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Chapter 6: Problem 103

Let \(Y_{1}\) and \(Y_{2}\) be independent, standard normal random variables. Find the probability density $$\text { function of } U=Y_{1} / Y_{2}$$

Short Answer

Expert verified
The pdf of \( U = \frac{Y_1}{Y_2} \) is \( f_U(u) = \frac{1}{\pi(1+u^2)} \).

Step by step solution

01

Understanding the Problem

We need to find the probability density function (pdf) of the random variable \( U \), where \( U = \frac{Y_1}{Y_2} \). Here, \( Y_1 \) and \( Y_2 \) are independent standard normal random variables, meaning \( Y_1 \sim N(0,1) \) and \( Y_2 \sim N(0,1) \).
02

Recognize the Distribution

The random variable \( U = \frac{Y_1}{Y_2} \) is known to have a particular distribution. This type of distribution is recognized as a Cauchy distribution due to the ratio of two independent standard normal variables.
03

Identify the Cauchy Distribution

When \( U\) is expressed as a ratio of two independent standard normal variables, its probability density function follows a standard Cauchy distribution. The form of the Cauchy distribution is given by: \[ f_U(u) = \frac{1}{\pi(1+u^2)} \], for \( u \in \, \mathbb{R} \).
04

Conclude with the Probability Density Function

We conclude that the probability density function of \( U = \frac{Y_1}{Y_2} \) is that of a standard Cauchy distribution: \[ f_U(u) = \frac{1}{\pi(1+u^2)} \], valid for all real numbers \( u \).

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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 probability theory and statistics. It is a function that describes the likelihood of a random variable taking on a specific value. For a continuous random variable, the probability that it takes on an exact value is zero, but the pdf allows us to determine the probability within a certain range.
For example, if you have a continuous random variable representing height, the pdf tells you how probable it is that a randomly chosen individual's height falls within a particular range. The area under the curve of the pdf between two points represents the probability that the random variable falls between those two points.
  • The pdf must be non-negative for all values of the random variable.
  • The total area under the pdf curve over all possible values of the variable is equal to 1.
The Cauchy distribution, which often arises from the pdf of the ratio of two standard normal variables, is an example where the pdf does not have moments like mean and variance defined, making it quite unique.
Standard Normal Distribution
The standard normal distribution is one of the most important distributions in statistics. It is a normal distribution with a mean of 0 and a standard deviation of 1. This means it is centered around zero and follows the familiar bell-shaped curve that symmetrically spreads out on both sides.
The standard normal distribution is fundamental for many statistical methods and distributions. When a variable is said to be standard normal, it is denoted as \( N(0, 1) \).
  • The total area under the standard normal curve is 1, representing the complete probability space.
  • It is used to convert any normal distribution to a standard normal by using Z-scores.
  • Z-scores help to understand how many standard deviations a particular value is from the mean, which is essential for hypothesis testing and constructing confidence intervals.
For the variables \( Y_1 \) and \( Y_2 \) in the exercise, both being standard normal, their ratio leads to a very intriguing result鈥攖he Cauchy distribution.
Independent Random Variables
Independence in random variables is a crucial concept in probability and statistics. When two random variables are independent, it means that the occurrence or value of one does not affect the other.
This independence simplifies the analysis of combined probability distributions and allows for the application of powerful statistical theorems and techniques. If \( Y_1 \) and \( Y_2 \) are independent random variables, it implies that knowing the outcome of one provides no information about the outcome of the other.
  • The joint probability distribution of independent random variables is the product of their individual distributions.
  • Independence is a key assumption in many statistical models and techniques, such as those used in regression and ANOVA.
  • For the exercise, noting \( Y_1 \) and \( Y_2 \) are independent standard normal variables, it was essential to determining that their ratio results in a Cauchy distribution, as independence ensures each variable behaves according to its defined distribution without influencing the other.
This concept underlines why the ratio \( Y_1/Y_2 \) has the properties it does鈥攆reedom from dependences enhances the predictability of such interactions.

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Most popular questions from this chapter

The manager of a construction job needs to figure prices carefully before submitting a bid. He also needs to account for uncertainty (variability) in the amounts of products he might need. To oversimplify the real situation, suppose that a project manager treats the amount of sand, in yards. needed for a construction project as a random variable \(Y_{1}\), which is normally distributed with mean 10 yards and standard deviation. 5 yard. The amount of cement mix needed, in hundreds of pounds, is a random variable \(Y_{2}\), which is normally distributed with mean 4 and standard deviation .2. The sand costs \(\$ 7\) per yard, and the cement mix costs \(\$ 3\) per hundred pounds. Adding \(\$ 100\) for other costs, he computes his total cost to be $$U=100+7 Y_{1}+3 Y_{2}$$. If \(Y_{1}\) and \(Y_{2}\) are independent, how much should the manager bid to ensure that the true costs will exceed the amount bid with a probability of only. \(01 ?\) Is the independence assumption reasonable here?

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent, normal random variables, each with mean \(\mu\) and variance \(\sigma^{2}.\) Let \(a_{1}, a_{2}, \ldots, a_{n}\) denote known constants. Find the density function of the linear combination \(U=\sum_{i=1}^{n} a_{i} Y_{i}\).

The time until failure of an electronic device has an exponential distribution with mean 15 months. If a random sample of five such devices are tested, what is the probability that the first failure among the five devices occurs a. after 9 months? b. before 12 months?

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent and identically distributed random variables such that for \(0

Suppose that the number of occurrences of a certain event in time interval \((0, t)\) has a Poisson distribution. If we know that \(n\) such events have occurred in \((0, t),\) then the actual times, measured from \(0,\) for the occurrences of the event in question form an ordered set of random variables, which we denote by \(W_{(1)} \leq W_{(2)} \leq \cdots \leq W_{(n)} .\left[W_{(i)} \text { actually is the waiting time from } 0\) until the \right. occurrence of the \(i \text { th event. }]\) It can be shown that the joint density function for \(W_{(1)}, W_{(2)}, \ldots, W_{(n)}\) is given by $$f\left(w_{1}, w_{2}, \ldots, w_{n}\right)=\left\\{\begin{array}{ll} \frac{n !}{t^{n}}, & w_{1} \leq w_{2} \leq \cdots \leq w_{n} \\ 0, & \text { elsewhere } \end{array}\right.$$ [This is the density function for an ordered sample of size \(n\) from a uniform distribution on the interval (0,t).] Suppose that telephone calls coming into a switchboard follow a Poisson distribution with a mean of ten calls per minute. A slow period of two minutes' duration had only four calls. Find the a. probability that all four calls came in during the first minute; that is, find \(P(W_{(4)} \leq 1\) ). b. expected waiting time from the start of the two-minute period until the fourth call.
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