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Chapter 6: Problem 6

Write a computer program to calculate the mean and variance of a distribution which you specify as data. Use the program to compare the variances for the following densities, both having expected value 0 : $$\begin{array}{l}p_{X}=\left(\begin{array}{ccccc} -2 & -1 & 0 & 1 & 2 \\\3 / 11 & 2 / 11 & 1 / 11 & 2 / 11 & 3 / 11\end{array}\right) \\\p_{Y}=\left(\begin{array}{ccccc} -2 & -1 & 0 & 1 & 2 \\\1 / 11 & 2 / 11 & 5 / 11 & 2 / 11 & 1 / 11\end{array}\right)\end{array}$$

Short Answer

Expert verified
The variances are \( \frac{20}{11} \) for \( p_X \) and \( \frac{10}{11} \) for \( p_Y \), with \( p_X \) having the greater variance.

Step by step solution

01

Understanding the Probability Distributions

The problem provides two probability distributions, \( p_X \) and \( p_Y \), both defined over the values \(-2, -1, 0, 1, 2\). For \( p_X \), the probabilities are \( \frac{3}{11}, \frac{2}{11}, \frac{1}{11}, \frac{2}{11}, \frac{3}{11} \) respectively, and for \( p_Y \), they are \( \frac{1}{11}, \frac{2}{11}, \frac{5}{11}, \frac{2}{11}, \frac{1}{11} \). Each distribution has an expected value of 0.
02

Calculating the Mean (Expected Value)

The mean of a distribution is given by the sum of the products of each value and its probability. The mean for \( p_X \) is computed as follows: \[ \mu_X = (-2)\times \frac{3}{11} + (-1)\times \frac{2}{11} + 0\times \frac{1}{11} + 1\times \frac{2}{11} + 2\times \frac{3}{11} = 0. \] Similarly, for \( p_Y \): \[ \mu_Y = (-2)\times \frac{1}{11} + (-1)\times \frac{2}{11} + 0\times \frac{5}{11} + 1\times \frac{2}{11} + 2\times \frac{1}{11} = 0. \] Both means are 0, as given.
03

Calculating the Variance

The variance is calculated using the formula \( \text{Var}(X) = \sum (x_i - \mu)^2 \cdot P(x_i) \). Since the mean is 0, \( \text{Var}(p_X) = \sum (x_i)^2 \cdot P(x_i) \). Therefore: \[ \text{Var}(p_X) = (-2)^2 \cdot \frac{3}{11} + (-1)^2 \cdot \frac{2}{11} + 0^2 \cdot \frac{1}{11} + 1^2 \cdot \frac{2}{11} + 2^2 \cdot \frac{3}{11} = \frac{20}{11}. \] Similarly for \( p_Y \): \[ \text{Var}(p_Y) = (-2)^2 \cdot \frac{1}{11} + (-1)^2 \cdot \frac{2}{11} + 0^2 \cdot \frac{5}{11} + 1^2 \cdot \frac{2}{11} + 2^2 \cdot \frac{1}{11} = \frac{10}{11}. \]
04

Writing the Computer Program

To calculate these variances programmatically, write a function in a language like Python. Example: ```python # Define the data p_x_values = [-2, -1, 0, 1, 2] p_x_probs = [3/11, 2/11, 1/11, 2/11, 3/11] p_y_values = [-2, -1, 0, 1, 2] p_y_probs = [1/11, 2/11, 5/11, 2/11, 1/11] # Function to calculate variance def calculate_variance(values, probabilities): mean = sum(x * p for x, p in zip(values, probabilities)) return sum((x - mean) ** 2 * p for x, p in zip(values, probabilities)) # Calculate variances variance_x = calculate_variance(p_x_values, p_x_probs) variance_y = calculate_variance(p_y_values, p_y_probs) print(f"Variance of p_X: {variance_x}") print(f"Variance of p_Y: {variance_y}") ```
05

Comparing the Variances

By executing the program or directly from our manual calculations, \( p_X \) has a variance of \( \frac{20}{11} \) and \( p_Y \) has a variance of \( \frac{10}{11} \). This implies that \( p_X \) has a larger spread around the mean compared to \( p_Y \).

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Key Concepts

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

Probability Distribution
Probability distributions describe how probabilities are assigned to different values in a dataset. In this problem, we looked at two distributions, \( p_X \) and \( p_Y \), each defined over the values \(-2, -1, 0, 1, 2\). Probability distributions allow us to understand models of real-life phenomena by depicting how likely different outcomes are.

