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The joint probability distribution is $$ \begin{array}{llcll} x & -1 & 0 & 0 & 1 \\ y & 0 & -1 & 1 & 0 \\ f_{X Y}(x, y) & 1 / 4 & 1 / 4 & 1 / 4 & 1 / 4 \end{array} $$ Show that the correlation between \(X\) and \(Y\) is zero, but \(X\) and \(Y\) are not independent.

Short Answer

Expert verified
Correlation is zero, but variables are not independent because joint probabilities do not equal the product of marginals in general.

Step by step solution

01

Calculate Means

First, calculate the expected values of \(X\) and \(Y\). For \(X\), \(E[X] = \sum (x \cdot P(X=x)) = (-1 \times \frac{1}{4}) + (0 \times \frac{1}{4}) + (0 \times \frac{1}{4}) + (1 \times \frac{1}{4}) = 0\). Similarly, for \(Y\), \(E[Y] = (0 \times \frac{1}{4}) + (-1 \times \frac{1}{4}) + (1 \times \frac{1}{4}) + (0 \times \frac{1}{4}) = 0\). Thus, both means are \(0\).
02

Calculate Covariance

Next, calculate the covariance of \(X\) and \(Y\). \(\text{Cov}(X, Y) = E[(X - E[X])(Y - E[Y])] = E[XY] - E[X]E[Y]\). Since \(E[X] = 0\) and \(E[Y] = 0\), it simplifies to \(\text{Cov}(X, Y) = E[XY]\). Now calculate \(E[XY] = \sum (x \cdot y \cdot P(X=x, Y=y)) = (-1\cdot0\cdot\frac{1}{4}) + (0\cdot(-1)\cdot\frac{1}{4}) + (0\cdot1\cdot\frac{1}{4}) + (1\cdot0\cdot\frac{1}{4}) = 0\). So \(\text{Cov}(X, Y) = 0\).
03

Calculate Variances

Calculate the variance of both \(X\) and \(Y\). For \(X\), \(\text{Var}(X) = E[X^2] - (E[X])^2 = E[X^2] = (-1)^2\cdot\frac{1}{4} + 0^2\cdot\frac{1}{4} + 0^2\cdot\frac{1}{4} + 1^2\cdot\frac{1}{4} = \frac{1}{2}\). Similarly for \(Y\), \(\text{Var}(Y) = E[Y^2] = 0^2\cdot\frac{1}{4} + (-1)^2\cdot\frac{1}{4} + 1^2\cdot\frac{1}{4} + 0^2\cdot\frac{1}{4} = \frac{1}{2}\).
04

Calculate Correlation

Calculate the correlation coefficient \(\rho(X,Y)\). Using \(\rho(X,Y) = \frac{\text{Cov}(X, Y)}{\sqrt{\text{Var}(X)} \cdot \sqrt{\text{Var}(Y)}}\), substitute the known values: \(\rho(X,Y) = \frac{0}{\sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2}}} = 0\). Thus, the correlation is zero.
05

Check Independence

To show independence, verify if \(P(X=x, Y=y) = P(X=x)P(Y=y)\). For \((-1, 0)\), \(P(X=-1, Y=0) = \frac{1}{4}\), but \(P(X=-1) = \frac{1}{4}\times\frac{1}{4}\), which means the joint probability is equal to the product of marginals only by chance. Check the other pairs to find that the condition fails for other \((x,y)\) pairs, showing \(X\) and \(Y\) are not independent.

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Key Concepts

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

Correlation Coefficient
The correlation coefficient is a measure that tells us about the strength and direction of a linear relationship between two variables. It is represented by the Greek letter \( \rho \) (rho), and its value ranges between -1 and 1. A value of 1 implies a perfect positive linear relationship, -1 indicates a perfect negative linear relationship, and 0 suggests no linear relationship. In our exercise, we calculated the correlation coefficient between \( X \) and \( Y \) using the formula:\[\rho(X,Y) = \frac{\text{Cov}(X, Y)}{\sqrt{\text{Var}(X)} \cdot \sqrt{\text{Var}(Y)}}\]Given that the covariance is zero, as computed earlier, the result comes out to be zero, which suggests that there is no linear correlation between \( X \) and \( Y \). However, it does not mean there is no relationship at all. It merely indicates an absence of linearity in their relationship.
Covariance
Covariance is a statistical measure that indicates the extent to which two random variables change together. If the covariance is positive, it implies that as one variable increases, the other tends to increase as well. If it's negative, as one variable increases, the other tends to decrease. When the covariance is zero, it suggests that there is no linear relationship between the variables.

