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)?
Any help is much appreciated, thank you!
This is what I think I need to know for the problem anyway. For anyone who's interested, the whole problem is: Alice and her n friends (a total of n+1 people) decide the following rule for sharing a cash prize totaling K dollars. The K dollars will be divided by the number of people who get a head tossing a p-coin. What is expected prize for Alice?
Thank you, but I still don't quite understand. K is constant, so you can pull that out, but then I am still left with the problem of E(1/(1+x)), which is similar to what I asked initially of how to solve E(1/X). My answer should be in terms of n,p, and K.
When I initially tried to solve this problem, I set X~Binom(n+1, p) to be the number of heads obtained by her and her friends, found the expected value of that, and divided K by E(X). However, that is K(1/E(X)), and I think the answer should be K(E(1/X)). Can I say these two values are equal? I'm not sure...
which makes the expectation infinite.
be the number of heads obtained by her friends.
![(1-p)\times 0 + p \times E\left[\frac{K} {1+X}\right]](/~talkmath/tex/img/0ef7aa2fa65cee80aaa4b0d4466e7d94-1.gif "(1-p)\times 0 + p \times E\left[\frac{K} {1+X}\right]")
can be calculated directly by definition:
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Reply With Quote![E\left[\frac {1} {1+X}\right] = \sum_{x=0}^n \frac {1} {1+x}
\frac {n!} {x!(n-x)!} p^x(1-p)^{n-x}](/~talkmath/tex/img/313ce671bba0a8f0673f40c92afd3355-1.gif "E\left[\frac {1} {1+X}\right] = \sum_{x=0}^n \frac {1} {1+x}
\frac {n!} {x!(n-x)!} p^x(1-p)^{n-x}")
![= \frac {1} {n+1} \sum_{x=0}^n \frac {(n+1)!} {(x+1)![n+1-(x+1)]!} p^x(1-p)^{n-x}](/~talkmath/tex/img/21611d4cf51aaee2d6923275793e8bc0-1.gif "= \frac {1} {n+1} \sum_{x=0}^n \frac {(n+1)!} {(x+1)![n+1-(x+1)]!} p^x(1-p)^{n-x}")
![= \frac {1} {(n+1)p} \left[ \sum_{x=0}^{n+1} \frac {(n+1)!} {x!(n+1-x)!} p^x(1-p)^{n+1-x} - (1 - p)^{n+1} \right]](/~talkmath/tex/img/751ec7665736dd9c9c4294e78282424c-1.gif "= \frac {1} {(n+1)p} \left[ \sum_{x=0}^{n+1} \frac {(n+1)!} {x!(n+1-x)!} p^x(1-p)^{n+1-x} - (1 - p)^{n+1} \right]")
