24th, 2017, 08:31 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 925 Thanks: 499 
how are we supposed to read that image?

February 24th, 2017, 08:38 PM  #3 
Newbie Joined: Aug 2013 Posts: 14 Thanks: 0 
Hi, sorry I've corrected the image! The original file got downscaled, I did not anticipate image compression, my apologies

February 26th, 2017, 03:33 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,204 Thanks: 486 
You must show W(a)+W(b)W(ac)W(bc) is normal. The mean = 0, but I am not sure what the variance is.


Tags 
brownian, motion, normal, proving, random, variable, variance 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Max Transformation of a normal distributed random variable  hansherman  Advanced Statistics  0  February 6th, 2014 05:25 AM 
Let Z be a standard normal random variable  Exfactor  Algebra  1  January 13th, 2014 10:30 PM 
Standard Normal Random Variable  Ethnikation  Algebra  0  November 8th, 2012 02:58 AM 
Finding standard normal random variable  football  Algebra  1  September 11th, 2011 05:25 AM 
Standard normal random variable  desiclub07  Algebra  3  June 21st, 2010 02:49 PM 