Problem 196 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose that \(r\) chips are drawn with replacement from an urn containing \(n\) chips, numbered 1 through \(n\). Let \(V\) denote the sum of the numbers drawn. Find \(E(V)\).

Step-by-Step Solution

Answer

The expected sum of the numbers drawn, \(E(V)\), is \(r \times ((n+1)/2)\).

Step 1: Define the random variable

There exists a random variable \(X_i\) for \(i = 1, 2, ..., r\), which represents the number on the \(i\)-th draw. \(V = X_1 + X_2 + ... + X_r\). It's important to note that the draws are with replacement, so each \(X_i\) ranges from 1 to \(n\) and they are independent.

Step 2: Find the Expected Value of a Single Random Variable \(X_i\)

The expected value of a single random variable \(E(X_i)\) can be found using the formula for the expected value of a discrete random variable which is \(\sum (value \times probability)\). Since the draws are with replacement, the probability of each number from 1 to \(n\) is the same, or \(1/n\). So, \(E(X_i) = \sum_{k=1}^{n} k * (1/n) = (1/n) \times (1 + 2 + ... + n) = (1/n) \times n*(n+1)/2 = (n+1)/2\)

Step 3: Calculate the Expected Value of \(V\)

Since \(V\) is the sum of \(r\) independent random variables, the expected value of \(V\) is the sum of the expected values of the \(r\) variables. From Step 2, we know that each \(E(X_i) = (n+1)/2\). So, \(E(V) = E(X_1 + X_2 + ... + X_r) = E(X_1) + E(X_2) + ... + E(X_r) = r \times ((n+1)/2)\).

Key Concepts

Random Variable

In probability and statistics, a random variable is a variable that can take on different values randomly, each with a certain probability. Imagine you're drawing chips from an urn. Each time you draw a chip, the number on the chip is a random outcome. This number can be any value between 1 and \(n\).

In this exercise, the random variable is denoted by \(X_i\), where \(i\) represents the \(i\)-th draw. Since this is a random process, \(X_i\) can vary from one draw to the next. Consequently, the value of \(X_i\) is not fixed until a draw occurs.

One core feature of random variables is that they come with a probability distribution – this means each possible value has a particular chance of being the outcome. In our case, because draws are conducted with replacement, each number is equally likely to appear in each draw.

Independent Events

When two or more events are independent, the outcome of one event does not influence the outcomes of the others. This principle is crucial when dealing with random variables in probability.

For our chip drawing exercise, each draw is performed with replacement. This means every time a chip is drawn, it is put back into the urn before the next draw. As a result, the outcome of drawing a chip doesn’t impact future draws. Each draw is an independent event.

Knowing that the events are independent helps simplify the process of finding the expected value of the sum of all draws, since the expected value of independent events can be calculated just by adding their expected values. This is why we can sum up the expected value of each individual draw to get the expected value of the entire sum \(V\).

Discrete Probability Distribution

A discrete probability distribution is used when dealing with events that have distinct and separate outcomes. In our exercise, we're dealing with the numbers on the chips, which are discrete and finite, ranging from 1 to \(n\).

This distribution tells us how the probabilities are spread over the possible outcomes. For this particular distribution, each number from 1 to \(n\) has an equal chance of being drawn because each draw is independent and with replacement. Hence, the probability for each number is \(1/n\).

In the context of expected value, knowing the discrete probability distribution allows us to calculate the average or expected value of these numbers. We sum up the products of each possible value and its probability – a straightforward task given our uniform distribution here. This comprehensive approach enables us to understand the likelihood of different outcomes and derive a meaningful average from them.