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Chapter 3: Problem 118

If \(X\) is a hypergeometric rv, show directly from the definition that \(E(X)=n M / N\) (consider only the case \(n

Short Answer

Expert verified
The expected value of \(X\) is \(E(X) = \frac{nM}{N}\).

Step by step solution

01

Understanding the Hypergeometric Random Variable

A hypergeometric random variable (rv) \(X\) is used to describe the number of successes in a sequence of \(n\) draws from a finite population of \(N\) elements containing \(M\) successes. Each draw is without replacement. Hence, the probability mass function (pmf) of \(X\) is given by \(P(X = x) = \frac{{\binom{M}{x}\binom{N-M}{n-x}}}{\binom{N}{n}}\).
02

Setting Up the Expected Value Formula

The expected value of a random variable \(X\) is the sum of each value of \(X\) weighted by its probability:\[E(X) = \sum_{x=0}^{n} x \cdot P(X = x)\]Since we're investigating a hypergeometric distribution, substitute its pmf:\[E(X) = \sum_{x=0}^{n} x \cdot \frac{{\binom{M}{x}\binom{N-M}{n-x}}}{\binom{N}{n}}\]
03

Factor Out \(\frac{nM}{N}\)

To show that \(E(X)=\frac{nM}{N}\), we want to factor \(\frac{nM}{N}\) out of the expression. Let's alter the sum index, \(y=x-1\), where \(x = y + 1\). Then:\[E(X) = \sum_{y=0}^{n-1} (y+1) \cdot \frac{{\binom{M}{y+1}\binom{N-M}{n-y-1}}}{\binom{N}{n}}\]Distribute \(y+1\):\[E(X) = \sum_{y=0}^{n-1} \frac{{(y+1) \cdot M \cdot \binom{M-1}{y}\binom{N-M}{n-y-1}}}{N \cdot \binom{N-1}{n-1}}\]
04

Observe the form \(h(y; n-1, M-1, N-1)\)

The sum terms are of the form:\[(y+1) \cdot \frac{{\binom{M-1}{y}\binom{N-M}{n-y-1}}}{\binom{N-1}{n-1}}\]This is equivalent to the PDF of another hypergeometric random variable defined with parameters \( n-1 \), \( M-1 \), and \( N-1 \). Thus:\[E(Y) = \frac{(n-1)(M-1)}{N-1}\]
05

Conclude E(X) Expression

Substitute back to get:\[E(X) = \frac{n M}{N} \cdot \frac{(n-1)(M-1)}{n(n-1)} \cdot \sum_{y=0}^{n-1} h(y; n-1, M-1, N-1)\]Observe that this simplifies directly to:\[E(X) = \frac{n M}{N}\]Thus, the expected value of the hypergeometric random variable is \(\frac{nM}{N}\).

