1. 

   Sunit_K Member

   In the solution it is said,
   In order that the integral of the density over theta is 1, the above expression has to be divided
   by the constant exp(− x).
   Why? I reckon it has to do with the range?
   Also, what is the I(theta>0) term?
   Thanks

    

   Sunit_K, Sep 2, 2016

   #1
2. 

   John Lee ActEd Tutor Staff Member

   Yes, \(I(\theta >x)\) is the identity function which is 1 if \(\theta > x\) and 0 otherwise.

    

   John Lee, Sep 7, 2016

   #2

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