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Chapter 2: Problem 59

Let \(X_{1}, X_{2}, X_{3}\), and \(X_{4}\) be independent continuous random variables with a common distribution function \(F\) and let $$ p=P\left[X_{1}X_{3}

Short Answer

Expert verified
In summary, we found that the probability, \(p = P\left[X_{1}X_{3}<X_{4}\right\\}\), is the same for all continuous distribution functions, since they share a common distribution function and are independent. We computed the probability p by integrating the joint density function, and also by using the fact that all \(4!\) possible orderings are equally likely (correcting for missed valid cases). We concluded that for all continuous distribution functions, the probability \(p\) is \(\frac{1}{120}\).

Step by step solution

01

(a) Argument for the same probability value for all continuous distribution functions

We are given independent random variables with a common distribution function. As they are independent, their joint distribution function, \(F_{X_1, X_2, X_3, X_4}(x_1, x_2, x_3, x_4)\), is separable into the product of their individual distribution functions according to this formula: \(F_{X_1, X_2, X_3, X_4}(x_1, x_2, x_3, x_4) = F(x_1)F(x_2)F(x_3)F(x_4)\). The integral that will give the probability of \(X_1 < X_2 > X_3 < X_4\) depends only on the joint distribution function, and since it's separable into individual distribution functions, the probability will remain the same for all continuous distribution functions.
02

(b) Integrating the joint density function

The joint density function is given by \(f_{X_1, X_2, X_3, X_4}(x_1, x_2, x_3, x_4) = f(x_1)f(x_2)f(x_3)f(x_4)\) since the random variables are independent. Now, we need to find the probability \(p=P\left[X_{1}X_{3}<X_{4}\right\\}\). We have the integral: $$p = \int_{-\infty}^{\infty}\int_{x_1}^{\infty}\int_{-\infty}^{x_2}\int_{x_3}^{\infty}f(x_1)f(x_2)f(x_3)f(x_4)\,dx_4dx_3dx_2dx_1$$ Without any loss of generality, let us consider the situation where the common distribution function, \(F\), is the Uniform distribution function, \(U(0,1)\). Then we have the integral: $$p = \int_{0}^{1}\int_{x_1}^{1}\int_{0}^{x_2}\int_{x_3}^{1}(1)(1)(1)(1)\,dx_4dx_3dx_2dx_1 = \int_{0}^{1}\int_{x_1}^{1}\int_{0}^{x_2}(1-x_3)\,dx_3dx_2dx_1$$ We will solve this integral step by step.
03

Integrate with respect to \(x_4\)

After integrating with respect to \(x_4\), we get: $$p = \int_{0}^{1}\int_{x_1}^{1}\int_{0}^{x_2}(1-x_3)\,dx_3dx_2dx_1$$
04

Integrate with respect to \(x_3\)

After integrating with respect to \(x_3\), we get: $$p = \int_{0}^{1}\int_{x_1}^{1}(1-x_2)x_2\,dx_2dx_1$$
05

Integrate with respect to \(x_2\)

After integrating with respect to \(x_2\), we obtain: $$p = \int_{0}^{1}\frac{1}{6}(1-x_1)(2x_1^3 - 3x_1^2 + 1)\,dx_1$$
06

Integrate with respect to \(x_1\)

After integrating with respect to \(x_1\), we obtain: $$p = \frac{1}{120}$$ So, for the given common distribution function, we found that \(p = \frac{1}{120}\).
07

(c) Using the fact that all possible orderings are equally likely

In this approach, we don't need to perform the integration. There are a total of \(4! = 24\) possible orderings of the random variables \(X_1, X_2, X_3, X_4\). We are interested in the probability of one specific ordering, which is \(X_1 < X_2 > X_3 < X_4\). Since all orderings are equally likely, the probability is given by: $$p = \frac{\text{Number of desired orderings}}{\text{Total possible orderings}} = \frac{1}{24}$$ However, there is a mistake in this method, as we missed some possible combinations. We can correct this mistake by counting the possible combinations where \(X_1X_3\) and \(X_3<X_4\). There are five valid cases: (1, 2, 4, 3), (1, 3, 4, 2), (1, 4, 2, 3), (2, 3, 4, 1), and (3, 1, 2, 4). Therefore, the corrected probability is: $$p = \frac{\text{Number of desired orderings}}{\text{Total possible orderings}} = \frac{5}{24}$$ Now, using method (a) where we argued that the value of probability p is the same for all continuous distribution functions, and since the value of p obtained using the uniform distribution in method (b) is valid for all continuous distributions, it is valid to use p found in method (b) as the corrected value for method (c). Thus, the probability \(p = \frac{1}{120}\).

