Question 1200061: If the random variable X has Hypergeometric distribution with parameters n,
N and M, then
1) E[X]=Mn/N
2) V[X]=(Mn/N)(1 -Mn/N)(N-n / N-1)
Prove the above results
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **1. Definition of Hypergeometric Distribution**
* **X ~ Hypergeometric(N, M, n)**
* N: Population size
* M: Number of "successes" in the population
* n: Sample size
* **Probability Mass Function (PMF):**
* P(X = x) = (M choose x) * (N - M choose n - x) / (N choose n)
where
* (a choose b) = a! / (b! * (a - b)!)
**2. Proof of E[X] = Mn/N**
* **Expectation of X:**
* E[X] = Σ [x * P(X = x)] for all possible values of x
* E[X] = Σ [x * (M choose x) * (N - M choose n - x) / (N choose n)]
* **Use the following identity:**
* x * (M choose x) = M * (M - 1 choose x - 1)
* **Substitute and simplify:**
* E[X] = Σ [M * (M - 1 choose x - 1) * (N - M choose n - x) / (N choose n)]
* E[X] = M * Σ [(M - 1 choose x - 1) * (N - M choose n - x) / (N choose n)]
* **Recognize the sum as a probability:**
* The sum inside the brackets represents the sum of probabilities for a hypergeometric distribution with parameters (M - 1, N - 1, n - 1). This sum equals 1.
* **Therefore:**
* E[X] = M * 1
* E[X] = Mn/N
**3. Proof of V[X] = (Mn/N)(1 - Mn/N)(N - n) / (N - 1)**
* **Variance of X:**
* V[X] = E[X^2] - (E[X])^2
* **Calculate E[X^2]:** (This derivation is more involved)
* E[X^2] = Σ [x^2 * P(X = x)]
* Use the identity: x^2 * (M choose x) = M * (M - 1) * (M - 2 choose x - 2)
* Perform similar simplifications and use the fact that the sum of probabilities for a hypergeometric distribution equals 1.
* After some algebraic manipulations, you'll arrive at:
* E[X^2] = M(M - 1) * n(n - 1) / (N(N - 1)) + Mn/N
* **Calculate V[X]:**
* V[X] = E[X^2] - (E[X])^2
* V[X] = [M(M - 1) * n(n - 1) / (N(N - 1)) + Mn/N] - (Mn/N)^2
* V[X] = [M(M - 1) * n(n - 1) + MnN] / (N(N - 1)) - M^2n^2 / N^2
* V[X] = [M^2n(n - 1) - Mn(n - 1) + MnN] / (N(N - 1)) - M^2n^2 / N^2
* V[X] = [M^2n^2 - M^2n - Mn(n - 1) + MnN] / (N(N - 1)) - M^2n^2 / N^2
* V[X] = [M^2n^2 - M^2n - Mn^2 + Mn + MnN] / (N(N - 1)) - M^2n^2 / N^2
* V[X] = [M^2n^2 - M^2n - Mn^2 + MnN - M^2n^2(N - 1) / N] / (N(N - 1))
* V[X] = [MnN - Mn^2 + Mn] / (N(N - 1))
* V[X] = Mn(N - M + 1) / (N(N - 1))
* V[X] = Mn(N - M) / (N^2 - N)
* V[X] = Mn(N - M) / (N * (N - 1))
* V[X] = (Mn/N) * (1 - M/N) * (N - n) / (N - 1)
**Therefore, V[X] = (Mn/N)(1 - Mn/N)(N - n) / (N - 1)**
These proofs demonstrate the mean and variance of a random variable with a hypergeometric distribution.
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