Problem 9 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Suppose that the joint distribution function of \(X\) and \(Y\) is given by $$ F(x, y)=1-\mathrm{e}^{-2 x}-\mathrm{e}^{-y}+\mathrm{e}^{-(2 x+y)} \quad \text { if } x>0, y>0, $$ and \(F(x, y)=0\) otherwise. a. Determine the marginal distribution functions of \(X\) and \(Y\). b. Determine the joint probability density function of \(X\) and \(Y\). c. Determine the marginal probability density functions of \(X\) and \(Y\). d. Find out whether \(X\) and \(Y\) are independent.

Step-by-Step Solution

Answer

The marginal distribution functions are \(F_X(x) = 1 - \mathrm{e}^{-2x}\) and \(F_Y(y) = 1 - \mathrm{e}^{-y}\). The joint density is \(2\mathrm{e}^{-(2x+y)}\). The marginals are \(f_X(x) = 2\mathrm{e}^{-2x}\) and \(f_Y(y) = \mathrm{e}^{-y}\). \(X\) and \(Y\) are independent.

Step 1: Determine the Marginal Distribution of X

To find the marginal distribution function of \(X\), we need to differentiate the joint distribution function \(F(x, y)\) with respect to \(y\) and then let \(y \to \infty\). The marginal distribution of \(X\) is given by: \[ F_X(x) = F(x, \, \infty) = 1 - \mathrm{e}^{-2x} - 0 + 0 = 1 - \mathrm{e}^{-2x} \] So, \(F_X(x) = 1 - \mathrm{e}^{-2x}\) for \(x > 0\).

Step 2: Determine the Marginal Distribution of Y

Similarly, for the marginal distribution of \(Y\), differentiate the joint distribution function \(F(x, y)\) with respect to \(x\) and let \(x \to \infty\). The marginal distribution of \(Y\) is: \[ F_Y(y) = F(\infty, y) = 1 - 0 - \mathrm{e}^{-y} + 0 = 1 - \mathrm{e}^{-y} \]Thus, \(F_Y(y) = 1 - \mathrm{e}^{-y}\) for \(y > 0\).

Step 3: Find the Joint Probability Density Function

To find the joint probability density function \(f(x, y)\), differentiate the joint distribution function \(F(x, y)\) with respect to both \(x\) and \(y\). Thus: \[ f(x, y) = \frac{\partial^2 F(x, y)}{\partial x \partial y} = \frac{\partial}{\partial y} \left( \frac{\partial}{\partial x} (1 - \mathrm{e}^{-2x} - \mathrm{e}^{-y} + \mathrm{e}^{-(2x+y)}) \right)\]First, find \(\frac{\partial F}{\partial x}\): \[ \frac{\partial F}{\partial x} = 2\mathrm{e}^{-2x} - 2\mathrm{e}^{-(2x+y)} \]Then, \(\frac{\partial^2 F}{\partial x \partial y} = -\mathrm{e}^{-y}\cdot (-2)e^{-2x}=2\mathrm{e}^{-(2x+y)}\). Thus, \(f(x, y) = 2\mathrm{e}^{-(2x+y)}\) for \(x > 0, y > 0\).

Step 4: Determine the Marginal Probability Density Function of X

The marginal density function of \(X\) is obtained by integrating over \(y\): \[ f_X(x) = \int_0^\infty 2\mathrm{e}^{-(2x+y)} \, dy \]Thus: \[ f_X(x) = 2\mathrm{e}^{-2x} \int_0^\infty \mathrm{e}^{-y} \, dy = 2\mathrm{e}^{-2x}\cdot [ -\mathrm{e}^{-y} ]_0^\infty = 2\mathrm{e}^{-2x} \]So, \(f_X(x) = 2\mathrm{e}^{-2x}\) for \(x > 0\).

