- #1
kwy
- 17
- 0
Hi members,
Hope someone can help with this assignment question? I need to proof:
E(1/1+X) = [1-(1-p)^n+1]/p(n+1) where X ~ Bi(n,p)
Below are my steps and I'm not sure where I went wrong:
1. sum(x=0 to n) (1/1+x)*(n choose x)*p^x*(1-p)^n-x
2. sum(x=0 to n) (1/1+x)*[n!/x!*(n-x)!]*p^x*(1-p)^n-x
3. sum(x=0 to n) [n!/(x+1)!*(n-x)!]*p^x*(1-p)^n-x
4. sum(x=0 to n) [n!/(x+1)!*(n-x)!]*[(n+1)/(n+1)]*p^x*(p/p)*(1-p)^n-x
5. 1/p(n+1) * sum(x=0 to n) [(n+1)!/(x+1)!*(n-x)!]*p^x+1*(1-p)^n-x
let y = x+1
6. 1/p(n+1) * sum(y=0 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1
Think now the whole summation part equates to 1, leaving me with 1/p(n+1) only.
Any help will be much appreciated. Thanks in advance.
Hope someone can help with this assignment question? I need to proof:
E(1/1+X) = [1-(1-p)^n+1]/p(n+1) where X ~ Bi(n,p)
Below are my steps and I'm not sure where I went wrong:
1. sum(x=0 to n) (1/1+x)*(n choose x)*p^x*(1-p)^n-x
2. sum(x=0 to n) (1/1+x)*[n!/x!*(n-x)!]*p^x*(1-p)^n-x
3. sum(x=0 to n) [n!/(x+1)!*(n-x)!]*p^x*(1-p)^n-x
4. sum(x=0 to n) [n!/(x+1)!*(n-x)!]*[(n+1)/(n+1)]*p^x*(p/p)*(1-p)^n-x
5. 1/p(n+1) * sum(x=0 to n) [(n+1)!/(x+1)!*(n-x)!]*p^x+1*(1-p)^n-x
let y = x+1
6. 1/p(n+1) * sum(y=0 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1
Think now the whole summation part equates to 1, leaving me with 1/p(n+1) only.
Any help will be much appreciated. Thanks in advance.