My Math Forum  

Go Back   My Math Forum > College Math Forum > Advanced Statistics

Advanced Statistics Advanced Probability and Statistics Math Forum


Reply
 
LinkBack Thread Tools Display Modes
November 22nd, 2014, 08:29 AM   #1
Newbie
 
Joined: Nov 2014
From: Rome, Italy

Posts: 1
Thanks: 0

Lightbulb Simple equivalence to verify.

Hi everyone,
I am new in here and I hope some of you may help me with this.
Let A,B and C be three events such that the following three inequalities are verified:
P(A|BC)>P(A|\BC)
P(A|B\C)>P(A|\B\C)
P(A|B)<P(A|\B)
where "\B"= complementary of B, "\C"= complementary of C.
Now, if we define
A=(T>t+s)
B=(X=x)
C=(Y=y)
and
D=(x<=X<=x')$\displaystyle \cap$(y<=Y<=y')$\displaystyle \cap$(T>t)
and if we apply the previous inequalities to the conditional probability $\displaystyle P(\cdot|D)$, they become equivalent to say that
$\displaystyle P(T>t+s|T>t,X=x,Y=y)$ is strictly decreasing in x for every y
and
$\displaystyle P(T>t+s|T>t,X=x)$ is increasing in x.

I don't understand what it means to apply those inequalities to the conditional probability $\displaystyle P(\cdot|D)$ and, even if I did, I am not sure I'd be able to verify such equivalence.

Hope some of you want to help!
Thank you in advance
Vivi is offline  
 
November 29th, 2014, 12:03 AM   #2
Senior Member
 
Joined: Aug 2012

Posts: 229
Thanks: 3

Hey Vivi.

Can you please re-summarize what you are trying to do and what you are trying to understand?
chiro is offline  
Reply

  My Math Forum > College Math Forum > Advanced Statistics

Tags
equivalence, inequality, probability, simple, verify



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Verify the following identity roketmike Trigonometry 3 July 15th, 2014 08:51 AM
Verify the following Rakshasa Applied Math 7 May 7th, 2012 01:37 PM
A simple question, just need to verify mathslog Algebra 2 March 26th, 2012 06:17 AM
can you verify this.... sivela Calculus 1 April 13th, 2010 11:28 AM
Could someone please verify this proof pilfer00 Applied Math 0 June 8th, 2009 08:33 PM





Copyright © 2019 My Math Forum. All rights reserved.