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Magazine Chapter 6: Problem 51
Derive the distribution of the range of a sample of size 2 from a distribution
having density function \(f(x)=2 x, 0
Short Answer
Expert verified
The distribution of the range of a sample of size 2 from a distribution having density function \(f(x)=2x\) with \(0<x<1\) is given by \(f_Y(y) = \frac{4}{3}(1+y)(1-y)^3\) for \(0 \leq y < 1\).
Step by step solution
01
Find the Joint Distribution of X_1 and X_2
Since \(X_1\) and \(X_2\) both have the same density function \(f(x) = 2x\) and are independent, the joint probability density function (pdf) of \(X_1\) and \(X_2\) can be found by multiplying their individual pdfs:
\[f_{X_1, X_2}(x_1, x_2) = f_{X_1}(x_1) \cdot f_{X_2}(x_2) = (2x_1)(2x_2) = 4x_1x_2\]
02
Find the Joint Distribution of Y and One of the Variables
Let us denote one of the sample points as \(Z\). We want to find the joint distribution of \(Y\) and \(Z\), where \(Y = \max(X_1, X_2) - \min(X_1, X_2)\) and \(Z = X_1\) or \(Z = X_2\).
There are 4 possible cases for the values of \(X_1\) and \(X_2\):
1. \(0 \leq Y \leq Z \leq 1-Y\): In this case, \(Z = X_1\), so \(X_2 = X_1 + Y\).
2. \(0 \leq Z \leq 1\): In this case, \(Z = X_1\) and \(X_2 = X_1 - Y\).
3. \(0 \leq Y \leq Z \leq 1-Y\): In this case, \(Z = X_2\), so \(X_1 = X_2 + Y\).
4. \(0 \leq Z \leq 1\): In this case, \(Z = X_2\) and \(X_1 = X_2 - Y\).
03
Find the Joint Distribution of Y and Z for Each of the 4 Cases
For the first case \(0 \leq Y \leq Z \leq 1-Y\), we have:
\[f_{Y, Z}(y, z) = f_{X_1, X_2}(z, z + y)\]
We apply the Jacobian transformation method to find the joint distribution. Let us denote \(u = x_1 = z\) and \(v = x_2 - x_1 = y\). The Jacobian can be found by calculating the determinant of the matrix formed by the partial derivatives of \(x_1\) and \(x_2\) with respect to \(u\) and \(v\):
\[J = \det \begin{bmatrix} \frac{\partial x_1}{\partial u} & \frac{\partial x_1}{\partial v}\\ \frac{\partial x_2}{\partial u} & \frac{\partial x_2}{\partial v} \end{bmatrix} = \det \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = 1\]
Now, we can find the joint distribution of \(Y\) and \(Z\):
\[f_{Y, Z}(y, z) = f_{X_1, X_2}(z, z + y)|J| = 4z^2(1 + y)\]
We will now find the joint distributions for the other cases similarly.
04
Summarize the Joint Distribution of Y and Z
Putting all the cases together, we get the joint distribution of \(Y\) and \(Z\):
\[f_{Y, Z}(y, z) = 4z^2(1 + y)\]
However, we want to find the distribution of the range \(Y = \max(X_1, X_2) - \min(X_1, X_2)\), which implies that we should find the marginal distribution of \(Y\).
05
Find the Marginal Distribution of Y
To find the marginal distribution of \(Y\), we need to integrate the joint distribution of \(Y\) and \(Z\) with respect to \(Z\). Let's do so:
\[f_Y(y) = \int_0^{1-y} f_{Y, Z}(y, z) dz\]
Substituting the joint distribution we found earlier:
\[f_Y(y) = \int_0^{1-y} 4z^2(1 + y) dz\]
Now, integrate with respect to \(z\):
\[f_Y(y) = 4(1 + y)\Big[\frac{1}{3}z^3\Big]_0^{1-y} = 4(1+y)\Big[\frac{1}{3}(1-y)^3 - \frac{1}{3}(0)^3\Big]\]
Simplifying, we get:
\[f_Y(y) = \frac{4}{3}(1+y)(1-y)^3\]
Thus, the distribution of the range of a sample of size 2 from the given density function is \(f_Y(y) = \frac{4}{3}(1+y)(1-y)^3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Density Function
The probability density function (pdf) is a fundamental concept in understanding various phenomena in probability and statistics. A pdf is a function that describes the likelihood of a continuous random variable taking on a specific value. It is a powerful mathematical tool used to model the distribution of continuous outcomes.
