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 Advanced Statistics Advanced Probability and Statistics Math Forum

 November 22nd, 2014, 08:29 AM #1 Newbie   Joined: Nov 2014 From: Rome, Italy Posts: 1 Thanks: 0 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)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 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? Tags equivalence, inequality, probability, simple, verify Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post roketmike Trigonometry 3 July 15th, 2014 08:51 AM Rakshasa Applied Math 7 May 7th, 2012 01:37 PM mathslog Algebra 2 March 26th, 2012 06:17 AM sivela Calculus 1 April 13th, 2010 11:28 AM pilfer00 Applied Math 0 June 8th, 2009 08:33 PM

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