1. Nov 26, 2011 #1

   torquerotates

   

   let X1 and X2 be two randomly selected random variables from the same discrete distribution. Let Y=min(X1,X2). Find P(Y=1)
   
   according to the book. P(Y=1)=P[(X1=1)and(X2>=1)]+P[(X2=1)and(X1>=2)]
   
   I don't understand why is the asymmetry is even possible. The way I look at it, it's
   P(Y=1)=P[(X1=1)and(X2>1)]+P[(X2=1)and(X1>1)]

    

   torquerotates, Nov 26, 2011
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3. Nov 26, 2011 #2

   Stephen Tashi

   

   Science Advisor

   In the case of discrete random variables, your way omits the event "X1 = 1 and X2 = 1". In the case of continuous random variables (where that event has zero probability), your method is correct.
   
   The book's way is correct for the case of discrete random variables that only take integer values. It wouldn't be correct for continuous random variables.

    

   Stephen Tashi, Nov 26, 2011

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