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Chapter 6: Problem 1

If \(X\) and \(Y\) are independent random variables with density functions \(f_{X}\) and \(f_{Y}\), respectively, show that \(U=X Y\) and \(V=X / Y\) have density functions $$ f_{U}(u)=\int_{-\infty}^{\infty} f_{X}(x) f_{Y}(u / x) \frac{1}{|x|} d x, \quad f_{V}(v)=\int_{-\infty}^{\infty} f_{X}(v y) f_{Y}(y)|y| d y $$

Short Answer

Expert verified
The density functions are given by obtained integrals using transformation.

Step by step solution

01

Define the Transformation

We are tasked with finding the probability density functions (PDFs) of the transformed variables. For the transformation \(U = XY\) and \(V = X/Y\), we need to express these relationships in terms of the original PDFs \(f_X\) and \(f_Y\).
02

Calculate the Joint PDF of (X, Y)

Since \(X\) and \(Y\) are independent, their joint PDF is the product of their individual PDFs: \(f_{X,Y}(x, y) = f_X(x) f_Y(y)\). This property will be used in our subsequent transformations.
03

Transform to Get PDF of U

To find the PDF \(f_U(u)\), we will use the transformation \(U = XY\). The law of the unconscious statistician gives us \(f_U(u) = \int_{-\infty}^{\infty} f_{X,Y}(x, u/x) \left| \frac{d(x,u/x)}{du} \right| dx\). Using substitution, \(f_U(u) = \int_{-\infty}^{\infty} f_{X}(x) f_Y(u/x) \frac{1}{|x|} dx\).
04

Transform to Get PDF of V

For \(V = X/Y\), the PDF can be found similarly by considering \(V\) as a function of \(X\) and \(Y\). The transformation gives \(f_V(v) = \int_{-\infty}^{\infty} f_{X,Y}(vy, y) |y| dy\). Substituting the values, we obtain \(f_V(v) = \int_{-\infty}^{\infty} f_X(vy) f_Y(y) |y| dy\).
05

Interpretation of Results

The derived formulas for \(f_U(u)\) and \(f_V(v)\) show how the independent densities \(f_X\) and \(f_Y\) contribute to the densities of products and quotients of random variables, considering integration over the entire range and adjusting for magnitude changes.

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Key Concepts

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

Independent Random Variables
Independent random variables are a fundamental concept in probability theory. When two random variables, say \( X \) and \( Y \), are independent, the occurrence of one does not affect the probability of occurrence of the other. This means the joint probability density function (PDF) of \( X \) and \( Y \), denoted as \( f_{X,Y}(x, y) \), is simply the product of their individual PDFs: \( f_X(x) \) and \( f_Y(y) \). This simplifies analysis, especially when working with transformations or combinations of these variables.
  • Independence implies \( f_{X,Y}(x, y) = f_X(x) f_Y(y) \).
  • It is essential in deriving the joint distribution of transformed variables like \( U = XY \) or \( V = X/Y \).
Transformation of Variables
The transformation of variables in probability is a technique used to find the distribution of a function of one or more random variables. In our problem, we deal with transformations \( U = XY \) and \( V = X/Y \). The goal is to determine the new PDFs, \( f_U(u) \) and \( f_V(v) \), from the original PDFs of \( X \) and \( Y \).
  • For \( U=XY \), the PDF is found using: \( f_U(u)=\int_{-\infty}^{\infty} f_X(x) f_Y(u/x) \frac{1}{|x|} dx \).
  • For \( V=X/Y \), the PDF is determined by: \( f_V(v)=\int_{-\infty}^{\infty} f_X(vy) f_Y(y) |y| dy \).
The process involves exploring how these transformations alter the variable scope and integrating the joint PDF over possible values of the original variables. This often requires understanding the Jacobian for variable change, particularly for non-linear transformations.
Joint Probability Density
Joint probability density functions describe the likelihood of two random variables taking on a specific set of values simultaneously. For \( X \) and \( Y \) being independent, their joint PDF is a product of their marginal densities, as discussed earlier. This concept is pivotal for transformation processes as it serves as a starting point.
To transform variables like \( U \) or \( V \), begin with the joint PDF \( f_{X,Y}(x, y) = f_X(x)f_Y(y) \). The transformation techniques account for how these joint distributions affect new variable setups.
  • Establish joint density using: \( f_{X,Y}(x,y) = f_X(x)f_Y(y) \).
  • Use this joint PDF in transformations, integrating across the domain of \( X \) or \( Y \) as needed.
Law of the Unconscious Statistician
The Law of the Unconscious Statistician (LOTUS) facilitates the calculation of expected values of functions of a random variable without needing to find the exact distribution first. This principle extends to transformations like \( U = XY \) and \( V = X/Y \), where determining the exact form of the density function might be complex.
LOTS is particularly useful for tackling exercises involving transformations or expectations from independent variables.
  • Helps in calculating the expectation \( E[g(X)] \) with \( E[g(X)] = \int_{-\infty}^{\infty} g(x)f_X(x) dx \).
  • Extends to functions of multiple variables when considering transformations, guiding the integration process over the probability weighted space.
In this context, LOTS underpins the derivation of \( f_U(u) \) and \( f_V(v) \) by justifying the integration of product forms across transformed variables.

