- #1

- 17

- 0

## Main Question or Discussion Point

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.