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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(ABC)>P(A\BC) P(AB\C)>P(A\B\C) P(AB)<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 $\displaystyle P(\cdotD)$, they become equivalent to say that $\displaystyle P(T>t+sT>t,X=x,Y=y)$ is strictly decreasing in x for every y and $\displaystyle P(T>t+sT>t,X=x)$ is increasing in x. I don't understand what it means to apply those inequalities to the conditional probability $\displaystyle P(\cdotD)$ 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 resummarize what you are trying to do and what you are trying to understand? 

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equivalence, inequality, probability, simple, verify 
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