Expected Value - Help PLEASE!!!

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Aria1

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Oct 30, 2012
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How do you find E(1/X)? X is distributed binomially with probability p and n+1 terms: X~Binom(n+1, p). So E(X) = p(n+1), but how do I find E(1/X)?
 
Have you considered Basic Principles?

\(\displaystyle E[Y] = \sum y\cdot p(y)\)

Of course, for 1/X, you will have to decide what to do with x = 0.
 
So, here:[FONT=MathJax_Math]E[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]Y[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Size1]∑(1/X)[/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Math]p[/FONT][FONT=MathJax_Main](1/x)[/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main]
For x=0 I would set to zero because the problem does give me that info (I didn't post that detail - sorry) However, what I am confused about is how to actually find the sum since I don't know how many terms there are(it is a sum from x=1 to x=n+1)
Since X is binomially distributed, can I say 1/X is as well? I currently have an expected value of (n+1)(p/x), but how do I get rid of that x?

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Well, that doesn't quite make sense, since you've no longer a valid Distribution:

\(\displaystyle \sum_{i=0}^{n}{{n}\choose{i}}\cdot p^{i}(1-p)^{n-i}\;=\;1\)

\(\displaystyle \sum_{i=1}^{n}{{n}\choose{i}}\cdot p^{i}(1-p)^{n-i}\) Doesn't!
 
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