# maths question about a modified binomial distribution

humy
Science 28 May '16 07:09
1. 28 May '16 07:096 edits
I know the required equation for this from the output of my computer program but what I want is the algebraic derivation of it, which I don't yet have;

The equation for the probability mass of the random discrete variable x in a binomial distribution can be expressed as;

P(x | p, x≤n) = C(n, x) p^x (1 – p)^(n – x)
( x, n ∈ ℕ
p ∈ [0, 1] )

where:
n is the number of Bernoulli trials conducted to form the distribution.
x is a possible total number of yes results from that n number of Bernoulli trials.
p is the probability of a yes result from each of the Bernoulli trials.
C(n, x) = n!/( x!(n – x)! ) n, x ∈ ℕ
and is an application of the binomial coefficient.

(see https://en.wikipedia.org/wiki/Binomial_distribution )

But suppose we didn't actual know what the value of p is here but we do know that all its possible values of p within [0, 1] are equally likely i.e. the probability density of p is:

probability_density(p | 0 ≤ p ≤ 1) = 1 else 0

( So this is a probability density OF a probability; what I like to call a 'meta-probability' )
So how to algebraically derive the the probability mass of an x value given the above condition?
Here we are treating x as the constant (because we know the x we are currently considering) and p as if it was the random variable along a continuous distribution from 0 to 1.

I have made a computer program to find out that uses a numerical approach to get ever closer approximations and its output clearly shows the answer is that the probability mass of x given these conditions is this simple continuous distribution equation of:

P(x | 0≤x≤n) = 1 / (n + 1)

which also intuitively looks about right to me. The problem here is that I don't yet see how to algebraically show it.
I assume the derivation would involve an integral but all the ones I tried so far somehow don't seem to work.
2. 28 May '16 08:26
The "C(n, x) = n!/( x!(n – x)! ) n, x ∈ ℕ " was supposed to be:

C(n, x) = n!/( x!(n – x)! )
where n, x ∈ ℕ
3. 28 May '16 09:064 edits
Forgot to mention:

As I said, I checked this with a computer program using a numerical method. But that numerical method works by getting the approximation of;

P(x|n) = lim{t → ∞} (( ∑[i=0, t] P(x | p=i/t, x≤n) ) / t )
= lim{t → ∞} (( ∑[i=0, t] C(n, x) (i/t)^x (1 – (i/t))^(n – x) ) / t )
= 1/(n + 1) (according to the program's output)
where i, t, x, n ∈ ℕ

This works by getting the average of P(x) as t increases (thus as the range of evenly-speed p values increases).
Don't know if that helps.
4. 28 May '16 13:39
Originally posted by humy
Forgot to mention:

As I said, I checked this with a computer program using a numerical method. But that numerical method works by getting the approximation of;

P(x|n) = lim{t → ∞} (( ∑[i=0, t] P(x | p=i/t, x≤n) ) / t )
= lim{t → ∞} (( ∑[i=0, t] C(n, x) (i/t)^x (1 – (i/t))^(n – x) ) / t )
= 1/(n + 1) (according to the program's output)
where i, t, ...[text shortened]... s t increases (thus as the range of evenly-speed p values increases).
Don't know if that helps.

5. DeepThought
28 May '16 17:14
Originally posted by humy
I know the required equation for this from the output of my computer program but what I want is the algebraic derivation of it, which I don't yet have;

The equation for the probability mass of the random discrete variable x in a binomial distribution can be expressed as;

P(x | p, x≤n) = C(n, x) p^x (1 – p)^(n – x)
( x, n ∈ ℕ
p ∈ [0, 1 ...[text shortened]... derivation would involve an integral but all the ones I tried so far somehow don't seem to work.
You want to show that:

P(m | 0≤m≤n) = 1 / (n + 1)

where I'm using your notation, but replacing x with m throughout.

Given:

P(m | p, m≤n) = C(n, m) p^m (1 – p)^(n – m)

and

probability_density(p | 0 ≤ p ≤ 1) = 1 else 0

What I think you have to do is integrate the probability over p using the constant density as a weighting:

P(m | 0≤m≤n) = Integral in range [0, 1] dp probability_density(p) P(m | p, m≤n)

= C(n, m) Integral in range [0, 1] dp p^m (1 – p)^(n – m)
= C(n, m) I(n, m)

I(n, m) = Integral in range [0, 1] dp p^m (1 - p)^(n - m)

We can do the integral for m = 0, when the integrand is (1 - p)^N and it integrates simply to give 1/(1 + n).

I(n, 0) = 1/(1 + n)

Use integration by parts to get the recurrence relation:

I(n, m) = [m/(N - m + 1)] I(N, m - 1)

Working out the product we get:

I(n, m) = 1/C(N, m) I(N, 0) = 1/((1 + n)C(n, m))

So:

P(m | 0≤m≤n) = C(n, m) I(n, m) = 1/(1 + n)
6. 28 May '16 18:284 edits
Originally posted by DeepThought
You want to show that:

P(m | 0≤m≤n) = 1 / (n + 1)

where I'm using your notation, but replacing x with m throughout.

Given:

P(m | p, m≤n) = C(n, m) p^m (1 – p)^(n – m)

and

probability_density(p | 0 ≤ p ≤ 1) = 1 else 0

What I think you have to do is integrate the probability over p using the constant density as a weighting:
...[text shortened]... m) = 1/C(N, m) I(N, 0) = 1/((1 + n)C(n, m))

So:

P(m | 0≤m≤n) = C(n, m) I(n, m) = 1/(1 + n)
fantastic! You are sure better at maths than I am.

