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Chapter 5: Problem 1

Determine the covariance and correlation for the following joint probability distribution: $$ \begin{array}{lllll} x & 1 & 1 & 2 & 4 \\ y & 3 & 4 & 5 & 6 \\ f_{x y}(x, y) & 1 / 8 & 1 / 4 & 1 / 2 & 1 / 8 \end{array} $$

Short Answer

Expert verified
Covariance is \(\frac{1}{4}\) and correlation is approximately \(0.227\).

Step by step solution

01

Calculate Expected Values

First, we need to calculate the expected values of the random variables \( X \) and \( Y \). The expected value of \( X \) is:\[ E(X) = \sum_{i=1}^{n} x_i \cdot f_{x,y}(x_i, y_i) = 1 \cdot \frac{1}{8} + 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{2} + 4 \cdot \frac{1}{8} = \frac{1}{8} + \frac{2}{8} + \frac{8}{8} + \frac{4}{8} = \frac{15}{8} \]. Similarly, for \( Y \), \[ E(Y) = 3 \cdot \frac{1}{8} + 4 \cdot \frac{1}{4} + 5 \cdot \frac{1}{2} + 6 \cdot \frac{1}{8} = \frac{3}{8} + \frac{4}{8} + \frac{10}{8} + \frac{6}{8} = \frac{23}{8} \].
02

Calculate Covariance

The covariance \( \text{Cov}(X,Y) \) is computed as \[ \text{Cov}(X,Y) = \sum_{i=1}^{n} (x_i - E(X))(y_i - E(Y))f_{x, y}(x_i, y_i) \]. Applying this formula, we find \( \text{Cov}(X,Y) = (1 - \frac{15}{8})(3 - \frac{23}{8})\frac{1}{8} + (1 - \frac{15}{8})(4 - \frac{23}{8})\frac{1}{4} + (2 - \frac{15}{8})(5 - \frac{23}{8})\frac{1}{2} + (4 - \frac{15}{8})(6 - \frac{23}{8})\frac{1}{8} \). Solving this gives \( \text{Cov}(X,Y) = \frac{1}{4} \).
03

Calculate Variance of X and Y

We calculate \( \, \text{Var}(X) \, \) as \[ \text{Var}(X) = \sum_{i=1}^{n} (x_i - E(X))^2 f_{x, y}(x_i, y_i) \]. Substitute the values: \( \text{Var}(X) = (1 - \frac{15}{8})^2\frac{1}{8} + (1 - \frac{15}{8})^2\frac{1}{4} + (2 - \frac{15}{8})^2\frac{1}{2} + (4 - \frac{15}{8})^2\frac{1}{8} \) gives \( \frac{23}{32} \). Similarly, calculate \( \text{Var}(Y) = \frac{15}{32} \).
04

Calculate Correlation

The correlation coefficient \( \rho(X,Y) \) is \( \frac{\text{Cov}(X,Y)}{\sqrt{\text{Var}(X) \cdot \text{Var}(Y)}} \). Substitute the values: \[ \rho(X,Y) = \frac{\frac{1}{4}}{\sqrt{\frac{23}{32} \cdot \frac{15}{32}}} = \frac{1}{4} \cdot \frac{32}{\sqrt{345}} \approx 0.227 \].

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Key Concepts

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

Joint Probability Distribution
Joint probability distribution is a crucial concept in statistics and probability which describes the simultaneous distribution of two random variables. Consider the example of variables \( X \) and \( Y \), where the distribution is given as pairs of \( x \) and \( y \) values along with their corresponding probabilities, as outlined in our exercise. The joint probability distribution helps us understand how the combination of these variables behave collectively.

Key aspects include:
  • The sum of all joint probabilities must equal 1, which reflects the certainty of one of the potential outcomes occurring.
  • Each pair \( (x,y) \) has an associated probability \( f_{x,y}(x, y) \), indicating the probability of \( X \) taking a value \( x \), while \( Y \) takes value \( y \).
Understanding joint probability distributions allow us to explore interactions between random variables, a foundation for calculating covariance and correlation.
Expected Value
Expected value is an essential concept when working with probability distributions. For a random variable, the expected value, often referred to as the mean, provides a measure of the center or average of the distribution.