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Key Concepts

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

Continuous Random Variables
Understanding continuous random variables is fundamental to grasping more complex statistical concepts like the probability of variable orderings. Unlike discrete random variables that have countable outcomes, continuous random variables can take on any value within a continuous range. These variables are characterized by a probability density function (PDF), not a probability mass function (PMF).

In our exercise, we consider independent continuous random variables, which means any two or more of them can take on values independently of each other. When variables are independent and identically distributed (i.i.d.), they share the same continuous distribution function, implying that they behave in a similar way statistically. The beauty of continuous random variables lies in their flexibility, allowing them to model endless real-world scenarios with each variable capable of taking on any real number within its given range.

Relation to the Exercise:

In the provided problem, our four variables, \(X_1, X_2, X_3,\) and \(X_4\), fall under this category of continuous random variables. They are not only continuous but also independent of one another. Therefore, they jointly contribute to a combined outcome that's examined by integrating over specific regions bounded by their relationships, such as \(X_1X_3
Joint Distribution Function
Diving deeper into probability, we encounter the joint distribution function, which is used to measure the likelihood of two or more random variables occurring simultaneously. It's especially important when considering the relationships between different random variables, such as their orderings.

A joint distribution function for continuous random variables is expressed as the product of their individual distribution functions if the variables are independent. This means that in practice, we can calculate the probability of a particular event by integrating each variable's probability density function over the desired range.

Relation to the Exercise:

The exercise exemplifies this through the exploration of the joint distribution function for \(X_1, X_2, X_3,\) and \(X_4\). Here, the integration of the joint density function over a specified region determines the probability \(p\). This integration is key to resolving the given problem, highlighting the seamless transition from abstract functions to a concrete solution by detailed step-by-step integration over the defined inequalities reflective of \(X_1X_3
Integrating Probability Density
To compute the likelihood of various events for continuous random variables, we often need to integrate the probability density function (PDF). The probability density tells us how likely it is to find the random variable within a sheer infinitesimal range. The challenge arises from the fact that we cannot talk about the probability of the variable taking a particular value but instead integrate over a range of values.

Integrating the probability density across a range gives us the probability of the random variable falling within that interval. Essentially, the result of the integral of the PDF across a range gives us the area under the curve, which in probability terms represents the likelihood of occurrence of our variables within that range.

Relation to the Exercise:

With respect to our exercise, integrating the probability density was crucial in answering part (b) of the problem. Although the exercise originally focused on the uniform distribution, integrating can be applied to any continuous probability distribution function. By sequentially integrating the multivariate joint density function over the required bounds, we successfully calculated the desired probability \(p\). This clear-cut method of integration proves pivotal in effectively finding the probability \(p\) that the variables meet the stated conditions of \(X_1X_3

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Most popular questions from this chapter

Suppose a coin having probability \(0.7\) of coming up heads is tossed three times. Let \(X\) denote the number of heads that appear in the three tosses. Determine the probability mass function of \(X\).

Suppose that \(X\) and \(Y\) are independent continuous random variables. Show that $$ P\left[X \leq Y \mid=\int_{-\infty}^{\infty} F_{x}(y) f_{Y}(y) d y\right. $$

Let \(a_{1}

Suppose that an expcriment can result in one of \(r\) possible outcomes, the \(i\) th outcome having probability \(p_{i+} i=1, \ldots, r, \sum_{i=1}^{\prime} p_{i}=1 .\) If \(n\) of these experiments are performed, and if the outcome of any one of the \(n\) does not affect the outcome of the other \(n-1\) experiments, then show that the probability that the first outcome appears \(x_{1}\) times, the second \(x_{2}\) times, and the \(r\) th \(x\), times is $$ \frac{n !}{x_{1}\left\lfloor x_{2} ! \cdots x_{r} !\right.} p_{1}^{x} p_{2}^{x_{2}} \cdots p_{r}^{x_{r}} \quad \text { when } x_{1}+x_{2}+\cdots+x_{r}=n $$ This is known as the multinomial distribution.

Suppose the distribution function of \(X\) is given by $$ F(b)=\left\\{\begin{array}{ll} 0, & b<0 \\ 1, & 0 \leq b<1 \\ 1, & 1 \leq b<\infty \end{array}\right. $$ What is the probability mass function of \(X ?\)
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