Expected value of function in binomial distribution

[kwy]

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.

 

[I like Serena]

When you shifted from x to y you forgot to adjust the lower bound of the summation.
You need to compensate for the term you lost.

 

[kwy]

Thanks, but doesn't the summation still equates to 1?

6. 1/p(n+1) * sum(y=1 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1

Which would still leave me with 1/p(n+1) instead of [1-(1-p)^n+1]/p(n+1)?

 

[I like Serena]

Thanks, but doesn't the summation still equates to 1?

6. 1/p(n+1) * sum(y=1 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1

Which would still leave me with 1/p(n+1) instead of [1-(1-p)^n+1]/p(n+1)?

The regular expectation of a binomial distribution with parameters n and p is given by:
EX = np

This means that:
sum(y=0 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1 = EY = (n+1)p

Apart from that we have a first term for which must be compensated.
This would be:
(1-p)^(n+1)

Can you put those together?

 

[kwy]

Oh dear, if EY is p(n+1), I think I somehow got the entire equation to equal to 1. I believe I went off the wrong track after line 3. Would you agree? Sorry, it's been close to 20 years since I've done this kind of maths. My brain just cannot interpret things as fast. I'm really appreciating your assistance. Thanks.

 

[I like Serena]

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)

Now that I look at your assignment question again, I think the question is wrong.
I suspect it should be:

$$E(\frac 1 {1+X}) = 1- \frac {(1-p)^{n+1}} {p(n+1)} \text{ where } X \sim Bi(n,p)}$$

Can you confirm?

And as far as I can tell your calculation is entirely correct until step 6 where you added (1-p)ⁿ⁺¹ without noticing.

 

[kwy]

Hi I like Serena

The question on the sheet definitely says

1 - (1-p)^n+1
-----------------
p(n+1)

Cheers and thanks.

 

[I like Serena]

Hi I like Serena

The question on the sheet definitely says

1 - (1-p)^n+1
-----------------
p(n+1)

Cheers and thanks.

My bad, the expectation EY is not in the picture.

You're quite right when you say:

sum(y=0 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1 = 1

This basically states that the sum of the chances on all possible outcomes is 1.

That still leaves the first term for which must be compensated.
More specifically:

1 = sum(y=0 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1 =
(1-p)^(n+1) + sum(y=1 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1

 

[kwy]

I got it. You are right about line 6 where the summation's lower bound should y=1. This means the summation part is:

sum(y=1 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1

which equates to say A - B where

A = sum(y=0 to n+1) [(n+1)!/y!*(n-y+1)!]*p^y*(1-p)^n-y+1 = 1
B where y is 0 giving [(n+1)!/0!*(n-0+1)!]*p^0*(1-p)^n-0+1 = (1-p)^n+1

Many thanks for the reminder.

Cheers