Question

Two continuous random variables \(X\) and \(Y\) have a joint probability distribution $$ f(x, y)=A\left(x^{2}+y^{2}\right) $$ where \(A\) is a constant and \(0 \leq x \leq a, 0 \leq y \leq a\). Show that \(X\) and \(Y\) are negatively correlated with correlation coefficient \(-15 / 73 .\) By sketching a rough contour map of \(f(x, y)\) and marking off the regions of positive and negative correlation, convince yourself that this (perhaps counter-intuitive) result is plausible.

Short answer

The random variables are negatively correlated with a correlation coefficient of \(-\frac{15}{73}\). The symmetry of the joint distribution supports this result.

Step-by-step solution

Understand the joint probability density function

Given the joint probability density function: \[ f(x, y) = A (x^2 + y^2) \] where \(A\) is a constant and \(0 \leq x \leq a\), \(0 \leq y \leq a\).

Determine the value of the constant \(A\)

The total probability must be 1. Therefore, integrate the joint probability density function over the region: \[ \int_{0}^{a} \int_{0}^{a} A (x^2 + y^2) \, dx \, dy = 1 \] Solve this to find the value of \(A\).

Perform the double integration

Perform the integration: \[ \int_{0}^{a} \int_{0}^{a} A (x^2 + y^2) \, dx \, dy = A \int_{0}^{a} \int_{0}^{a} (x^2 + y^2) \, dx \, dy \] \[ = A \left[ \int_{0}^{a} x^2 \, dx \int_{0}^{a} \, dy + \int_{0}^{a} \int_{0}^{a} y^2 \, dy \, dx \right]\] Calculate the individual integrals and sum them.

Solve the integrals for \( x \) and \( y \)

First, compute the inner integrals: \[ \int_{0}^{a} x^2 \, dx = \left( \frac{x^3}{3} \right) \Big|_0^a = \frac{a^3}{3} \] Similarly, \[ \int_{0}^{a} y^2 \, dy = \frac{a^3}{3} \] Then: \[ A \left( \frac{a^3}{3} \cdot a + a \cdot \frac{a^3}{3} \right) = 1 \] \[ A \left( \frac{a^4}{3} + \frac{a^4}{3} \right) = 1 \] \[ A \cdot \frac{2a^4}{3} = 1 \] \[ A = \frac{3}{2a^4} \]

Compute marginal distributions

To find the marginal distributions \( f_X (x) \) and \( f_Y (y) \), integrate out the other variable: \[ f_X (x) = \int_0^a f(x, y) \, dy = \int_0^a \frac{3}{2a^4} (x^2 + y^2) \, dy \] \[ f_Y (y) = \int_0^a f(x, y) \, dx = \int_0^a \frac{3}{2a^4} (x^2 + y^2) \, dx \] Compute these integrals.

Integrate to find marginal PDFs

Perform the integration: \[ f_X (x) = \frac{3}{2a^4} \left[ x^2 y + \frac{y^3}{3} \right]_0^a = \frac{3}{2a^4} \left( x^2 a + \frac{a^3}{3} \right) = \frac{3}{2a^4} \left( x^2 a + \frac{a^3}{3} \right) = \frac{3x^2 a}{2a^4} + \frac{a^3}{2a^4} = \frac{3x^2}{2a^3} + \frac{1}{2a} \] Do the same for \(f_Y(y)\).

Find the expectations

Calculate the expectations using marginal PDFs: \[ E[X] = \int_0^a x f_X(x) \, dx \] \[ E[Y] = \int_0^a y f_Y(y) \, dy \] Compute these integrals.