I take it what you mean by;

" C(n, m) Integral in range [0, 1] dp p^m (1 – p)^(n – m) "

is, using m for x in your notation, the same as:

C(n, m) * ∫[0, 1] p^m (1 – p)^(n – m) dp

That was the very first integral I tried but somehow must have done it all wrong because I just got nonsense when I tried it so I had assumed I made a conceptual error somewhere with that. Now I think I actually started on the right track after all but then went all wrong after that point.

I don't yet get your "...recurrence relation" thing but I will study this and I am sure one way or another I will be able to give a satisfactory explanation of it in my book so, cheers.
7. DeepThought
28 May '16 20:18
Originally posted by humy
fantastic! You are sure better at maths than I am.

I take it what you mean by;

" C(n, m) Integral in range [0, 1] dp p^m (1 – p)^(n – m) "

is, using m for x in your notation, the same as:

C(n, m) * ∫[0, 1] p^m (1 – p)^(n – m) dp

That was the very first integral I tried but somehow must have done it all wrong because I just got nonsense when I ...[text shortened]... ne way or another I will be able to give a satisfactory explanation of it in my book so, cheers.
Yes, that is what I meant - I was being lazy and not looking up the unicode for an integral sign. I was trying to produce a proof by induction but ended up with a direct method. I'll fill out the working for the integration:

I(m, n) = ∫[0, 1] p^m (1 – p)^(n – m) dp

Integration by parts is based on the identity:

∫[0, 1] u(x) dv(x) = ∫[0, 1] d(u(x)v(x)) - ∫[0, 1] v(x) du(x)
=(u(1)v(1) - u(0)v(0)) - ∫[0, 1] v(x) du(x)

If u(p) = p^m then du = m p^(m-1) dp

If dv(p) = (1 - p)^(n-m) dp then v(p) = -(1 - p)^(n - m + 1)/(n - m + 1)

So:

I(n, m) = (p^m (1 - p)^(n - m + 1)/(n - m + 1))|[0, 1] - ∫[0, 1] m p^(m - 1) (-1)(1 - p)^(n - m + 1)/(n - m + 1)

In the first term when p = 0 then p^m = 0, except when m = 0. When p = 1 (1 - p) ^(n - m + 1) = 0. So the first term is zero for all m st 0 < m < n + 1. We are left with:

I(n, m > 0) = [m/(n - m + 1)] ∫[0, 1] p^(m - 1) (1 - p)^(n - m + 1)

The integral on the right hand side is just I(n, m - 1), so we have:

I(n, m) = (m / (n - m + 1)) I(n, m - 1)

since this is true for all values of m in the range [1, N] we also have:

I(n, m - 1) = (m - 1)/(n - m + 2) I(n, m - 2)
...
I(n, 1) = 1/(n - 1 + 1) I(n, 0) = I(n, 0)/n

So we can recursively substitute these expressions until we have I(n, 0) on the right hand side:

I(n, m) = (m/(n - m + 1))((m - 1)/(n - m + 2)) I(n, m - 2)
= (m/(n - m + 1))((m - 1)/(n - (m - 1) + 1)). ... 1/n) I(n, 0)
= (m (m - 1) ... 1)/[(n - m + 1)(n - m + 2) . ... n] I(n, 0)
= [m! (n - m)! / n!] I(n, 0)
= I(n, 0)/C(n, m)

let x = (1 - p)

I(n, 0) = ∫[0, 1] (1 - p)^n dp = ∫[x = 1, x = 0] x^n -dx = ∫[0, 1] x^n dx = 1/(1 + n)

So we have the result:

I(n, m) = 1/[(n + 1) C(n,m)]
8. DeepThought
30 May '16 22:23
Originally posted by humy
fantastic! You are sure better at maths than I am.

I take it what you mean by;

" C(n, m) Integral in range [0, 1] dp p^m (1 – p)^(n – m) "

is, using m for x in your notation, the same as:

C(n, m) * ∫[0, 1] p^m (1 – p)^(n – m) dp

That was the very first integral I tried but somehow must have done it all wrong because I just got nonsense when I ...[text shortened]... ne way or another I will be able to give a satisfactory explanation of it in my book so, cheers.
If I was better at maths I'd have noticed that your integral:

I(N,m) = ∫[0, 1] p^m (1 – p)^(n – m) dp

Is the Euler Beta Function, given by:

B(x, y) = ∫[0, 1] t^(x - 1) (1 – t)^(y - 1) dt

with:

I(N, m) = B(m + 1, N - m + 1)

The Euler Beta function for positive integer arguments is (x - 1)!(y - 1)!/(x + y - 1)! which gives us the desired result.

https://en.wikipedia.org/wiki/Beta_function
9. 31 May '16 07:281 edit
Originally posted by DeepThought
If I was better at maths I'd have noticed that your integral:

I(N,m) = ∫[0, 1] p^m (1 – p)^(n – m) dp

Is the Euler Beta Function, given by:

B(x, y) = ∫[0, 1] t^(x - 1) (1 – t)^(y - 1) dt

with:

I(N, m) = B(m + 1, N - m + 1)

The Euler Beta function for positive integer arguments is (x - 1)!(y - 1)!/(x + y - 1)! which gives us the desired result.

https://en.wikipedia.org/wiki/Beta_function
thanks for that. I think I might use this Euler Beta Function often so I better study it at some time.