/*! 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 121 If \(X\) is uniformly distribute... [FREE SOLUTION] | 91Ó°ÊÓ

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If \(X\) is uniformly distributed on \([-1,3]\), find the pdf of \(Y=X^{2}\).

Short Answer

Expert verified
The pdf of \(Y\) is \(f_Y(y) = \frac{1}{4\sqrt{y}}\) for \(0 < y \leq 9\).

Step by step solution

01

Define the Original PDF

If a random variable \(X\) is uniformly distributed over \([-1, 3]\), then its probability density function (pdf) is given by \(f_X(x) = \frac{1}{b-a}\), where \(a = -1\) and \(b = 3\). Hence, \(f_X(x) = \frac{1}{3-(-1)} = \frac{1}{4}\) for \(-1 \leq x \leq 3\).
02

Transform the Variable

We are given \(Y = X^2\). The possible values of \(Y\) will range from the square of the smallest value of \(X\) to the square of the largest value. Therefore, \(Y\) ranges from \(0\) to \(9\), because \((-1)^2 = 1\) and \(3^2 = 9\).
03

Find the Corresponding Values of X

For a given \(y \geq 0\), solve for \(x\) such that \(y = x^2\). This gives \(x = \sqrt{y}\) or \(x = -\sqrt{y}\). These solutions provide the set of \(x\) values that map to each \(y\) value.
04

Use the Change of Variable Formula

The pdf of \(Y\) is found by taking the absolute value of the derivative of the inverse function (from Step 3) and multiplying it by the pdf of \(X\). Therefore, the pdf \(f_Y(y)\) is \( f_Y(y) = \sum \left| \frac{dx}{dy} \right| f_X(x) \). Since \(x = \pm \sqrt{y}\), \( \frac{dx}{dy} = \frac{1}{2\sqrt{y}}\), hence \( f_Y(y) = \left| \frac{1}{2\sqrt{y}} \right| \frac{1}{4} + \left| \frac{1}{(-2\sqrt{y})} \right| \frac{1}{4} = \frac{1}{4\sqrt{y}} \). This is valid for \(0 < y \leq 1\).
05

Adjust for Full Range of Y

For \(1 < y \leq 9\), \(x\) can only take positive values, which is \(x = \sqrt{y}\). For these values of \(y\), the derivative \( \frac{dx}{dy} = \frac{1}{2\sqrt{y}} \), and after adjustments, we find that the full pdf of \(Y\) is \[ f_Y(y) = \begin{cases} \frac{1}{4\sqrt{y}}, & 0 < y \leq 9 \end{cases} \].

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Key Concepts

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

Uniform Distribution
Uniform distribution is one of the simplest probability distributions in statistics. It describes a situation where all outcomes are equally likely within a certain range. This range is defined by two parameters, the lower limit and the upper limit.

In mathematical terms, if a random variable \(X\) is uniformly distributed over an interval \([a, b]\), its probability density function (pdf) is a constant given by \(f_X(x) = \frac{1}{b-a}\). This means that the likelihood of \(X\) taking any value within the interval \([a, b]\) is the same. For all values outside this interval, the probability is zero.
  • The shape of the pdf is a rectangle, with the height representing the constant probability.
  • The area under this rectangle must equal 1, ensuring total probability sums to one.
  • Uniform distribution is typically expressed over a finite interval where \(a < b\).
Remember, the uniform distribution assumes each outcome in the interval has identical probability. This simple distribution is a great starting point for understanding more complex probabilistic models.
Change of Variable Formula
The Change of Variable Formula is an essential tool in probability and statistics, especially when dealing with transformations of random variables. This mathematical technique helps us find the probability distribution of a transformed random variable by changing from the original variable to a new one.

The Change of Variable Formula, in essence, involves three main steps:
  • **Identify the transformation:** If \(Y\) is a function of \(X\), say \(Y = g(X)\), you first express \(X\) in terms of \(Y\), if possible.
  • **Calculate the derivative:** Find the derivative of \(x\) with respect to \(y\), \(\frac{dx}{dy}\), for each possible solution of \(x\).
  • **Express the new pdf:** Insert this derivative into the formula \(f_Y(y) = | \frac{dx}{dy} | f_X(x)\), which accounts for how the "stretching" or "shrinking" of regions under the transformation affects the density of probabilities.
In the context of our exercise, using this formula allows us to determine how the uniform distribution of \(X\) over \([-1, 3]\) transforms into a distribution of \(Y = X^2\). This transformation changes the shape of the distribution due to the non-linear relationship between \(X\) and \(Y\).
Transformation of Variables in Probability
The transformation of variables is a powerful concept in probability that allows us to understand and compute the probabilities of functions of random variables. Often starting with a known distribution, this method lets us discover how a transformation affects the distribution's characteristics.

When transforming a random variable, it includes the following steps:
  • **Determine the range:** Identify the range of the transformed variable by applying the transformation to the range of the original variable.
  • **Find inverse transformations:** Determine which values of the original variable relate to each value of the transformed variable.
  • **Adjust probabilities:** Using the pdf of the original variable, adjust for changes brought by the transformation, ensuring the pdf reflects accurate probabilities over the new variable.
For instance, transforming \(X\) to \(Y = X^2\) creates a range from 0 to 9, assuming \(X\) ranges from -1 to 3. This changes from a simple linear interval into a quadratic interval. Understanding this approach helps address a wide array of real-world problems, including risk assessment, signal processing, and more.

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Most popular questions from this chapter

The Rockwell hardness of a metal is determined by impressing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of an alloy is normally distributed with mean 70 and standard deviation 3. (Rockwell hardness is measured on a continuous scale.) a. If a specimen is acceptable only if its hardness is between 67 and 75 , what is the probability that a randomly chosen specimen has an acceptable hardness? b. If the acceptable range of hardness is \((70-c\), \(70+c\) ), for what value of \(c\) would \(95 \%\) of all specimens have acceptable hardness? c. If the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimens is independently determined, what is the expected number of acceptable specimens among the ten? d. What is the probability that at most 8 of 10 independently selected specimens have a hardness of less than \(73.84\) ? [Hint: \(Y=\) the number among the ten specimens with hardness less than \(73.84\) is a binomial variable; what is \(p\) ?]

Determine \(z_{\alpha}\) for the following: a. \(\alpha=.0055\) b. \(\alpha=.09\) c. \(\alpha=.663\)

Let \(X\) denote the distance \((\mathrm{m})\) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose that for bannertailed kangaroo rats, \(X\) has an exponential distribution with parameter \(\lambda=.01386\) (as suggested in the article "Competition and Dispersal from Multiple Nests,"Ecology, 1997: 873–883). a. What is the probability that the distance is at most \(100 \mathrm{~m}\) ? At most \(200 \mathrm{~m}\) ? Between 100 and \(200 \mathrm{~m}\) ? b. What is the probability that distance exceeds the mean distance by more than 2 standard deviations? c. What is the value of the median distance?

Let \(X\) have the pdf \(f_{X}(x)=2 / x^{3}, x>1\). Find the pdf of \(Y=\sqrt{X}\).

An ecologist wishes to mark off a circular sampling region having radius \(10 \mathrm{~m}\). However, the radius of the resulting region is actually a random variable \(R\) with pdf $$ f(r)=\left\\{\begin{array}{cl} \frac{3}{4}\left[1-(10-r)^{2}\right] & 9 \leq r \leq 11 \\ 0 & \text { otherwise } \end{array}\right. $$ What is the expected area of the resulting circular region?

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