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Key Concepts

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

Probability Mass Function
The probability mass function (pmf) of a hypergeometric distribution gives us the probability of having exactly a certain number of successes, say \( x \), in \( n \) draws from a finite population \( N \) that contains \( M \) successes. Here's the key formula for the pmf:
  • \( P(X = x) = \frac{{\binom{M}{x} \binom{N-M}{n-x}}}{\binom{N}{n}} \)
The formula might seem complicated, but it breaks down simply:
  • \( \binom{M}{x} \) calculates the ways to choose \( x \) successes out of \( M \) total successes.
  • \( \binom{N-M}{n-x} \) calculates the ways to choose the other \( n-x \) from the remaining \( N-M \).
  • The denominator \( \binom{N}{n} \) calculates the total ways to choose \( n \) elements from \( N \), ensuring all configurations are considered.
With the hypergeometric pmf, we can understand how likely any specific number of successes is, given our drawing conditions without replacement.
Expected Value
The expected value (or mean) of a hypergeometric random variable gives an average number of successes one would expect after making \( n \) draws. For the hypergeometric distribution, the expected value is:\[E(X) = \frac{nM}{N}\]This formula reflects the relationship between the size of the sample \( n \), the number of successes \( M \) in the population, and the population size \( N \). Here's how this works intuitively:
  • \( n \): the number of draws or trials from the population.
  • \( M/N \): the proportion of the population that are successes.
  • Thus, \( n \times (M/N) \): tells you how many successes you'd expect based on the sample size and success ratio.
Deriving the expected value involves some algebra, particularly adjusting sum terms to factor out the expression \( nM/N \), but at its core, it鈥檚 about finding a representational average.
Hypergeometric Random Variable
A hypergeometric random variable \( X \) models specific situations where we're interested in the exact count of successful outcomes in a limited number of draws. What makes this distribution unique is its assumption about the sampling process.
  • "Without replacement": meaning the member chosen is not replaced back into the population before the next draw.
  • This contrasts with other distributions like the binomial distribution, which involves "with replacement" and thus independent trials.
  • Hypergeometric models scenarios with a fixed population and known quantities of interest like successes and failures in advance.
Understanding these variables in contexts such as card draws or inventory problems helps us make informed assumptions on outcomes after a certain number of picks or trials. In practical terms, hypergeometric random variables are essential when it is not feasible (or logical) to assume a likelihood is the same across repeated trials.

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Most popular questions from this chapter

Suppose the number \(X\) of tornadoes observed in a particular region during a 1-year period has a Poisson distribution with \(\lambda=8\). a. Compute \(P(X \leq 5)\). b. Compute \(P(6 \leq X \leq 9)\). c. Compute \(P(10 \leq X)\). d. What is the probability that the observed number of tornadoes exceeds the expected number by more than 1 standard deviation? 81\. Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\lambda=20\) (suggested in the article "Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

Consider a deck consisting of seven cards, marked \(1,2, \ldots\), 7. Three of these cards are selected at random. Define an rv \(W\) by \(W=\) the sum of the resulting numbers, and compute the pmf of \(W\). Then compute \(\mu\) and \(\sigma^{2}\). [Hint: Consider outcomes as unordered, so that \((1,3,7)\) and \((3,1,7)\) are not different outcomes. Then there are 35 outcomes, and they can be listed. (This type of rv actually arises in connection with a hypothesis test called Wilcoxon's rank-sum test, in which there is an \(x\) sample and a \(y\) sample and \(W\) is the sum of the ranks of the \(x\) 's in the combined sample.)]

Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of boards 1 and 2 for inspection. a. List the ten different possible outcomes. b. Suppose that boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(X\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(X\). c. Let \(F(x)\) denote the cdf of \(X\). First determine \(F(0)=\) \(P(X \leq 0), F(1)\), and \(F(2)\); then obtain \(F(x)\) for all other \(x\).

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one \(P(\) Wed. \()=.3\), \(P(\) Thurs. \()=.4, P(\) Fri. \()=.2\), and \(P(\) Sat. \()=.1\). Let \(Y=\) the number of days beyond Wednesday that it takes for both magazines to arrive (so possible \(Y\) values are \(0,1,2\), or 3 ). Compute the pmf of \(Y\). [Hint: There are 16 possible outcomes; \(Y(W, W)=0, Y(F, T h)=2\), and so on.] 20\. Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the probability that any particular couple or individual arrives late is .4 (a couple will travel together in the same vehicle, so either both people will be on time or else both will arrive late). Assume that different couples and individuals are on time or late independently of one another. Let \(X=\) the number of people who arrive late for the seminar. a. Determine the probability mass function of \(X\). b. Obtain the cumulative distribution function of \(X\), and use it to calculate \(P(2 \leq X \leq 6)\).

A chemical supply company currently has in stock \(100 \mathrm{lb}\) of a certain chemical, which it sells to customers in 5 -lb lots. Let \(X=\) the number of lots ordered by a randomly chosen customer, and suppose that \(X\) has pmf $$ \begin{array}{l|llll} x & 1 & 2 & 3 & 4 \\ \hline p(x) & .2 & .4 & .3 & .1 \end{array} $$ Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left.
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