As an example, let's consider the pdf given in the exercise: \(f(x)=2x\) for \(0
In our exercise, the pdf is used to derive the joint probability distribution of two random variables and later to calculate the distribution of their range, showcasing its critical role in the problem-solving process.
As an example, let's consider the pdf given in the exercise: \(f(x)=2x\) for \(0
In our exercise, the pdf is used to derive the joint probability distribution of two random variables and later to calculate the distribution of their range, showcasing its critical role in the problem-solving process.
Joint Probability Distribution
Moving deeper into probability theory, the joint probability distribution is an extension of the probability density function concept but applied to two or more random variables simultaneously. It defines the probability of two variables, \(X_1\) and \(X_2\) in our case, taking on certain values in relation to each other.
In the solution provided, the joint distribution is found by multiplying the individual pdfs of \(X_1\) and \(X_2\), since they are independent. The result, \(f_{X_1, X_2}(x_1, x_2) = 4x_1x_2\), represents how the density of probability is distributed across the plane where both \(X_1\) and \(X_2\) can vary. In other words, it gives us a 'map' of the likelihood of various outcomes of the two variables occurring simultaneously.
In the solution provided, the joint distribution is found by multiplying the individual pdfs of \(X_1\) and \(X_2\), since they are independent. The result, \(f_{X_1, X_2}(x_1, x_2) = 4x_1x_2\), represents how the density of probability is distributed across the plane where both \(X_1\) and \(X_2\) can vary. In other words, it gives us a 'map' of the likelihood of various outcomes of the two variables occurring simultaneously.
Jacobian Transformation Method
The Jacobian transformation method is quite the toolbox essential when it comes to changing variables in probability distributions. It's named after the mathematician Carl Gustav Jacobi and is used when we want to switch from one set of variables to another. This method involves the use of partial derivatives, which gives us a matrix known as the Jacobian matrix.
As the exercise unfolds, we see the Jacobian used to transform from the variables \(X_1\) and \(X_2\) to new variables \(u\) and \(v\), which represent the original variables \(Z\) and their range \(Y\). The determinant of the Jacobian matrix serves as a scaling factor for the transformation, ensuring that the total probability is conserved. In our solution, the determinant is equal to 1, implying that the scale is unchanged during the transformation, which simplifies the calculation of the joint distribution for \(Y\) and \(Z\).
As the exercise unfolds, we see the Jacobian used to transform from the variables \(X_1\) and \(X_2\) to new variables \(u\) and \(v\), which represent the original variables \(Z\) and their range \(Y\). The determinant of the Jacobian matrix serves as a scaling factor for the transformation, ensuring that the total probability is conserved. In our solution, the determinant is equal to 1, implying that the scale is unchanged during the transformation, which simplifies the calculation of the joint distribution for \(Y\) and \(Z\).
Marginal Distribution
And finally, the marginal distribution is what we seek when we are interested in the behavior of a single random variable regardless of the values of others. It is essentially derived from the joint probability distribution by integrating out the other variables.
In the context of our textbook problem, we already have the joint distribution of the range \(Y\) and auxiliary variable \(Z\), but we are specifically interested in the range. To get its distribution, we integrate the joint distribution over the possible values of \(Z\). This process produces the marginal distribution of \(Y\), \(f_Y(y)\), providing us with a direct understanding of the probable distribution of the range all by itself 鈥� the final goal in the problem-solving journey we've taken.
In the context of our textbook problem, we already have the joint distribution of the range \(Y\) and auxiliary variable \(Z\), but we are specifically interested in the range. To get its distribution, we integrate the joint distribution over the possible values of \(Z\). This process produces the marginal distribution of \(Y\), \(f_Y(y)\), providing us with a direct understanding of the probable distribution of the range all by itself 鈥� the final goal in the problem-solving journey we've taken.
