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A random variable \(X\) has the following probability distribution: $$ f_{X}(x)=e^{-x}, \quad x \geq 0 $$ (a) Find the probability distribution for \(Y=X^{2}\). (b) Find the probability distribution for \(Y=X^{1 / 2}\). (c) Find the probability distribution for \(Y=\ln X\).

Short Answer

Expert verified
(a) \(f_Y(y) = \frac{1}{2\sqrt{y}} e^{-\frac{y}{2}}\), \(y \geq 0\). (b) \(f_Y(y) = 2y e^{-y^2}\), \(y \geq 0\). (c) \(f_Y(y) = e^{-y-e^y}\), \(y \in \mathbb{R}\).

Step by step solution

01

Identify the transformation for Y=X^2

To find the probability distribution of \(Y = X^2\), we first need to derive the transformation from \(x\) to \(y\). We have \(y = x^2\), which implies \(x = \sqrt{y}\).
02

Calculate the derivative for the transformation

The transformation \(x = \sqrt{y}\) gives us \(\frac{dx}{dy} = \frac{1}{2\sqrt{y}}\). This derivative is necessary to find the probability density function of \(Y\).
03

Determine the PDF of Y=X^2

The probability density function of \(Y\) can be found using \(f_Y(y) = f_X(x) \left|\frac{dx}{dy}\right|\). Substitute \(x = \sqrt{y}\) and \(\frac{dx}{dy} = \frac{1}{2\sqrt{y}}\). Hence, \[f_Y(y) = e^{- rac{y}{2}} \frac{1}{2\sqrt{y}} = \frac{1}{2\sqrt{y}} e^{- rac{y}{2}}, \quad y \geq 0.\]
04

Identify the transformation for Y=X^{1/2}

For \(Y = X^{1/2}\), we have \(y = \sqrt{x}\), which means \(x = y^2\).
05

Calculate the derivative for the transformation

Using \(x = y^2\), the derivative \(\frac{dx}{dy} = 2y\) is found, and it's needed for the transformed PDF.
06

Determine the PDF of Y=X^{1/2}

The probability density function of \(Y\) is found by substitution:\[f_Y(y) = f_X(x) \left|\frac{dx}{dy}\right| = e^{-y^2} \cdot 2y = 2y e^{-y^2}, \quad y \geq 0.\]
07

Identify the transformation for Y=ln X

For \(Y = \ln X\), we have \(y = \ln x\), implying \(x = e^y\).
08

Calculate the derivative for the transformation

The derivative \(\frac{dx}{dy} = e^y\) is necessary for determining the probability distribution of \(Y\).
09

Determine the PDF of Y=ln X

The probability density function of \(Y\) can be calculated as:\[f_Y(y) = e^{-e^y} \cdot e^y = e^{-y - e^y}, \quad y \in \mathbb{R}.\]

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Key Concepts

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

Probability Density Function
A Probability Density Function (PDF) describes the likelihood of a random variable taking on a particular value in a continuous data set. It specifies the probabilities of all possible outcomes. Unlike probabilities in discrete distributions that sum up to 1, the integral of a PDF over its range equals 1. This is because the probability of a continuous random variable being exactly any given value is theoretically zero, which means we focus on ranges or intervals.
For example, if we have a random variable with an exponential distribution, its PDF could express how likely it is for the variable to fall within a specific range. The formula for the exponential distribution is typically shown as \(f(x) = \lambda e^{-\lambda x}\), where \(\lambda > 0\) is called the rate parameter. In this exercise, however, the distribution is \(f_{X}(x)=e^{-x}\) for \(x \geq 0\). This is a special case where \(\lambda = 1\).
When transforming random variables, we also transform their PDFs. This requires using derivatives to adjust for changes in the scale or domain of the function, ensuring the transformed PDF correctly represents the new distribution.
Random Variables
Random Variables are fundamental to probability theory, acting as functions that assign a real number to each outcome of a random experiment. There are two primary types: discrete and continuous random variables. Discrete random variables have countable outcomes, like the sum of dice rolls, while continuous ones can take any value within a range, like the height of individuals.
In the given problem, \(X\) is a continuous random variable with an exponential distribution. The transformations considered, like \(Y = X^2\), \(Y = X^{1/2}\), and \(Y = \ln X\), demonstrate how random variables can be manipulated to create new variables with different statistical properties. This involves finding a new PDF for \(Y\) by utilizing the relationship between \(X\) and \(Y\), for example, calculating derivatives like \(\frac{dx}{dy}\), which help express how probabilities scale during transformation.
By understanding these transformations, you gain insight into the flexibility and adaptability of random variables in various probabilistic scenarios.
Exponential Distribution
The Exponential Distribution is a widely used continuous probability distribution applicable to various processes, often modeling time until an event occurs, like waiting times or lifespans of certain products. An exponential distribution's defining characteristic is its constant hazard rate, meaning the event's likelihood doesn't change over time.
The PDF for an exponential distribution with rate parameter \(\lambda\) is given by \(f(x) = \lambda e^{-\lambda x}\), where \(x \geq 0\). In solving problems involving transformations of exponential distributions, identifying the nature of the transformed distribution is crucial. This involves employing calculus to transform the original PDF into the altered context.
For instance, in the exercise given, the transformations of \(X\) such as squaring or taking the square root result in new distributions that may answer different probabilistic questions. By mastering exponential distributions and their transformations, it is possible to address a wide array of real-world problems effectively.

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Most popular questions from this chapter

Contamination problems in semiconductor manufacturing can result in a functional defect, a minor defect, or no defect in the final product. Suppose that \(20,50,\) and \(30 \%\) of the contamination problems result in functional, minor, and no defects, respectively. Assume that the defects of 10 contamination problems are independent. (a) What is the probability that the 10 contamination problems result in two functional defects and five minor defects? (b) What is the distribution of the number of contamination problems that result in no defects? (c) What is the expected number of contamination problems that result in no defects?

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