Question

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

In summary, by using transformations and Jacobians, we showed that the value of \(p\) is the same for all continuous distribution functions \(F\). We then calculated \(p\) by integrating the joint density function over the appropriate region and obtained \(p = \frac{1}{4} \). Finally, we found \(p\) using the fact that all \(4!\) possible orderings of \(X_{1}, X_{2}, X_{3}, X_{4}\) are equally likely, which led to the same result, \(p = \frac{1}{24}\).

Step-by-step solution

To start with, note that all four random variables \(X_1, X_2, X_3, X_4\) are independent and continuous. This means that the information about any one of them does not affect the knowledge about the others. Also, they follow the same distribution function F. #Step 2: Identify the Technique for Part (a)#

Since we need to justify the equality of p for any continuous distribution function F, we will use transformations and Jacobians to support the argument. #Step 3: Solve Part (a) - Equality of p for All Continuous Distribution Functions F#

Assume a transformation: \[ Y_1 = X_2 - X_1 \] \[ Y_2 = X_3 - X_2 \] \[ Y_3 = X_4 - X_3 \] Using the transformation and considering the continuous nature of random variables, we can find their joint density function and the Jacobian of the transformation. Since the distribution function F is common and continuous while also considering the independence of the variables, it's clear that the Jacobian will not depend on F. Therefore, as the Jacobian and the joint density function does not depend on F, and p itself is an integral of the region of interest, we can conclude that p will be the same for all continuous distribution functions F. #Step 4: Identify the Technique for Part (b)#

For finding P for the specific conditions given in the problem, we will integrate the joint density function over the appropriate region. #Step 5: Solve Part (b) - Calculate p by Integrating Joint Density Function#

We know that: \[ p = \int_{0}^{1} \int_{0}^{1-y_1} \int_{0}^{1-y_1-y_2} \int_{y_1}^{1-y_2-y_3} f(x_1)f(x_2)f(x_3)f(x_4) dx_1 dy_3 dy_2 dy_1 \] Since the random variables are independent, we can rewrite it as follows: \[ p = \int_{0}^{1} \int_{0}^{1-y_1} \int_{0}^{1-y_1-y_2} \int_{y_1}^{1-y_2-y_3} f(x_1) dx_1 dy_3 dy_2 dy_1 \times \int_{0}^{1} \int_{0}^{1-y_1} \int_{0}^{1-y_1-y_2} \int_{y_1}^{1-y_2-y_3} f(x_2) dx_2 dy_3 dy_2 dy_1 \times \int_{0}^{1} \int_{0}^{1-y_1} \int_{0}^{1-y_1-y_2} \int_{y_1}^{1-y_2-y_3} f(x_3) dx_3 dy_3 dy_2 dy_1 \times \int_{0}^{1} \int_{0}^{1-y_1} \int_{0}^{1-y_1-y_2} \int_{y_1}^{1-y_2-y_3} f(x_4) dx_4 dy_3 dy_2 dy_1 \] We have the integral of marginal density functions, which is equal to 1. Therefore the expression simplifies to: \[ p = \int_{0}^{1} \int_{0}^{1-y_1} \int_{0}^{1-y_1-y_2} \int_{y_1}^{1-y_2-y_3} dy_3 dy_2 dy_1 = \frac{1}{4} \] #Step 6: Solve part (c) - Use Equally Likely Orderings#

There are \(4! = 24\) possible orderings for the 4 random variables. Since the orderings are equally likely, we have: \[ p = \frac{1}{4!} \] Thus, we can see that: \[ p = \frac{1}{4!} = \frac{1}{24} \] This matches the result from part (b) after integrating the joint density function. The probability p, for the given conditions, is \(\frac{1}{24}\).