In our joint probability distribution exercise, we found the covariance \( \text{Cov}(X,Y) \) by using the formula:\[\text{Cov}(X, Y) = E[XY] - E[X]E[Y]\]Considering that the expected values were zero for both \( X \) and \( Y \), this simplifies to calculating \( E[XY] \), which also turns out to be zero. Hence, the covariance is zero, indicating no linear relationship between \( X \) and \( Y \). However, this does not automatically imply independence.
Independence in Probability
Independence between two variables in probability implies that the occurrence of one event does not affect the probability of the occurrence of the other. Mathematically, \( X \) and \( Y \) are independent if and only if the joint probability distribution can be expressed as the product of the individual marginal distributions:\[P(X=x, Y=y) = P(X=x) \cdot P(Y=y)\]

In our exercise, we checked the joint probabilities for different pairs of \( (x, y) \) that revealed the joint probability \( P(X=x, Y=y) \) does not always equal \( P(X=x) \times P(Y=y) \). This failure to meet the independence condition for all pairs confirms that \( X \) and \( Y \) are not independent, providing a crucial insight often behind zero correlation. Hence, it is essential to remember that zero covariance or correlation does not imply independence.
Variance
Variance is a statistic that measures the dispersion or spread in a set of values. Essentially, it helps us understand how far the data points in a distribution deviate from the mean. The variance of a random variable \( X \) is symbolized as \( \text{Var}(X) \) and is calculated by:\[\text{Var}(X) = E[X^2] - (E[X])^2\]The square root of the variance is known as the standard deviation.

In the exercise, the variances \( \text{Var}(X) \) and \( \text{Var}(Y) \) were computed. Using the joint distribution values, we determined both variances to be 0.5. This tells us that the data values of \( X \) and \( Y \) have similar deviations from their means, given they are inherently symmetric around the origin. Understanding variance is crucial as it lays the foundation for more complex statistics like the covariance and correlation coefficient.

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Most popular questions from this chapter

The percentage of people given an antirheumatoid medication who suffer severe, moderate, or minor side effects are \(10,20,\) and \(70 \%,\) respectively. Assume that people react independently and that 20 people are given the medication. Determine the following: (a) The probability that \(2,4,\) and 14 people will suffer severe, moderate, or minor side effects, respectively (b) The probability that no one will suffer severe side effects (c) The mean and variance of the number of people who will suffer severe side effects (d) What is the conditional probability distribution of the number of people who suffer severe side effects given that 19 suffer minor side effects? (e) What is the conditional mean of the number of people who suffer severe side effects given that 19 suffer minor side effects?

Suppose the random variables \(X, Y\), and \(Z\) have the following joint probability distribution. $$ \begin{array}{cccc} \hline x & y & z & f(x, y, z) \\ \hline 1 & 1 & 1 & 0.05 \\ 1 & 1 & 2 & 0.10 \\ 1 & 2 & 1 & 0.15 \\ 1 & 2 & 2 & 0.20 \\ 2 & 1 & 1 & 0.20 \\ 2 & 1 & 2 & 0.15 \\ 2 & 2 & 1 & 0.10 \\ 2 & 2 & 2 & 0.05 \\ \hline \end{array} $$ Determine the following: (a) \(P(X=2)\) (b) \(P(X=1, Y=2)\) (c) \(P(Z < 1.5)\) (d) \(P(X=1\) or \(Z=2)\) (e) \(E(X)\) (f) \(P(X=1 \mid Y=1)\) (g) \(P(X=1, Y=1 \mid Z=2)\) (h) \(P(X=1 \mid Y=1, Z=2)\) (i) Conditional probability distribution of \(X\) given that \(Y=1\) and \(Z=2\)

The weight of adobe bricks for construction is normally distributed with a mean of 3 pounds and a standard deviation of 0.25 pound. Assume that the weights of the bricks are independent and that a random sample of 25 bricks is chosen. (a) What is the probability that the mean weight of the sample is less than 2.95 pounds? (b) What value will the mean weight exceed with probability \(0.99 ?\)

Suppose that the joint probability function of the continuous random variables \(X\) and \(Y\) is constant on the rectangle \(0 < x < a, 0 < y < b\). Show that \(X\) and \(Y\) are independent.

In the manufacture of electroluminescent lamps, several different layers of ink are deposited onto a plastic substrate. The thickness of these layers is critical if specifications regarding the final color and intensity of light are to be met. Let \(X\) and \(Y\) denote the thickness of two different layers of ink. It is known that \(X\) is normally distributed with a mean of 0.1 millimeter and a standard deviation of 0.00031 millimeter, and \(Y\) is normally distributed with a mean of 0.23 millimeter and a standard deviation of 0.00017 millimeter. The value of \(\rho\) for these variables is equal to zero. Specifications call for a lamp to have a thickness of the ink corresponding to \(X\) in the range of 0.099535 to 0.100465 millimeter and \(Y\) in the range of 0.22966 to 0.23034 millimeter. What is the probability that a randomly selected lamp will conform to specifications?

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