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 November 26th, 2017, 06:24 AM #1 Member   Joined: Jan 2016 From: Blackpool Posts: 98 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: 2,124 Thanks: 1103 what do you think the answer is just by intuition?
November 26th, 2017, 01:28 PM   #3
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Quote:
 Originally Posted by Jaket1 If E(X)=3, find E(1/x)
I depends entirely on the distribution of X. 2 Examples.
1. P(X=3)=1: E(1/X)=1/3
2. P(X=0)=1/2, P(X=6)=1/2: E(1/X) is infinite.

 November 26th, 2017, 02:49 PM #4 Member   Joined: Jan 2016 From: Blackpool Posts: 98 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
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Quote:
 Originally Posted by Jaket1 so would the answer just be E(1/x)=1/3??? Can you do this though?
no, in general as Mathman has pointed out, you can't.

It's a strange question as without more information about the density you can't give any definite answer.

 November 27th, 2017, 02:25 AM #6 Member   Joined: Jan 2016 From: Blackpool Posts: 98 Thanks: 2 Hi Romsek, the actual question is: X~Gam(3,1) and (y|X=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(Y|X=x)=1/x and therefore E(Y)=E(E(Y|X)=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: 6,613 Thanks: 617 You left out the definition of Y in the original problem statement.

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