/*! 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 122 If \(X\) is distributed as \(N(0... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If \(X\) is distributed as \(N(0,1)\), find the pdf of \(|X|\).

Short Answer

Expert verified
The pdf of \(|X|\) is \(\sqrt{\frac{2}{\pi}}e^{-y^2/2}\) for \(y \geq 0\).

Step by step solution

01

Understand the properties of the standard normal distribution

The random variable\(X\) is distributed as \(N(0,1)\), which means it is a standard normal distribution with a mean \(\mu=0\) and variance \(\sigma^2=1\). Its probability density function (pdf) is given by \(f_X(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\) for \(-\infty < x < \infty\).
02

Analyze the absolute value transformation

We are interested in the distribution of the random variable \(|X|\). The absolute value transformation implies that all negative values of \(X\) map to their positive counterparts. Therefore, \(|X|\) takes on non-negative values only, \([0, \infty)\).
03

Find the CDF of \(|X|\)

The cumulative distribution function (CDF) of \(Y = |X|\) at a point \(y\) is given by \(F_Y(y) = P(|X| \leq y)\). Since \(|X|\leq y\) is equivalent to \(-y \leq X \leq y\), we have \(P(-y \leq X \leq y) = P(X \leq y) - P(X < -y)\). Using symmetry, this simplifies to \(F_Y(y) = 2\Phi(y) - 1\), where \(\Phi\) is the CDF of \(X\).
04

Differentiate to find the pdf of \(|X|\)

To find the probability density function (pdf) of \(|X|\) from the CDF, we differentiate \(F_Y(y)\) with respect to \(y\). The derivative of \(2\Phi(y) - 1\) is \(2\phi(y)\), where \(\phi(y) = \frac{1}{\sqrt{2\pi}}e^{-y^2/2}\) is the pdf of \(X\). Thus, the pdf of \(|X|\) is \(f_Y(y) = 2\phi(y) = \sqrt{\frac{2}{\pi}}e^{-y^2/2}\) for \(y \geq 0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Standard Normal Distribution
The Standard Normal Distribution is an essential concept in statistics and probability, serving as a quintessential model for quantitative data analysis. It is represented by the symbol \(N(0,1)\) because it has a mean (average) of \(0\) and a standard deviation and variance of \(1\). This unique distribution is characterized by its symmetrical bell-shaped curve, making it a probability distribution where the majority of data points are concentrated around the mean.

Some important properties of the standard normal distribution include:
  • The total area under the curve is equal to \(1\), indicating that it accounts for the entire probability space.
  • It is symmetric about the mean \(0\), meaning that values below and above the mean are equally likely.
  • Approximately \(68\%\) of the data lies within one standard deviation from the mean, \(95\%\) within two, and \(99.7\%\) within three standard deviations it aligns the empirical rule.
Understanding the standard normal distribution is fundamental, as it extends to use in complex concepts and transformations in probability, such as the Absolute Value Transformation discussed below.
Probability Density Function (pdf)
The Probability Density Function (pdf) is a fundamental component allowing us to understand how probabilities are distributed across different random values of a continuous random variable.

For a standard normal distribution, the pdf is defined mathematically as: \[ f_X(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \] Here, the pdf describes the likelihood of a random variable \(X\) taking on a specific value.

Key features of the pdf include:
  • The pdf is only defined for continuous random variables, excluding probabilities at discrete points.
  • It indicates how probable it is for a random variable to assume a particular value, and when plotted, it shows the shape of the distribution.
  • The area under the pdf curve over an interval corresponds to the probability that the random variable lies within that interval.
For example, the pdf of the standard normal distribution reflects its symmetrical bell shape and spreads densities symmetrically from the center across the horizontal axis.
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) allows us to understand the probability that a continuous random variable takes on a value less than or equal to a given point. It accumulates probabilities from left to right across the distribution.

For a standard normal distribution, denoted as \(X\), the CDF is represented by \(\Phi(x)\). This function can be used to calculate probabilities of the form \(P(X \leq x)\), representing the area under the pdf curve from \(\-\infty\) to \(x\).

When considering the transformation into \(|X|\), the CDF changes. The transformed variable \(|X|\) has a CDF given by: \[ F_Y(y) = 2\Phi(y) - 1 \] where \(F_Y(y)\) provides the probability that the absolute value of \(X\) is less than or equal to \(y\). Here, the symmetry of the normal distribution allows for simplifying the CDF using the properties of \(X\) and \(-X\). Utilizing the CDF effectively helps derive probabilities and solve problems related to the distribution of transformed variables.
Absolute Value Transformation
The Absolute Value Transformation concerns changing a random variable, \(X\), into its absolute value, \(|X|\). This transformation turns any negative outcomes into positive ones, significantly impacting the range and probability distribution of the variables involved.

