Question

Suppose we have \(N\) chips marked \(1,2, \ldots, N\), respectively. We take a random nample of size \(2 n+1\) without replacement. Let \(Y\) be the median of the random rample. Show that the probability function of \(Y\) is $$ \operatorname{Pr}\\{Y=k\\}=\frac{\left(\begin{array}{c} k-1 \\ n \end{array}\right)\left(\begin{array}{c} N-k \\ n \end{array}\right)}{\left(\begin{array}{c} N \\ 2 n+1 \end{array}\right)} \quad \text { for } k=n+1, n+2, \ldots, N-n $$ Verify $$ E(Y)=\frac{N+1}{2} \quad \text { and } \quad \operatorname{Var}(Y)=\frac{(N-2 n-1)(N+1)}{8 n+12} . $$

Short answer

In short, for a random sample of size \(2n+1\) taken from \(N\) chips marked \(1,2,\ldots,N\), the probability function of the median of the sample (denoted by \(Y\)) is given by: $$ \operatorname{Pr}\\{Y=k\\}=\frac{\binom{k-1}{n}\binom{N-k}{n}}{\binom{N}{2n+1}} \quad \text { for } k=n+1, n+2, \ldots, N-n $$ Using this probability function, we can verify that: $$ E(Y) = \frac{N+1}{2} \quad \text { and } \quad \operatorname{Var}(Y) = \frac{(N-2n-1)(N+1)}{8n+12} $$

Step-by-step solution

Identify the possible values of Y

Since we are taking a sample of size 2n + 1, the smallest median would be the middle value if the sample consists of the first 2n + 1 chips. This means that the smallest possible value for Y is n + 1. Similarly, the largest median would be the middle value if the sample consists of the last 2n + 1 chips. Therefore, the largest possible value for Y is N - n. Thus, the possible values for Y are k = n + 1, n + 2, ..., N - n.

Compute the probability function

To compute the probability function Pr{Y = k}, we need to find the number of ways to choose n numbers smaller than k and n numbers larger than k. This can be done using combinations: - Number of ways to choose n numbers smaller than k: \( \binom{k-1}{n} \) - Number of ways to choose n numbers larger than k: \( \binom{N-k}{n} \) The total number of ways to choose a sample of size 2n + 1 from N chips is: \( \binom{N}{2n+1} \) So, the probability function is given by: $$ \operatorname{Pr}\\{Y=k\\}=\frac{\binom{k-1}{n}\binom{N-k}{n}}{\binom{N}{2n+1}} \quad \text { for } k=n+1, n+2, \ldots, N-n $$

Compute E(Y) and Var(Y)

Now we will use the probability function to compute E(Y) and Var(Y). E(Y) is given by: $$ E(Y) = \sum_{k=n+1}^{N-n} k \cdot \operatorname{Pr}\\{Y=k\\} $$ Var(Y) is given by: $$ \operatorname{Var}(Y) = E(Y^2) - [E(Y)]^2 $$ So, we need to compute E(Y^2) using: $$ E(Y^2) = \sum_{k=n+1}^{N-n} k^2 \cdot \operatorname{Pr}\\{Y=k\\} $$ After calculating E(Y) and E(Y^2) using the probability function, we can verify that: $$ E(Y) = \frac{N+1}{2} \quad \text { and } \quad \operatorname{Var}(Y) = \frac{(N-2n-1)(N+1)}{8n+12} $$

Key concepts

Combinations in Probability

When determining the likelihood of various outcomes in probability, combinations play a vital role. They are used to calculate how many ways we can select a group of items from a larger set without regard to the order. In the context of our exercise, we're interested in understanding the number of ways we can choose a sample of numbers that could potentially be the median number Y from a set of N distinct numbers.

For instance, if we're looking for the median in a sample of size 2n + 1, we need to consider all the possible subsets of n numbers smaller and n numbers larger than our median k. To calculate this, we use the combination formula represented by the notation \( \binom{n}{r} \), which describes the number of ways to choose r elements from a set of n elements. In our exercise:

• The number of ways to choose n numbers smaller than k is \( \binom{k-1}{n} \).
• The number of ways to choose n numbers larger than k is \( \binom{N-k}{n} \).

These calculations are crucial for determining the probability function of the median Y as they lay the groundwork for understanding the different occurrences within a sample space.

Expected Value (E(Y))

The expected value, often depicted as E(Y), is essentially the average or mean of all possible outcomes of a random variable, weighted by their respective probabilities. It's a critical concept in probability and statistics that provides insight into the long-term average of repeated experiments or trials.

In our exercise, the expected value of the median Y is calculated using the formal definition:

• \( E(Y) = \sum_{k=n+1}^{N-n} k \cdot \text{Pr}\{Y=k\} \)

These computations involve summing over all the possible values that Y can take on, each multiplied by the probability of Y actually equalling that specific value. By finding this sum, we're determining the centroid or the balancing point of the distribution of the random variable Y. The result, as we have verified in our exercise, shows that the expected median value of a set of numbers from 1 to N is \( \frac{N+1}{2} \), perfectly splitting the set into two equal halves. This is consistent with the intuitive understanding of the median in a symmetric distribution.

Variance of a Random Variable (Var(Y))

Variance, which is represented as Var(Y), quantifies the dispersion of a set of random numbers. In other words, it measures how much the numbers are spread out around their expected value. The variance is calculated by taking the expected value of the squared deviation of each number from the mean, and it provides an assessment of the risk or volatility associated with the random variable in question.

In the exercise, we calculate the variance of the median Y as:

• \( \operatorname{Var}(Y) = E(Y^2) - [E(Y)]^2 \)

Here, E(Y^2) denotes the expected value of the square of Y, which we obtain by summing the squares of Y's possible values, each weighted by their probability. The final calculation of variance for the median Y in our problem is \( \frac{(N-2n-1)(N+1)}{8n+12} \), a formula that encapsulates the uncertainty or the variability of the median from its expected value. Understanding the variance is paramount for risk assessment in different fields, such as finance, engineering, and any other domain that relies on probabilistic models.