91Ó°ÊÓ

Between two distinct methods for manufacturing certain goods, the quality of goods produced by method \(i\) is a continuous random variable having distribution \(F_{i}, i=\) 1,2. Suppose that \(n\) goods are produced by method 1 and \(m\) by method \(2 .\) Rank the \(n+m\) goods according to quality, and let$$X_{j}=\left\\{\begin{array}{ll}1 & \text { if the } j \text { th best was produced from } \\\& \text { method } 1 \\\2 & \text { otherwise }\end{array}\right.$$ For the vector \(X_{1}, X_{2}, \ldots, X_{n+m},\) which consists of \(n\) 1's and \(m\) 2's, let \(R\) denote the number of runs of \(1 .\) For instance, if \(n=5, m=2,\) and \(X=1,2,1,1,1,1,2,\) then \(R=2 .\) If \(F_{1}=F_{2}\) (that is, if the two methods produce identically distributed goods), what are the mean and variance of \(R ?\)

Step by step solution

01

Calculate the probability of R runs occurring

Let C(n, m, R) be the number of ways to form a sequence of length (n + m) with n 1's and m 2's and R runs of 1. The probability of any given sequence is: \[\frac{1}{\binom{n+m}{n}}\] And the probability `Pr(R)` of R runs occurring is: \[\Pr(R) = \frac{C(n, m, R)}{\binom{n+m}{n}}\]

02

Derive a formula for the expected value of R

The expected value of R is the sum of the product of R and its probability of occurrence: \[E[R] = \sum_{R=1}^{n} R \cdot \Pr(R)\]

03

Calculate the expected value of R

Using the formula derived in Step 2 and the probability of R in Step 1, we can write the expected value of R as: \[E[R] = \sum_{R=1}^{n} R \cdot \frac{C(n, m, R)}{\binom{n+m}{n}}\] Since the two methods produce identically distributed goods (meaning that each \")run of 1s\" has the same probability), the expected value of R (mean) is: \[E[R] = \frac{2nm}{n+m}\]

04

Derive a formula for the variance of R

The variance of R can be calculated as: \[Var(R) = E[R^{2}] - (E[R])^{2}\]

05

Calculate the variance of R

Now we have to find a formula to calculate E[R^2]. Each term in the sum E[R^2] will have the form R^2 * Pr(R). Thus, we can write E[R^2] as: \[E[R^{2}] = \sum_{R=1}^{n} R^{2} \cdot \frac{C(n, m, R)}{\binom{n+m}{n}}\] After calculating E[R^2] and substituting the values of E[R^2] and E[R] into the formula for variance, the variance of R (when the two methods produce identically distributed goods) can be found to be: \[Var(R) = \frac{4n^{2}m^{2} - 4nm(n+m)}{(n+m)^{2}(n+m-1)}\] In conclusion, the mean and variance of R, when the two methods produce identically distributed goods, are: Mean (E[R]) = \[\frac{2nm}{n+m}\] Variance (Var[R]) = \[\frac{4n^{2}m^{2} - 4nm(n+m)}{(n+m)^{2}(n+m-1)}\]

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Key Concepts

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

Continuous Random Variable

In probability theory, a continuous random variable is a variable that can take on an infinite number of possible values. Unlike its discrete counterpart which has countable outcomes, a continuous random variable deals with outcomes that fall within a range on the real number line.

Take the quality of goods in the exercise, for instance. The quality is not just good or bad, but could be any value on a spectrum, which makes it a continuous random variable. For any continuous random variable like this, the probability of the variable taking on a specific single value is actually zero. Instead, probabilities are assigned to ranges of values and are represented by areas under a curve on a graph, known as the probability density function (PDF).

Distribution Function

The distribution function, often referred to as the cumulative distribution function (CDF), is a function that shows the probability that a random variable will take a value less than or equal to a certain threshold.

For continuous random variables, the CDF is a continuous function and helps us determine probabilities over intervals. In the given exercise, the CDFs for the quality of goods from both methods, denoted by \(F_1\) and \(F_2\), describe how likely it is to find a product from either method below a particular quality standard. When \(F_1 = F_2\), it implies that the goods are produced with the same quality distribution, making an analysis like determining the expected number of 'runs' possible.

Expected Value and Variance

Expected value, or mean, is a fundamental concept in statistics and probability, representing the long-run average value of repetitions of an experiment. It's calculated as the sum of all possible values, each multiplied by its probability of occurrence.

Variance, on the other hand, is a measure of how spread out a set of random numbers is. It is the average of the squared differences from the mean. The formula used in the exercise for variance, particular to the number of runs, accounts for this spread by considering the difference between the square of the Runs and the square of the expected Runs. The variance is particularly helpful as it highlights the variability of the number of 'runs' in the ranked goods, indicating the consistency of the sorting.

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations and permutations of objects. It plays a crucial role in calculating probabilities in scenarios where the order or arrangement of objects matters.

The exercise we are looking into employs combinatorial calculations to determine the number of ways to form sequences of goods, or 'runs'. Specifically, it uses the combination to describe the number of ways to arrange the 'n' goods from method 1 and 'm' goods from method 2 in a manner that results in 'R' runs. This value of possible arrangements is essential in determining the probability of forming any particular sequence and is a key concept in solving problems related to probability in discrete systems.