Problem 176 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose \(f_{X, Y}(x, y)=x y e^{-(x+y)}, x>0, y>0\). Prove for any real numbers \(a, b, c\), and \(d\) that $$ P(a

Step-by-Step Solution

Answer

The independence of \(X\) and \(Y\) will be established by showing that the joint density function equals the product of the marginal density functions.

Step 1: Setting up the equation

The given joint density function of \(X\) and \(Y\) is \(f_{X, Y}(x, y)=x y e^{-(x+y)}\), with \(x>0\) and \(y>0\). To show that \(X\) and \(Y\) are independent, it is necessary to show that the product of the marginal densities equals the joint density.

Step 2: Finding the Marginal Densities

The marginal density functions of \(X\) and \(Y\) are obtained by integrating the joint density function over the other variable. The marginal density of \(X\) is \(f_X(x) = \int_{0}^{\infty} x y e^{-(x+y)} dy\), and the marginal density of \(Y\) is \(f_Y(y) = \int_{0}^{\infty} x y e^{-(x+y)} dx\). Solve these integrals to get the marginal densities.

Step 3: Proving Independence

After having calculated the marginal densities of \(X\) and \(Y\), it can be shown if they are independent by comparing the product of the marginal densities with the joint density. If \(f_{X, Y}(x, y) = f_X(x)f_Y(y)\) for all \(x, y\), then \(X\) and \(Y\) are independent.

Step 4: Validation

Once the calculations have been finished, it can be verified if, indeed, \(P(a

Key Concepts

Marginal Densities

Marginal densities are crucial when analyzing joint probability distributions. In simple terms, marginal densities allow us to examine one variable independently of the other, within a given joint distribution. This is achieved by integrating the joint probability to "remove" one of the variables.

For instance, if you are given a joint density function for two variables, like in our exercise, to find the marginal density of the variable \(X\), you would integrate over the other variable. In our case, for \(Y\), the calculation is \(f_X(x) = \int_{0}^{\infty} x y e^{-(x+y)} dy\). This concept helps to reveal the behavior of a single variable without considering the influence of the other.

• Marginal density of \(X\): Integrate \(f_{X,Y}(x, y)\) over \(y\).
• Marginal density of \(Y\): Integrate \(f_{X,Y}(x, y)\) over \(x\).

Ultimately, the marginal density represents the probability distribution of one variable independent of the other.

Independence of Random Variables

Random variables are said to be independent if the occurrence of one does not affect the occurrence of the other. Practically, this means that knowing the outcome of one variable provides no information about the other. In probabilistic terms, two variables \(X\) and \(Y\) are independent if their joint density \(f_{X,Y}(x, y)\) can be expressed as the product of their marginal densities \(f_X(x)\) and \(f_Y(y)\).

This property was crucial in our problem. We showed that the joint density \(f_{X,Y}(x, y) = x y e^{-(x+y)}\) could be split into the product of its marginal densities. Here are the steps summarized:

• Compute the marginal densities \(f_X(x)\) and \(f_Y(y)\).
• Check if \(f_{X,Y}(x, y) = f_X(x) f_Y(y)\).

If the equality holds for any \(x\) and \(y\), then the random variables are indeed independent.

Integration in Probability

Integration plays a pivotal role in probability to transform joint densities into marginal densities, among other things. When dealing with continuous random variables, integrating a function helps us "sum up" probabilities over a specified range.

In this context, integration focuses on handling continuous variables and finding probabilities over intervals, rather than precise values. To find a marginal density, we integrate the joint density function with respect to one variable. This action effectively "sums" the probabilities across all possible values of the integrated variable.

• Integration process for \(X\): \(f_X(x) = \int_{0}^{\infty} x y e^{-(x+y)} dy\)
• Integration process for \(Y\): \(f_Y(y) = \int_{0}^{\infty} x y e^{-(x+y)} dx\)

By solving these integrals, the marginal densities are extracted, simplifying the analysis of independence and showcasing important characteristics of each individual variable. Thus, integration in probability solidifies our understanding of each variable in the context of a joint distribution.

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Other exercises in this chapter

Problem 174

. Find the joint cdf of the independent random variables \(X\) and \(Y\), where \(f_{X}(x)=\frac{x}{2}, 0 \leq x \leq 2\), and \(f_{Y}(y)=2 y, 0 \leq y \leq 1 .

View solution

Problem 175

If two random variables \(X\) and \(Y\) are independent with marginal pdfs \(f_{X}(x)=2 x, 0 \leq x \leq 1\), and \(f_{Y}(y)=1\), \(0 \leq y \leq 1\), calculate

View solution

Problem 177

Given the joint pdf \(f_{X, Y}(x, y)=2 x+y-2 x y\), \(0

View solution

Problem 178

Prove that if \(X\) and \(Y\) are two independent random variables, then \(U=g(X)\) and \(V=h(Y)\) are also independent.

View solution

Key Concepts

Marginal Densities

Marginal densities are crucial when analyzing joint probability distributions. In simple terms, marginal densities allow us to examine one variable independently of the other, within a given joint distribution. This is achieved by integrating the joint probability to "remove" one of the variables.

For instance, if you are given a joint density function for two variables, like in our exercise, to find the marginal density of the variable \(X\), you would integrate over the other variable. In our case, for \(Y\), the calculation is \(f_X(x) = \int_{0}^{\infty} x y e^{-(x+y)} dy\). This concept helps to reveal the behavior of a single variable without considering the influence of the other.

• Marginal density of \(X\): Integrate \(f_{X,Y}(x, y)\) over \(y\).
• Marginal density of \(Y\): Integrate \(f_{X,Y}(x, y)\) over \(x\).

Ultimately, the marginal density represents the probability distribution of one variable independent of the other.

Independence of Random Variables

Random variables are said to be independent if the occurrence of one does not affect the occurrence of the other. Practically, this means that knowing the outcome of one variable provides no information about the other. In probabilistic terms, two variables \(X\) and \(Y\) are independent if their joint density \(f_{X,Y}(x, y)\) can be expressed as the product of their marginal densities \(f_X(x)\) and \(f_Y(y)\).

This property was crucial in our problem. We showed that the joint density \(f_{X,Y}(x, y) = x y e^{-(x+y)}\) could be split into the product of its marginal densities. Here are the steps summarized:

• Compute the marginal densities \(f_X(x)\) and \(f_Y(y)\).
• Check if \(f_{X,Y}(x, y) = f_X(x) f_Y(y)\).

If the equality holds for any \(x\) and \(y\), then the random variables are indeed independent.

Integration in Probability

Integration plays a pivotal role in probability to transform joint densities into marginal densities, among other things. When dealing with continuous random variables, integrating a function helps us "sum up" probabilities over a specified range.

In this context, integration focuses on handling continuous variables and finding probabilities over intervals, rather than precise values. To find a marginal density, we integrate the joint density function with respect to one variable. This action effectively "sums" the probabilities across all possible values of the integrated variable.

• Integration process for \(X\): \(f_X(x) = \int_{0}^{\infty} x y e^{-(x+y)} dy\)
• Integration process for \(Y\): \(f_Y(y) = \int_{0}^{\infty} x y e^{-(x+y)} dx\)

By solving these integrals, the marginal densities are extracted, simplifying the analysis of independence and showcasing important characteristics of each individual variable. Thus, integration in probability solidifies our understanding of each variable in the context of a joint distribution.