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Choose a number \(U\) from the interval [0,1] with uniform distribution. Find the cumulative distribution and density for the random variables (a) \(Y=|U-1 / 2|\) (b) \(Y=(U-1 / 2)^{2}\).

Short Answer

Expert verified
(a) CDF: 2y, PDF: 2 (for 0 ≤ y ≤ 1/2). (b) CDF: 2√y, PDF: 1/√y (for 0 < y ≤ 1/4).

Step by step solution

01

Understanding the Problem

We need to find the cumulative distribution function (CDF) and probability density function (PDF) for the given transformations of the uniformly distributed random variable \(U\) over the interval \([0,1]\) for two cases: \(Y = |U - 1/2|\) and \(Y = (U - 1/2)^2\).
02

Finding the CDF and PDF for Y = |U - 1/2|

For \(Y = |U - 1/2|\), the range of \(U\) is split into two parts: \([0, 1/2]\) and \((1/2, 1]\). Here, \(Y\) takes the values from \([0, 1/2]\) because the maximum value of \(|U - 1/2|\) is \(|0 - 1/2| = 1/2\) or \(|1 - 1/2| = 1/2\). To find the CDF \(F_Y(y)\), we calculate \(P(Y \leq y)\). For \(0 \leq y \leq 1/2\), since \(Y = |U - 1/2|\), \(U\) must be within \([1/2-y, 1/2 + y]\). The probability is then \(F_Y(y) = (1/2 + y) - (1/2 - y) = 2y\). Thus, the CDF is \(F_Y(y) = 2y\) for \(0 \leq y \leq 1/2\). To find the PDF, we differentiate the CDF: \(f_Y(y) = \frac{d}{dy}F_Y(y) = 2\) for \(0 \leq y \leq 1/2\).
03

Finding the CDF and PDF for Y = (U - 1/2)^2

For \(Y = (U - 1/2)^2\), \(Y\) can take values in \([0, 1/4]\) (the maximum for \((U-1/2)^2\) is when \(U\) is \(0\) or \(1\), resulting in a maximum of \(1/4\)). To find the CDF \(F_Y(y)\), we set \((U - 1/2)^2 \leq y\). Solving for \(U\), this means \(U \in [1/2 - \sqrt{y}, 1/2 + \sqrt{y}]\). Hence, \(F_Y(y) = (1/2 + \sqrt{y}) - (1/2 - \sqrt{y}) = 2\sqrt{y}\). Thus, for \(0 \leq y \leq 1/4\), the CDF is \(F_Y(y) = 2\sqrt{y}\). The PDF is found by differentiating the CDF: \(f_Y(y) = \frac{d}{dy}F_Y(y) = \frac{1}{\sqrt{y}}\) for \(0 < y \leq 1/4\). At \(y=0\), the PDF is undefined, as is usual in such cases where the CDF starts from zero.

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Key Concepts

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

Cumulative Distribution Function (CDF)
When discussing probability theory, the Cumulative Distribution Function (CDF) is a crucial concept to understand. It describes the probability that a random variable takes on a value less than or equal to a specific level. This function is essential in determining the likelihood of outcomes below a particular threshold within a probability distribution.
For example, let's walk through how CDF is used in different scenarios. In the case where \(Y = |U-1/2|\) (from the exercise), we want to find the probability of Y being less than or equal to a specific value, y. We look at the intervals in which Y can exist: 0 to 1/2. Using the CDF formula, \(F_Y(y) = P(Y \leq y)\), we get \(F_Y(y) = 2y\) for 0 \leq y \leq 1/2.

Similarly, for \(Y = (U-1/2)^2\), the CDF \(F_Y(y)\) tells us the probability that the value of (U-1/2)^2 is less than or equal to y, and this is calculated as \(F_Y(y) = 2\sqrt{y}\) for 0 \leq y \leq 1/4. By integrating the concepts of intervals and transformations, you can find the CDF for various random variables.
Probability Density Function (PDF)
The Probability Density Function (PDF) is another fundamental concept in probability theory, acting as a counterpart to the CDF. It provides a relative likelihood of a random variable to assume a particular value by describing the distribution of probabilities per unit along the continuum.
In our given scenarios, after determining the CDFs, we obtain the PDFs by differentiating these CDFs. This differentiation reveals the rate of change or the probability density across different values.
For instance, given \(Y = |U-1/2|\), the PDF \(f_Y(y)\) can be determined by differentiating \(F_Y(y) = 2y\) producing \(f_Y(y) = 2\) for 0 \leq y \leq 1/2. This constant PDF indicates a uniform distribution in this interval.

In contrast, for \(Y = (U-1/2)^2\), a different scenario unfolds. Here, \(f_Y(y)\) is obtained by differentiating \(F_Y(y) = 2\sqrt{y}\), leading to \(f_Y(y) = \frac{1}{\sqrt{y}}\) for 0 < y \leq 1/4. The difficulty often arises at the edges of the distribution, such as at y = 0, where the PDF is undefined due to a singularity.
Uniform Distribution
Uniform Distribution is a core concept that refers to a type of probability distribution where all outcomes are equally likely. It is like having a perfectly balanced die where each face has an equal chance of showing up after a roll.
The uniform distribution is often used as a simple model for random selection, exemplified in our exercise by the variable U, uniformly distributed over \[0, 1\]. This distribution implies that any subinterval \([a, b]\) of \[0, 1\] has the same likelihood relative to its length. This is essential when analyzing transformations of U, such as in \(Y = |U-1/2|\) and \(Y = (U-1/2)^2\).

In the context of transformations:
  • For \(Y = |U-1/2|\), it’s about identifying parts of U that equate absolute differences with respect to 1/2.
  • For \(Y = (U-1/2)^2\), the focus shifts to squaring these differences.
Both scenarios preserve the uniform distribution's characteristic of treating each transformed outcome within feasible bounds equitably. This makes uniform distribution a foundational model in probability facilitating many theoretical and practical applications.

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Most popular questions from this chapter

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