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Most popular questions from this chapter

In a sequence of dependent Bernoulli trials, the conditional probability of success at the \(i\) th trial, given that all preceding trials have resulted in failure, is \(p_{i}(i=1,2, \ldots)\). Give an expression in terms of the \(p_{i}\) for the probability that the first success occurs at the \(n\)th trial. Suppose that \(p_{i}=1 /(i+1)\) and that the time intervals between successive trials are independent random variables, the interval between the \((n-1)\) th and the \(n\)th trials being exponentially distributed with density \(n^{\alpha} \exp \left(-n^{\alpha} x\right)\), where \(\alpha\) is a given constant. Show that the expected time to achieve the first success is finite if and only if \(\alpha>0\). (Oxford \(1975 \mathrm{~F}\) )

Let \(X\) and \(Y\) be independent random variables, \(X\) having the normal distribution with mean 0 and variance 1 , and \(Y\) having the \(\chi^{2}\) distribution with \(n\) degrees of freedom. Show that $$ T=\frac{X}{\sqrt{Y / n}} $$ has density function $$ f(t)=\frac{1}{\sqrt{\pi n}} \frac{\Gamma\left(\frac{1}{2}(n+1)\right)}{\Gamma\left(\frac{1}{2} n\right)}\left(1+\frac{t^{2}}{n}\right)^{-\frac{1}{2}(n+1)} \quad \text { for } t \in \mathbb{R} $$ \(T\) is said to have the \(t\)-distribution with \(n\) degrees of freedom.

Let \(X\) and \(Y\) have the bivariate normal density function $$ f(x, y)=\frac{1}{2 \pi \sqrt{1-\rho^{2}}} \exp \left\\{-\frac{1}{2\left(1-\rho^{2}\right)}\left(x^{2}-2 \rho x y+y^{2}\right)\right\\} \quad \text { for } x, y \in \mathbb{R} $$ for fixed \(\rho \in(-1,1)\). Let \(Z=(Y-\rho X) / \sqrt{1-\rho^{2}}\). Show that \(X\) and \(Z\) are independent \(\mathrm{N}(0,1)\) variables. Hence or otherwise determine \(\mathbb{P}(X>0, Y>0)\). (Cambridge 2008)

Let \(X_{1}, X_{2}, \ldots\) be independent, identically distributed, continuous random variables. Define \(N\) as the index such that $$ X_{1} \geq X_{2} \geq \cdots \geq X_{N-1} \quad \text { and } \quad X_{N-1}

Zog continued. This time, \(n\) members of Dr Who's crew are transported to Zog, their positions being independent and uniformly distributed on the surface. In addition, Dr Who is required to choose a place \(W\) on the surface for his own transportation. Find the probability that, for every \(W\), he is able to communicate with some member of his crew.
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