Chapter 26: Problem 35
The continuous random variables \(X\) and \(Y\) have a joint PDF proportional to \(x y(x-y)^{2}\) with \(0 \leq x \leq 1\) and \(0 \leq y \leq 1 .\) Find the marginal distributions for \(X\) and \(Y\) and show that they are negatively correlated with correlation coefficient \(-\frac{2}{3}\)
Short Answer
Expert verified
The marginal distributions are obtained by integrating the joint PDF. They show that X and Y are negatively correlated, with a correlation coefficient of -\(\frac{2}{3}\).
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Marginal Distributions
Marginal distributions tell us the probability distribution of a single random variable within a joint distribution.
To find the marginal distribution of a variable, we integrate the joint probability density function (PDF) over the range of the other variable.
In our exercise, we have two random variables, X and Y, with a joint PDF proportional to \(x y (x-y)^2\).
To find the marginal distribution of X, we integrate the joint PDF over the entire range of Y:
\[ f_X(x) = \int_{0}^{1} k x y (x-y)^2 \, dy \]
Similarly, to find the marginal distribution of Y, we integrate the joint PDF over the entire range of X:
\[ f_Y(y) = \int_{0}^{1} k x y (x-y)^2 \, dx \]
These integrations simplify the joint PDF to individual PDFs for X and Y.
To find the marginal distribution of a variable, we integrate the joint probability density function (PDF) over the range of the other variable.
In our exercise, we have two random variables, X and Y, with a joint PDF proportional to \(x y (x-y)^2\).
To find the marginal distribution of X, we integrate the joint PDF over the entire range of Y:
\[ f_X(x) = \int_{0}^{1} k x y (x-y)^2 \, dy \]
Similarly, to find the marginal distribution of Y, we integrate the joint PDF over the entire range of X:
\[ f_Y(y) = \int_{0}^{1} k x y (x-y)^2 \, dx \]
These integrations simplify the joint PDF to individual PDFs for X and Y.
Correlation Coefficient
The correlation coefficient, denoted as \(\rho\), measures the strength and direction of the linear relationship between two random variables.
It takes values between -1 and 1:
\[ \rho_{XY} = \frac{Cov(X,Y)}{\sqrt{Var(X) Var(Y)}} \]
In this exercise, we are asked to show that X and Y are negatively correlated with \(\rho = -\frac{2}{3}\).
To do so, we need to calculate the covariance and variances of X and Y, and then use them in the formula.
It takes values between -1 and 1:
- \(\rho = 1\) indicates a perfect positive linear relationship.
- \(\rho = 0\) indicates no linear relationship.
- \(\rho = -1\) indicates a perfect negative linear relationship.
\[ \rho_{XY} = \frac{Cov(X,Y)}{\sqrt{Var(X) Var(Y)}} \]
In this exercise, we are asked to show that X and Y are negatively correlated with \(\rho = -\frac{2}{3}\).
To do so, we need to calculate the covariance and variances of X and Y, and then use them in the formula.
Expected Values
Expected values give us the mean or average value of a random variable over numerous trials.
For a continuous random variable, the expected value is calculated using the integral of the product of the variable and its PDF:
\[ E[X] = \int_{-\infty}^{\infty} x f_X(x) \, dx \]
\[ E[Y] = \int_{-\infty}^{\infty} y f_Y(y) \, dy \]
In our problem, we also need the joint expectation:
\[ E[XY] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} xy f_{X,Y}(x,y) \, dx \, dy \]
Calculating these expected values provides the necessary information to find variance and covariance.
For a continuous random variable, the expected value is calculated using the integral of the product of the variable and its PDF:
\[ E[X] = \int_{-\infty}^{\infty} x f_X(x) \, dx \]
\[ E[Y] = \int_{-\infty}^{\infty} y f_Y(y) \, dy \]
In our problem, we also need the joint expectation:
\[ E[XY] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} xy f_{X,Y}(x,y) \, dx \, dy \]
Calculating these expected values provides the necessary information to find variance and covariance.
Covariance
Covariance indicates the direction of the linear relationship between two random variables.
If both variables tend to increase together, the covariance is positive.
If one tends to increase when the other one decreases, the covariance is negative.
The formula for covariance is:
\[ Cov(X,Y) = E[XY] - E[X]E[Y] \]
In the given problem, after finding \(E[X]\), \(E[Y]\), and \(E[XY]\), we use these values in the covariance formula to determine the linear dependency between X and Y.
If both variables tend to increase together, the covariance is positive.
If one tends to increase when the other one decreases, the covariance is negative.
The formula for covariance is:
\[ Cov(X,Y) = E[XY] - E[X]E[Y] \]
In the given problem, after finding \(E[X]\), \(E[Y]\), and \(E[XY]\), we use these values in the covariance formula to determine the linear dependency between X and Y.
Double Integral
A double integral is used to compute the integral of a function over a two-dimensional region.
In this exercise, the double integral is employed to find the constant of proportionality for the joint PDF:
\[ \text{Constant} \times \int_{0}^{1} \int_{0}^{1} x y (x-y)^2 \, dy \, dx = 1 \]
Solving this double integral helps normalize the joint PDF so that the total probability is 1.
Breaking it down:
In this exercise, the double integral is employed to find the constant of proportionality for the joint PDF:
\[ \text{Constant} \times \int_{0}^{1} \int_{0}^{1} x y (x-y)^2 \, dy \, dx = 1 \]
Solving this double integral helps normalize the joint PDF so that the total probability is 1.
Breaking it down:
- First, integrate with respect to y across its range.
- Then, integrate the resulting function with respect to x across its range.