Calculate covariance and variance

The covariance is given by: \[ \text{Cov}(X, Y) = E[XY] - E[X]E[Y] \] Find \(E[XY]\) by integrating the product \(xy\) with the joint density: \[ E[XY] = \int_0^a \int_0^a xy f(x, y) \, dy \, dx \] Calculate the variance: \[ \text{Var}(X) = E[X^2] - (E[X])^2 \] \[ \text{Var}(Y) = E[Y^2] - (E[Y])^2 \]

Determine the correlation coefficient

The correlation coefficient \(\rho\) is: \[ \rho = \frac{\text{Cov}(X, Y)}{\sqrt{\text{Var}(X) \cdot \text{Var}(Y)}} \] Substitute the values found for covariance and variance to show that \(\rho = -\frac{15}{73}\).

Sketch a rough contour map

Sketch the contour map of \(f(x, y)\) showing the regions of positive and negative correlation. Explain why the negative correlation is plausible based on the shape and symmetry of the distribution.

Key concepts

continuous random variables

Continuous random variables are variables that can take any value within a given range. For example, in a real-world scenario, this could be the height of students in a class. Unlike discrete random variables, which only take on specific values (like the number of students), continuous random variables can assume any value within an interval. The joint probability distribution you see here describes how two continuous random variables, X and Y, are related over a range. In this context, the joint probability density function (PDF) is continuous since it applies over a continuous area for both x and y (0 to a).

The function given, \( f(x, y) = A\left( x^2 + y^2 \right)\), shows all the possible pairs (x, y) and their corresponding probabilities within the defined limits.

marginal distributions

Marginal distributions help us understand the individual behavior of each random variable apart from its joint behavior. To find the marginal distribution of X, we integrate the joint probability distribution over all values of Y, and vice versa.

Here, to find the marginal distribution of X, we do:
\[ f_X(x) = \int_0^a f(x,y) \, dy \]
When we do the integration, we get: \oint \[ \frac{3}{2a^4}(x^2 + y^2) \, dy \]. The result gives us the probability density for X alone, which also helps in understanding X's individual impact apart from Y.
This approach decouples the mutual relationship between X and Y, giving insights into each variable independently.

correlation coefficient

The correlation coefficient, denoted by \(\rho\), measures the strength and direction of the linear relationship between two continuous random variables. It ranges between -1 and 1. \( \rho = 1 \) indicates a perfect positive correlation, \( \rho = -1 \) indicates a perfect negative correlation, and \( \rho = 0\) indicates no correlation.

In this problem, you calculated the correlation coefficient as \( -15 / 73 \), showing a negative correlation. This means that as X increases, Y tends to decrease, and vice versa. The formula to find the correlation coefficient involves both covariance and variance:
\[ \rho = \frac{ \text{Cov}(X, Y) }{ \text{Var}(X) \text{Var}(Y) }\]. This specific negative result may seem counter-intuitive at first but it makes sense when you consider the symmetric pattern of the joint distribution we calculated earlier. You will further explore this by plotting the contour map.

covariance

Covariance is another important concept. It measures how much two random variables vary together. The covariance between X and Y is given by: \[ \text{Cov}(X, Y) = E[XY] - E[X]E[Y] \].

The covariance will be positive if both variables tend to increase or decrease simultaneously. It will be negative if one variable tends to increase when the other decreases.

In this exercise, you found the covariance using the expected values:
\oint \[ E[XY] = \int_0^a \int_0^a xy f(x, y) \, dy \, dx \]

By finding these values and substituting into the covariance formula, you can determine whether X and Y exhibit simultaneous behavior, which they don't in this case.

expectation

Expectation, or expected value, is an important concept that helps to find the 'center' or 'mean' value of a random variable. For continuous random variables, we compute it by integrating the value of the variable times its probability density function over the variable's range. For example, for X:

\[ E[X] = \int_0^a x f_X(x) \, dx \].
In this problem, you need to find the expected values for both X and Y using their respective marginal probability density functions. The results, which represent the average values or mean positions of X and Y over the interval from 0 to a, are crucial for further computations such as variance and covariance.

By understanding how to calculate these expected values, you can get deeper insights into the behavior of these continuous random variables individually.