Binomial formula

By Mike Latham on Wednesday, April 10, 2002 - 07:55 pm:

Dear all

Can anyone please provide an urgent proof of:

The binomial distribution formula provides values between 0 and 1

ALSO: That the summation of the binomial distribution formula, between lower limit i = 0 and upper limit n, equals 1.

Many thanks
Mike

By David Loeffler on Wednesday, April 10, 2002 - 10:18 pm:

Well, I could urgently give you a proof but the proof itself wouldn't necessarily be urgent :-)

I'll give you some clues, and if you get stuck let me know and I'll tell you in full.

We want to prove that

$\sum_{i = 0}^{n} C_{i} p^{i} q^{n - i} = (p + q)^{n}$

for any $p$ , $q$ , and $n \geq 0$ .

Do you know about proof by induction? It's obviously true for $n = 1$ , so what happens if you take the sum on the left and multiply it by $(p + q)$ , and collect terms with like powers of $p$ and $q$ ?

You'll need the identity $^{n + 1} C_{i} =^{n} C_{i} +^{n} C_{i - 1}$ .

David

By Mike Latham on Thursday, April 11, 2002 - 06:35 pm:

David

Many many thanks for your answer - you cannot believe the relief of having someone to assist. I now know how my own students feel. My ability in Maths. only goes so far I just have a great interest.

PLEASE forward details in full to the two proofs I detailed.
The first that the formula always gives values between 0 and 1.

The second that the summation gives a value of 1.

I promise that I shall work my way through your answer as homework :-).

Thanks again, I do really appreciate your help.

Mike

By David Loeffler on Thursday, April 11, 2002 - 10:43 pm:

Suppose it's true that the binomial distribution formula sums to 1 for some particular value of

n

. I'm going to write

^{n} C_{r}

(n, r)

to save typing.

We then know that

$\sum_{r = 0}^{n} (n, r) p^{r} q^{n - r} = 1$

Then as $(p + q) = 1$ , we have

$(p + q) \sum_{r = 0}^{n} (n, r) p^{r} q^{n - r} = 1$

But this expression is

$\sum_{r = 0}^{n} (n, r) p^{r + 1} q^{n - r} + \sum_{r = 0}^{n} (n, r) p^{r} q^{n - r + 1}$

$= (p q^{n} + (n, 1) p^{2} q^{n - 1} + (n, 2) p^{3} q^{n - 2} \dots) + (q^{n + 1} + (n, 1) p q^{n} + (n, 2) p^{2} q^{n - 1} \dots)$

$= q^{n + 1} + [(n, 1) + 1] p q^{n} + [(n, 2) + (n, 1)] p^{2} q^{n - 1} + [(n, 3) + (n, 2)] p^{3} q^{n - 2} \dots$

Now we'll use the identity I mentioned above. Remember that $(n, 0)$ is always 1.

$= q^{n + 1} + (n + 1, 1) p q^{n} + (n + 1, 2) p^{2} q^{n - 1} + (n + 1, 3) p^{3} q^{n - 2} \dots$

$= \sum_{r = 0}^{n + 1} (n + 1, r) p^{r} q^{n + 1 - r}$

So this is also 1.

So if what we want to prove is true for that particular value of $n$ , it is true for the next one, and hence for theone after that, and so on.

But we know it's true for 1. So it follows that it's true for all $n$ . This is what mathematicians call a proof by induction.

Now, the first question clearly follows from this: if a lot of things have sum 1 and are positive, then clearly none of them can be greater than 1.

David

By Mike Latham on Friday, April 12, 2002 - 06:52 am:

Many thanks again. I am sure that I may be calling here again for help and advice. Who knows maybe one day I will be able to provide help. !!

Regards

By Mike Latham on Saturday, April 13, 2002 - 03:07 pm:

Sorry - extra help needed.

I have tried to follow the logic shown in the proof but have failed !! Is it possible (practical) to split it down into steps. I would relly love to be able to UNDERSTAND this proof.

I cannot even 'see' how the expression is split into two summations. Please set it out for an idiot to follow.
Sorry again but we all have to start somewhere.

Many many thanks

By David Loeffler on Saturday, April 13, 2002 - 09:41 pm:

I split it up because I multiplied it by

p + q

, and then expanded this out.

The step is

$1 = \sum_{r = 0}^{n} (n, r) p^{r} q^{n - r} = (p + q) (\sum_{r = 0}^{n} (n, r) p^{r} q^{n - r})$

(because $p + q = 1$ )

$= p (\sum_{r = 0}^{n} (n, r) p^{r} q^{n - r}) + q (\sum_{r = 0}^{n} (n, r) p^{r} q^{n - r})$

$= \sum_{r = 0}^{n} (n, r) p^{r + 1} q^{n - r} + \sum_{r = 0}^{n} (n, r) p^{r} q^{n - r + 1}$

Is that any clearer?

David

By Mike Latham on Sunday, April 14, 2002 - 07:58 am:

Thanks David

If you have time and the inclination would you please continue the step by step approach throughtout the proof you earlier provided. I have spent a long time plodding my way through but with little (read 'none') success.
The soul is willing but the mind is weak :-(.
Please be patitent and attempt to guide me through this one. It is important for me to understand.

What would I do without you?

Many thanks again, I am indebted.
Mike

By David Loeffler on Sunday, April 14, 2002 - 03:59 pm:

Perhaps it would help if I gave a few examples for small n.

I presume you are familiar with the formula itself, P(n,r) = (n,r)p^r q^n-r , where p + q = 1?

Well, let's take n=1. We have

P(0) = (1,0) p⁰ q¹ = q
P(1) = (1,1) p¹ q⁰ = p

So P(0) + P(1) = q + p = 1, from above.

Now, let's look at the case n=2. Now we have
P(0) = (2,0)p⁰ q² = q² ,
P(1) = (2,1)p¹ q¹ = 2pq,
P(2) = (2,2)p² q⁰ = p² .

Now, expand out (p+q)² . You'll find that it's equal to
p(p+q) + q(p+q)
=p² + pq + pq + q² = p² + 2pq + q² .

However, since p+q = 1, (p+q)² is obviously also 1. So we know that p² + 2pq + q² = 1, which is what we need.

So the strategy for the proof is that for any n, we will show that P(0) + P(1) + ... + P(n) = (p+q)ⁿ .

Then as we know that p+q = 1, (p+q)ⁿ = 1, so it follows that the sum is 1.

Do you understand so far?

David

By Mike Latham on Monday, April 15, 2002 - 06:21 am:

By jove, I think I've got it.

You have provided the necessary stepping stones for me to work through it. I really do appreciate the help given.

Mike