In the context of our joint probability distribution:
  • For X: The formula to calculate the expected value is \( E(X) = \sum_{i=1}^{n} x_i \cdot f_{x,y}(x_i, y_i) \). We plug in our values to determine \( E(X) = \frac{15}{8} \).
  • For Y: Similarly, \( E(Y) = \sum_{i=1}^{n} y_i \cdot f_{x,y}(x_i, y_i) \), resulting in \( E(Y) = \frac{23}{8} \).
Essentially, the expected value serves as a weighted average of possible values, with the weights being the probabilities.
Variance
Variance is a key measure in statistics that provides insight into the spread or dispersion of a set of data points. It helps to determine how much the values of the random variable deviate from the expected value.

When calculating the variance from a joint probability distribution:
  • Variance of X, \( \text{Var}(X) \), is computed using \( \text{Var}(X) = \sum_{i=1}^{n} (x_i - E(X))^2 f_{x, y}(x_i, y_i) \) resulting in \( \frac{23}{32} \).
  • Variance of Y, \( \text{Var}(Y) \), follows a similar approach providing \( \frac{15}{32} \) as the result.
The variance is critical for understanding how much variability exists, leading us to grasp the overall uncertainty and reliability of the mean.
Correlation Coefficient
Correlation coefficient, denoted as \( \rho(X,Y) \), is a statistical indicator that measures the degree to which two variables move in relation to one another. Ranging between -1 and 1, this measure shows the strength and direction of a linear relationship between variables.

Calculation in our exercise involves:
  • The formula \( \rho(X,Y) = \frac{\text{Cov}(X,Y)}{\sqrt{\text{Var}(X) \cdot \text{Var}(Y)}} \).
  • Applying values gives \( \rho(X,Y) \approx 0.227 \).
A positive correlation suggests that as one variable increases, the other tends to increase as well, which reflects our calculated correlation. Thus, the correlation coefficient provides a concise summary of the relationship between the two variables.

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Most popular questions from this chapter

The systolic and diastolic blood pressure values (mm Hg) are the pressures when the heart muscle contracts and relaxes (denoted as \(Y\) and \(X,\) respectively). Over a collection of individuals, the distribution of diastolic pressure is normal with mean 73 and standard deviation \(8 .\) The systolic pressure is conditionally normally distributed with mean \(1.6 x\) when \(X=x\) and standard deviation of \(10 .\) Determine the following: a. Conditional probability density function of \(Y\) given \(X=73\) b. \(P(Y<115 \mid X=73)\) c. \(E(Y \mid X=73)\) d. Recognize the distribution \(f_{X Y}(x, y)\) and identify the mean and variance of \(Y\)

Show that the following function satisfies the properties of a joint probability mass function: \(\begin{array}{ccc}x & y & f(x, y) \\ 0 & 0 & 1 / 4 \\ 0 & 1 & 1 / 8 \\ 1 & 0 & 1 / 8 \\ 1 & 1 & 1 / 4 \\ 2 & 2 & 1 / 4\end{array}\) Determine the following: a. \(P(X<0.5, Y<1.5)\) b. \(P(X \leq 1)\) c. \(P(X<1.5)\) d. \(P(X>0.5, Y<1.5)\) e. \(E(X), E(Y), V(X), V(Y)\) f. Marginal probability distribution of the random variable \(X\) g. Conditional probability distribution of \(Y\) given that \(X=1\) h. \(E(Y \mid X=1)\) i. Are \(X\) and \(Y\) independent? Why or why not? j. Correlation between \(X\) and \(Y\).

Suppose that \(X\) is a random variable with probability distribution $$f_{X}(x)=1 / 4, \quad x=1,2,3,4$$ Determine the probability distribution of \(Y=2 X+1\).

A small-business Web site contains 100 pages and \(60 \%, 30 \%,\) and \(10 \%\) of the pages contain low, moderate, and high graphic content, respectively. A sample of four pages is selected randomly without replacement, and \(X\) and \(Y\) denote the number of pages in the sample with moderate and high graphics output. Determine: a. \(f_{X Y}(x, y)\) b. \(f_{X}(x)\) c. \(E(X)\)

The permeability of a membrane used as a moisture barrier in a biological application depends on the thickness of two integrated layers. The layers are normally distributed with means of 0.5 and 1 millimeters, respectively. The standard deviations of layer thickness are 0.1 and 0.2 millimeters, respectively. The correlation between layers is \(0.7 .\) a. Determine the mean and variance of the total thickness of the two layers. b. What is the probability that the total thickness is less than I millimeter? c. Let \(X_{1}\) and \(X_{2}\) denote the thickness of layers 1 and 2 , respectively. A measure of performance of the membrane is a function of \(2 X_{1}+3 X_{2}\) of the thickness. Determine the mean and variance of this performance measure.
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