Suppose \(X\) follows a standard normal distribution \(N(0,1)\). When transformed into \(|X|\), it spans only non-negative values \([0, \infty)\). To deduce its probability, you shift from the original random variable to the new form, using their relations.

The pdf of \(|X|\) found through this transformation is derived by differentiating the CDF of \(|X|\), as previously explained using \[ f_Y(y) = 2\phi(y) = \sqrt{\frac{2}{\pi}}e^{-y^2/2} \] for \(y \geq 0\). When computing derived probabilities, you ensure that the transformed distribution accounts for both positive and negative domains of the original variable. This method enables a historical overview and recalibration of probabilities in studies involving measurement and constraints where only absolute results show beneficial or realistic outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article "Reliability of Domestic-Waste Biofilm Reactors" (J. Envir. Engrg., 1995: 785-790) suggests that substrate concentration \(\left(\mathrm{mg} / \mathrm{cm}^{3}\right)\) of influent to a reactor is normally distributed with \(\mu=.30\) and \(\sigma=.06\). a. What is the probability that the concentration exceeds \(.25 ?\) b. What is the probability that the concentration is at most. 10 ? c. How would you characterize the largest \(5 \%\) of all concentration values?

Stress is applied to a 20 -in. steel bar that is clamped in a fixed position at each end. Let \(Y=\) the distance from the left end at which the bar snaps. Suppose \(Y / 20\) has a standard beta distribution with \(E(Y)=10\) and \(V(Y)=100 / 7\). a. What are the parameters of the relevant standard beta distribution? b. Compute \(P(8 \leq Y \leq 12)\). c. Compute the probability that the bar snaps more than 2 in. from where you expect it to snap.

Let \(X\) denote the ultimate tensile strength (ksi) at \(-200^{\circ}\) of a randomly selected steel specimen of a certain type that exhibits "cold brittleness" at low temperatures. Suppose that \(X\) has a Weibull distribution with \(\alpha=20\) and \(\beta=100\). a. What is the probability that \(X\) is at most \(105 \mathrm{ksi}\) ? b. If specimen after specimen is selected, what is the long-run proportion having strength values between 100 and \(105 \mathrm{ksi}\) ? c. What is the median of the strength distribution?

Let \(X\) denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The article "Blade Fatigue Life Assessment with Application to VAWTS" (J. Solar Energy Engrg., 1982: 107-111) proposes the Rayleigh distribution, with pdf $$ f(x ; \theta)=\left\\{\begin{array}{cl} \frac{x}{\theta^{2}} \cdot e^{-x^{2} /\left(2 \theta^{2}\right)} & x>0 \\ 0 & \text { otherwise } \end{array}\right. $$ as a model for the \(X\) distribution. a. Verify that \(f(x ; \theta)\) is a legitimate pdf. b. Suppose \(\theta=100\) (a value suggested by a graph in the article). What is the probability that \(X\) is at most 200 ? Less than 200 ? At least 200 ? c. What is the probability that \(X\) is between 100 and 200 (again assuming \(\theta=100\) )? d. Give an expression for \(P(X \leq x)\).

Let the ordered sample observations be denoted by \(y_{1}, y_{2}, \ldots, y_{n}\left(y_{1}\right.\) being the smallest and \(y_{n}\) the largest). Our suggested check for normality is to plot the \(\left(\Phi^{-1}[(i-.5) / n], y_{i}\right)\) pairs. Suppose we believe that the observations come from a distribution with mean 0, and let \(w_{1}, \ldots, w_{n}\) be the ordered absolute values of the \(x_{i}\) 's. A halfnormal plot is a probability plot of the \(w_{i} ' s\). More specifically, since \(P(|Z| \leq w)=P(-w \leq\) \(Z \leq w)=2 \Phi(w)-1\), a half-normal plot is a plot of the \(\left(\Phi^{-1}\left[\left(p_{i}+1\right) / 2\right], w_{i}\right)\) pairs, where \(p_{i}=(i\) \(-5) / n\). The virtue of this plot is that small or large outliers in the original sample will now appear only at the upper end of the plot rather than at both ends. Construct a half-normal plot for the following sample of measurement errors, and comment: \(-3.78,-1.27,1.44,-.39,12.38\), \(-43.40,1.15,-3.96,-2.34,30.84\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.