Hi, i have struggled with this question: * why can we not do E(1-p/p)=1-E(p)/E(p) ? this gives wrong answer as a/b. the solutions integrate pdf*(1-p/p) which i understand method but why does what i said not work? * do i assume that this E(1-p/p) is a bayesian estimate? would any form of E(...) work as a bayesian estimate? (i know its not the usual quadratic/all or nothing/absolute) thank you in advance for help
Thanks Alfie E[(1-p)/p] is a function of p, you can't separate out the 1/p as E(1/p) is different to 1/E(p), you can work out the expected value as the integral of the function multiplied by the pdf. This is the expected value of this function of p, it is the prior distribution mean. A Bayesian estimate of p under quadratic loss would be the mean of the posterior distribution. You work that out in iv using ii and iii. Thanks Andrea
Hi Andrea, thank you for this! sorry, what do you mean by 'a function of p'? im finding it difficult to see how this is different to say E(\sumx/\lambda) which is another question that i have seen but this time it is okay to pull out lambda. How do we know when its a function of something? i would have said E(\sumx/lambda) was a function of x thank you in advance!
p is your variable of interest, so 1/p is a function of p as is p^2 say. I suspect the other question was a function of x so it is okay to remove lambda in that case, as you said. You would be guided by the question, here it was about p in i) conjugate prior of p, so that is our variable.
