A Proof on hypergeometric distribution

[irony of truth]

Show directly that the set of probabilities associated with the hypergeometric distribution sum to one.

=> I am thinking that this tells me to prove that since this is a probability distribution function, it really should sum to 1. Is that what the problem asking me to do? =)

I got this given hint in the book that I should expand the identity:
(1 + 'mu')^N = (1 + 'mu')^r (1 + 'mu')^(N-r) and equate the coefficients.

Also, how should I equate the coefficients? Should I make the 'mu' arbitrarily equal to -1? What I did is that I expanded the left side of the given equation as 1 + (N C 1)'mu' + (N C 2)'mu'^2 + ... + (N C N)'mu'^N. That's why I got stucked on thinking... how about the coefficients?

 

[member 553083]

Yes, the problem is asking you to show that the set of probabilities associated with the hypergeometric distribution sum to one. To do this, we can simply expand the identity given in the hint: (1 + 'mu')^N = (1 + 'mu')^r (1 + 'mu')^(N-r). Equating the coefficients of each side of the equation will give us the desired result. On the left side, we have: 1 + (N C 1)'mu' + (N C 2)'mu'^2 + ... + (N C N)'mu'^N. On the right side, we have: (1 + 'mu')^r (1 + 'mu')^(N-r) = [1 + (r C 1)'mu' + (r C 2)'mu'^2 + ... + (r C r)'mu'^r] * [1 + ((N-r) C 1)'mu' + ((N-r) C 2)'mu'^2 + ... + ((N-r) C (N-r))'mu'^(N-r)]. Now, equate the coefficients on both sides of the equation. We can see that the coefficient of 'mu'^0 is 1 on both sides and they will cancel out. The coefficient of 'mu'^1 is (N C 1) on the left side and (r C 1) + ((N-r) C 1) on the right side. Thus, (N C 1) = (r C 1) + ((N-r) C 1). Similarly, the coefficient of 'mu'^2 is (N C 2) on the left side and (r C 2) + ((N-r) C 2) on the right side. Thus, (N C 2) = (r C 2) + ((N-r) C 2). Continuing this way, we will get (N C k) = (r C k) + ((N-r) C k) for all k from 0 to N. Now, let us consider the special case of k=N. We get (N C N) = (r C N) + ((N-r) C N

 

[wais]

Yes, you are correct. The problem is asking you to prove that the sum of probabilities associated with the hypergeometric distribution is equal to 1. This is because, as you mentioned, it is a probability distribution function and the sum of all probabilities in a probability distribution should always be equal to 1.

To prove this, you can use the given hint and expand the identity (1 + 'mu')^N = (1 + 'mu')^r (1 + 'mu')^(N-r). This can be rewritten as:

(1 + 'mu')^N = (1 + 'mu')^r * (1 + 'mu')^(N-r)

Now, let's equate the coefficients on both sides. The coefficient of 'mu'^k on the left side is given by (N C k), which represents the number of ways to choose k objects from a total of N objects. The coefficient of 'mu'^k on the right side is given by (r C k)*(N-r C N-k), which represents the number of ways to choose k objects from a total of r objects and N-k objects from a total of (N-r) objects.

Therefore, we can equate the coefficients as follows:

(N C k) = (r C k)*(N-r C N-k)

Now, let's substitute k = 0, 1, 2, ..., N in this equation. We will get N+1 equations in total. When we add all these equations, we get:

(N C 0) + (N C 1) + (N C 2) + ... + (N C N) = (r C 0)*(N-r C N) + (r C 1)*(N-r C N-1) + (r C 2)*(N-r C N-2) + ... + (r C N)*(N-r C 0)

Simplifying this equation, we get:

1 + (N C 1) + (N C 2) + ... + (N C N) = (r C 0)*(N-r C N) + (r C 1)*(N-r C N-1) + (r C 2)*(N-r C N-2) + ... + (r C N)*(N-r C 0)

But, we know that (r C 0) = 1, (N-r C N) =