/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Let \(Y_{1}0\), provided that \(... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 16

Let \(Y_{1}0\), provided that \(x \geq 0\), and \(f(x)=0\) elsewhere. Show that the independence of \(Z_{1}=Y_{1}\) and \(Z_{2}=Y_{2}-Y_{1}\) characterizes the gamma pdf \(f(x)\), which has parameters \(\alpha=1\) and \(\beta>0 .\) That is, show that \(Y_{1}\) and \(Y_{2}\) are independent if and only if \(f(x)\) is the pdf of a \(\Gamma(1, \beta)\) distribution. Hint: Use the change-of-variable technique to find the joint pdf of \(Z_{1}\) and \(Z_{2}\) from that of \(Y_{1}\) and \(Y_{2}\). Accept the fact that the functional equation \(h(0) h(x+y) \equiv\) \(h(x) h(y)\) has the solution \(h(x)=c_{1} e^{c_{2} x}\), where \(c_{1}\) and \(c_{2}\) are constants.

Short Answer

Expert verified
The randomness of the variables \(Z_1 = Y_1\) and \(Z_2 = Y_2 - Y_1\) characterizes the gamma pdf \(f(x)\), under the condition that \(Y_1\) and \(Y_2\) are independent only if \(f(x)\) is a pdf of a 螕(1,尾) distribution.

Step by step solution

01

Identify the Variables

The given are two random variables \(Y_1\) and \(Y_2\), and their transformations \(Z_1 = Y_1\) and \(Z_2 = Y_2 - Y_1\). We want to show that \(Z_1\) and \(Z_2\) are independent if and only if the pdf \(f(x)\) is a Gamma \(螕(1, 尾)\) distribution.
02

Use the Change-of-Variable Technique

First, let's find the joint pdf of \(Z_1\) and \(Z_2\) starting from that of \(Y_1\) and \(Y_2\). We can use the change-of-variable technique for this. The Jacobian of the transformation is \(1\), so the joint pdf \(g(z_1, z_2)\) for the transformed variables is \(f(y_1, y_2)\). But \(y_1 = z_1\) and \(y_2 = z_1 + z_2\), so substituting these into the joint pdf \(f(y_1, y_2)\) gives \(g(z_1, z_2) = f(z_1, z_1 + z_2)\).
03

Analyze the Joint pdf and Show Independence

Now, let's examine the functional form of the joint pdf. The problem provided a hint: any function \(h(x)\) satisfying the functional equation \(h(0)h(x+y) = h(x)h(y)\) has the solution \(h(x) = c_1 e^{c_2 x}\), where \(c_1\) and \(c_2\) are constants. In order for \(Z_1\) and \(Z_2\) to be independent, the joint pdf \(g(z_1, z_2)\) must be the product of the two marginal probability density functions. This means that \(g(z_1, z_2) = g_1(z_1)g_2(z_2)\), so comparing this to the form \(h(x) = c_1 e^{c_2 x}\), we can immediately see that the marginal pdfs are exponential, meaning that the independence of \(Z_1\) and \(Z_2\) characterizes the gamma distribution.

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Most popular questions from this chapter

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