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November 26th, 2017, 06:24 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 97 Thanks: 2  expectation question
If E(X)=3, find E(1/x)

November 26th, 2017, 10:28 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,042 Thanks: 1064 
what do you think the answer is just by intuition?

November 26th, 2017, 01:28 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,560 Thanks: 605  
November 26th, 2017, 02:49 PM  #4 
Member Joined: Jan 2016 From: Blackpool Posts: 97 Thanks: 2 
so would the answer just be E(1/x)=1/3??? Can you do this though?

November 26th, 2017, 02:53 PM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 2,042 Thanks: 1064  Quote:
It's a strange question as without more information about the density you can't give any definite answer. Is there more information about the density?  
November 27th, 2017, 02:25 AM  #6 
Member Joined: Jan 2016 From: Blackpool Posts: 97 Thanks: 2 
Hi Romsek, the actual question is: X~Gam(3,1) and (yX=x)~Exp(x) Find E(Y) This is what I have done and also the reason I asked the first question: we know that E(X)=1/x for an exp(x) distribution and hence E(YX=x)=1/x and therefore E(Y)=E(E(YX)=E(1/x) we also know that E(X) of a gamma distribution is E(X)=$\frac{\alpha}{\beta}$ where gam~($\alpha,\beta$) hence E(X)=3 and this is where i got stuck. Perhaps I have made a mistake somewhere. 
November 27th, 2017, 01:17 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,560 Thanks: 605 
You left out the definition of Y in the original problem statement.


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