91Ó°ÊÓ

Let \(Y\) have a hypergeometric distribution $$p(y)=\frac{\left(\begin{array}{l} r \\ y \end{array}\right)\left(\begin{array}{l} N-r \\ n-y \end{array}\right)}{\left(\begin{array}{l} N \\ n \end{array}\right)} \quad y=0,1,2, \ldots, n$$ a. Show that $$P(Y=n)=p(n)=\left(\frac{r}{N}\right)\left(\frac{r-1}{N-1}\right)\left(\frac{r-2}{N-2}\right) \cdots\left(\frac{r-n+1}{N-n+1}\right)$$ b. Write \(p(y)\) as \(p(y | r) .\) Show that if \(r_{1}\frac{p\left(y+1 | r_{1}\right)}{p\left(y+1 | r_{2}\right)}$$ c. Apply the binomial expansion to each factor in the following equation: $$(1+a)^{N_{1}}(1+a)^{N_{2}}=(1+a)^{N_{1}+N_{2}}$$ Now compare the coefficients of \(a^{n}\) on both sides to prove that $$\left(\begin{array}{c} N_{1} \\ 0 \end{array}\right)\left(\begin{array}{c} N_{2} \\ n \end{array}\right)+\left(\begin{array}{c} N_{1} \\ 1 \end{array}\right)\left(\begin{array}{c} N_{2} \\ n-1 \end{array}\right)+\cdots+\left(\begin{array}{c} N_{1} \\ n \end{array}\right)\left(\begin{array}{c} N_{2} \\ 0 \end{array}\right)=\left(\begin{array}{c} N_{1}+N_{2} \\ n \end{array}\right)$$ d. Using the result of part ( \(\underline{\mathrm{c}}\) ), conclude that $$\sum_{y=0}^{n} p(y)=1$$

Step by step solution

01

Hypergeometric Distribution Definition

The probability function for a hypergeometric distribution is given by \(p(y)=\frac{\binom{r}{y}\binom{N-r}{n-y}}{\binom{N}{n}}\). For part a, we need to show the expression for \(P(Y=n)\) given these parameters.

02

Part A: Showing Expression for P(Y=n)

Use \(p(n)=\frac{\binom{r}{n}\binom{N-r}{0}}{\binom{N}{n}}\). The term \(\binom{N-r}{0}=1\), simplifying this to \(p(n) = \frac{\binom{r}{n}}{\binom{N}{n}}\). Write \(\binom{r}{n}= \frac{r(r-1)...(r-n+1)}{n!}\) and \(\binom{N}{n}= \frac{N(N-1)...(N-n+1)}{n!}\). Thus, \(p(n)=\frac{r(r-1)...(r-n+1)}{N(N-1)...(N-n+1)}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Function

The probability function is an essential tool in statistics and probability theory. In the context of the hypergeometric distribution, the probability function is used to calculate the probability of drawing a certain number of successful outcomes from a finite population without replacement. The function specifically is given by \[ p(y) = \frac{\binom{r}{y} \binom{N-r}{n-y}}{\binom{N}{n}} \]where

• \( r \) is the number of successes in the population,
• \( N \) is the total population size,
• \( n \) is the number of draws, and
• \( y \) is the number of successful outcomes drawn.

This expression for the probability function combines combinatorial coefficients to calculate exact probabilities for specific outcomes, by evaluating how many ways the desired outcome can occur relative to the total number of possible outcomes.

Binomial Expansion

Binomial expansion is a significant mathematical concept with numerous applications, including in combinatorics and probability. It refers to expanding an expression raised to a power, like \((1 + a)^n\). In the context of the exercise, the expansion is applied to simplify combinations.
When you apply binomial expansion, you use the binomial theorem which states:\[(1 + a)^n = \sum_{k=0}^{n} \binom{n}{k} a^k\]In the exercise, this principle is leveraged to compare coefficients in the expansion of terms like \((1+a)^{N_1}(1+a)^{N_2}=(1+a)^{N_1+N_2}\).
This allows us to establish relationships between different combinatorial expressions, showing how combinations combine and overlap, fundamental in proving equalities in probabilities and outcomes in hypergeometric contexts.

Combinatorics

Combinatorics is the branch of mathematics that deals with counting, arrangement, and combination of objects. In the context of hypergeometric distribution, combinatorics helps determine how many ways certain outcomes can occur.
Central to combinatorics is the binomial coefficient, written as \(\binom{n}{k}\), representing the number of ways to choose \(k\) elements from a set of \(n\).
These coefficients appear in the calculation of probabilities in the hypergeometric distribution formula. For example, in \(p(y)=\frac{\binom{r}{y}\binom{N-r}{n-y}}{\binom{N}{n}}\), each binomial coefficient counts the ways to select subsets of successes or failures from the population.
Combinatorial principles also aid in proving logical statements in probability problems, as seen in the original problem where combinations are compared and manipulated to show the desired equalities.

Discrete Probability Distributions

Discrete probability distributions describe the probability of each outcome in a finite or countably infinite set of possible outcomes. The hypergeometric distribution is one such distribution.
In a discrete probability distribution, each possible outcome is assigned a probability that falls within the range of 0 to 1, and the total probability of all outcomes is always 1.
For the hypergeometric distribution, the probability function \[ p(y)=\frac{\binom{r}{y}\binom{N-r}{n-y}}{\binom{N}{n}} \]assigns probabilities to different numbers of successes \( y \) in a sample of size \( n \) from a larger population of size \( N \) with \( r \) successes.
This concept is put into practice effectively in the given task to determine relationships within sets of probabilities, making sure that for all \( y \), the respective probabilities add up to 1, fulfilling the conditions of a true probability distribution.