Step 5: Determine the Marginal Probability Density Function of Y

The marginal density function of \(Y\) is obtained by integrating over \(x\): \[ f_Y(y) = \int_0^\infty 2\mathrm{e}^{-(2x+y)} \, dx \]Thus: \[ f_Y(y) = \mathrm{e}^{-y} \int_0^\infty 2\mathrm{e}^{-2x} \, dx = \mathrm{e}^{-y} \cdot [ -\frac{1}{2}\mathrm{e}^{-2x} ]_0^\infty = \mathrm{e}^{-y} \]So, \(f_Y(y) = \mathrm{e}^{-y}\) for \(y > 0\).

Step 6: Check Independence of X and Y

To check if \(X\) and \(Y\) are independent, verify if the joint density function is the product of the marginals, i.e., \(f(x, y) = f_X(x)f_Y(y)\).\[ f(x, y) = 2\mathrm{e}^{-(2x+y)} \]\[ f_X(x) = 2\mathrm{e}^{-2x}, \quad f_Y(y) = \mathrm{e}^{-y} \]\[ f_X(x)f_Y(y) = 2\mathrm{e}^{-2x}\cdot \mathrm{e}^{-y} = 2\mathrm{e}^{-(2x+y)} \]Since \(f(x, y) = f_X(x)f_Y(y)\), \(X\) and \(Y\) are independent.

Key Concepts

Marginal Distribution Functions

Marginal distribution functions help us understand the distribution of each individual random variable within a joint distribution. To find the marginal distribution of a random variable, we need to "sum out" or integrate over the other variables present in the joint distribution. This simplifies the problem by focusing on a single variable from the joint setup.

For instance, to determine the marginal distribution function for the variable \(X\), we look at the cumulative distribution function for \(X\), denoted as \(F_X(x)\). This function is obtained by integrating out the other variable \(Y\). The marginal distribution function \(F_X(x)\) provides us with the probability that \(X\) takes on a value less than or equal to \(x\).

From the exercise, we compute it by substituting \(y\) to \(\infty\) in the joint distribution function \(F(x, y)\), effectively ignoring \(Y\) and focusing on \(X\). This gives us the marginal distribution function for \(X\). A similar method is applied to find \(F_Y(y)\), the marginal distribution for \(Y\), by letting \(x\) approach infinity while evaluating the joint distribution function. These calculations simplify the joint examination to one variable at a time.

Joint Probability Density Function

The joint probability density function (pdf) characterizes the likelihood of two continuous random variables occurring simultaneously at particular values. It provides a more granular view than the cumulative distribution, focusing on the infinitesimal probabilities over an area rather than cumulative up to a point.

In practice, obtaining the joint pdf often requires differentiating the joint distribution function twice, once with respect to each variable. This double differentiation is akin to zooming in to see the minute chances of particular outcomes for \(X\) and \(Y\) together. The joint pdf \(f(x, y)\) is fully responsible for conveying information on how \(X\) and \(Y\) interact with each other.

Using the given joint distribution function \(F(x, y)\), we can find \(f(x, y)\) by taking partial derivatives concerning \(x\) and \(y\). This precision is crucial when we need to determine probabilities over specific intervals or to extract further characteristics such as correlations from data that are inherently multidimensional.

Independence of Random Variables

The concept of independence between random variables \(X\) and \(Y\) is fundamental in probability theory. Two random variables are independent if the occurrence of one does not affect the probability distribution of the other. This is determined by inspecting their joint and marginal distribution functions.

Mathematically, for \(X\) and \(Y\) to be independent, the joint pdf \(f(x, y)\) must equal the product of their marginal pdfs: \(f_X(x)\) and \(f_Y(y)\).

• If \(f(x, y) = f_X(x) \cdot f_Y(y)\) holds true for all values within their domains, then \(X\) and \(Y\) are independent.
• This implies that knowing the value of one provides no information about the other.

In the exercise, verification of \(f(x, y) = f_X(x)f_Y(y)\) confirms the independence of \(X\) and \(Y\). Understanding independence helps to break down complex joint distributions into simpler, manageable parts.