For example, in the distribution \( p_X \), the value \(-2\) has a probability of \( \frac{3}{11} \), while the value \(0\) has a probability of \( \frac{1}{11} \). Distributions are essential for determining other statistical properties, such as mean and variance, that help in understanding the data's characteristics and reliability.
Expected Value
The expected value, often known as the mean of a probability distribution, is the average outcome if you were to repeat an experiment an infinite number of times. For the distributions \( p_X \) and \( p_Y \), calculating the mean helps determine the center of the distribution.

By multiplying each value by its probability and then summing these products, you find the expected value. Mathematically, for \( p_X \), it was calculated as: \( \mu_X = (-2)\times \frac{3}{11} + (-1)\times \frac{2}{11} + 0\times \frac{1}{11} + 1\times \frac{2}{11} + 2\times \frac{3}{11} = 0. \)

Thus, the expected value for both distributions was 0, confirming what we already knew from the problem statement.
Computer Programming
Incorporating computer programming into statistical calculations allows for a more efficient, repeatable, and accurate method of finding statistical measures like mean and variance. The provided Python code showcases how a programmatic approach can simplify complex calculations.

The code defines functions to compute variance, utilizing loops and built-in Python functions. This minimizes errors seen in manual calculations. It permits experimentation with different datasets or altered probabilities to explore other distributions, showcasing the flexibility of using software when dealing with mathematical computations.
Mean
The mean is a measure of central tendency and tells us the balancing point of a distribution. It's crucial when comparing distributions to understand where the data tends to 'gather' around a central point. Both distributions \( p_X \) and \( p_Y \) had calculated means of 0, aligning with the details given in the problem statement.

The mean is calculated by the formula: \[ \text{Mean} = \sum_{i} (x_i \times P(x_i)) \]

This calculation ensures we're averaging the outcomes, considering how likely each one is (probabilities). Recognizing the mean helps in insights into how values are distributed around it and serves as a comparison basis for variance analysis.
Educational Program
Using an educational program to convey how calculations like variance and mean are performed helps build understanding beyond mere theory. With specific tools and code examples, learners get hands-on practice. This doubles as learning material and an applied scenario for students looking to apply mathematical concepts in computer programming.

Structured educational programs teach the 'why' and 'how' of concepts using practical examples, as seen in this exercise. By providing step-by-step progression from understanding definitions to implementing code, these programs build a strong foundation for students factoring in diverse learning styles.
  • Encourage experimentation
  • Promote practical application
  • Support the theoretical understanding

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Most popular questions from this chapter

Find the expected value and the variance for the number of boys and the number of girls in a royal family that has children until there is a boy or until there are three children, whichever comes first.

Suppose that \(n\) people have their hats returned at random. Let \(X_{i}=1\) if the ith person gets his or her own hat back and 0 otherwise. Let \(S_{n}=\sum_{i=1}^{n} X_{i}\) Then \(S_{n}\) is the total number of people who get their own hats back. Show that (a) \(E\left(X_{i}^{2}\right)=1 / n\) (b) \(E\left(X_{i} \cdot X_{j}\right)=1 / n(n-1)\) for \(i \neq j\). (c) \(E\left(S_{n}^{2}\right)=2\) (using (a) and (b)). (d) \(V\left(S_{n}\right)=1\).

The Pilsdorff Beer Company runs a fleet of trucks along the 100 mile road from Hangtown to Dry Gulch. The trucks are old, and are apt to break down at any point along the road with equal probability. Where should the company locate a garage so as to minimize the expected distance from a typical breakdown to the garage? In other words, if \(X\) is a random variable giving the location of the breakdown, measured, say, from Hangtown, and \(b\) gives the location of the garage, what choice of \(b\) minimizes \(E(|X-b|) ?\) Now suppose \(X\) is not distributed uniformly over \([0,100],\) but instead has density function \(f_{X}(x)=2 x / 10,000 .\) Then what choice of \(b\) minimizes \(E(|X-b|) ?\)

A professor wishes to make up a true-false exam with \(n\) questions. She assumes that she can design the problems in such a way that a student will answer the \(j\) th problem correctly with probability \(p_{j},\) and that the answers to the various problems may be considered independent experiments. Let \(S_{n}\) be the number of problems that a student will get correct. The professor wishes to choose \(p_{j}\) so that \(E\left(S_{n}\right)=.7 n\) and so that the variance of \(S_{n}\) is as large as possible. Show that, to achieve this, she should choose \(p_{j}=.7\) for all \(j\); that is, she should make all the problems have the same difficulty.

Suppose you are standing on the bank of a straight river. (a) Choose, at random, a direction which will keep you on dry land, and walk \(1 \mathrm{~km}\) in that direction. Let \(P\) denote your position. What is the expected distance from \(P\) to the river? (b) Now suppose you proceed as in part (a), but when you get to \(P\), you pick a random direction (from among all directions) and walk \(1 \mathrm{~km}\). What is the probability that you will reach the river before the second walk is completed?
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