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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) \(Y=1 /(U+1)\). (b) \(Y=\log (U+1)\).

Short Answer

Expert verified
(a) CDF: \(F_Y(y) = 1 - \max\left(0, \frac{1}{y} - 1\right)\) and density \(f_Y(y) = \frac{1}{y^2}\); (b) CDF: \(F_Y(y) = e^y - 1\) and density \(f_Y(y) = e^y\).

Step by step solution

01

Define the cumulative distribution function (CDF) for Y

To find the CDF of a transformed variable like \(Y\), we express \(P(Y \leq y)\) in terms of \(U\). Since \(U\) is uniformly distributed over \([0,1]\), its CDF is \(F_U(u) = u\).
02

CDF for Y = 1 / (U + 1)

We want to find \(P\left(\frac{1}{U+1} \leq y\right)\). This simplifies to \(P(U+1 \geq \frac{1}{y})\), and \(P(U \geq \frac{1}{y} - 1)\). Constraining this to the interval \([0,1]\), the CDF is \(F_Y(y) = 1 - \max\left(0, \frac{1}{y} - 1\right)\) for \(y \geq 1\).
03

Density function for Y = 1 / (U + 1)

Differentiate the CDF \(F_Y(y)\) with respect to \(y\) to find the density function \(f_Y(y)\). Since \(F_Y(y) = 1 - (1/y) + 1 = 2 - 1/y\), differentiating gives \(f_Y(y) = \frac{1}{y^2}\) for \(y \geq 1\).
04

CDF for Y = log(U + 1)

For \(Y = \log(U+1)\), express \(P(\log(U+1) \leq y)\), which means \(U+1 \leq e^y\) or \(U \leq e^y - 1\). Therefore, \(F_Y(y) = e^y - 1\) for \(-\infty < y \leq \log(2)\).
05

Density function for Y = log(U + 1)

Differentiate \(F_Y(y) = e^y - 1\) with respect to \(y\) to obtain the density function. We get \(f_Y(y) = e^y\) for \(-\infty < y \leq \log(2)\).

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Key Concepts

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

Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is a pivotal statistical tool that helps us understand the probability distribution of a random variable. It represents the probability that a random variable takes on a value less than or equal to a specific value. The CDF is expressed mathematically as \( F_Y(y) = P(Y \leq y) \).
For example, in the exercise, when we deal with the transformed variables \( Y = 1/(U+1) \) and \( Y = \log(U+1) \), we redefine the problem in terms of the original uniformly distributed variable \( U \).
  • For \( Y = 1/(U+1) \): The CDF is computed by recognizing that \( P\left( \frac{1}{U+1} \leq y \right) \) transforms to \( P(U \geq \frac{1}{y} - 1) \). This gives us the CDF formula \( F_Y(y) = 1 - \max(0, \frac{1}{y} - 1) \) for \( y \geq 1 \).
  • For \( Y = \log(U+1) \): Here \( P(\log(U+1) \leq y) \) changes into \( P(U \leq e^y - 1) \) resulting in the CDF \( F_Y(y) = e^y - 1 \) for \( -\infty < y \leq \log(2) \).
CDFs are useful for determining the probability that a random variable falls within a particular range, making them essential for statistical analysis and probabilistic modeling.
Uniform Distribution
Uniform distribution describes a situation where all outcomes are equally likely within a specified range. In mathematical terms, if a random variable \( U \) is uniformly distributed over an interval \([a, b]\), each number in this interval is just as likely to be picked as any other. When \([a, b] = [0, 1]\), as in our exercise, it is known as the standard uniform distribution.
  • Characteristics: A uniform distribution over \([0, 1]\) implies that the random variable can take any value in this interval with equal probability, forming the basis for generating random numbers or modeling uncertainties.
  • Importance: This distribution serves as a foundation for other statistical concepts. Many random number generators are based on it, leveraging its simplicity and balance.
  • Connection: The variable \( U \) in the exercise represents a uniform distribution over the interval \([0, 1]\). Its role is crucial in finding CDFs and density functions of transformed variables \( Y \).
Understanding uniform distribution allows for better comprehension of basic statistical models and aids in building more complex probability distributions from simpler ones.
Density Function
The density function, or Probability Density Function (PDF), is a crucial component in understanding continuous probability distributions. For a given random variable, the density function provides the probabilities for the outcomes within its range. Essentially, it describes the distribution of probabilities across different values of the variable.
In our exercise, we find the density function by differentiating the CDF of a transformed variable:
  • For \( Y = 1/(U+1) \): The density function \( f_Y(y) \) is derived as \( \frac{1}{y^2} \) for \( y \geq 1 \). This function tells us how the probabilities are spread, indicating that smaller values of \( Y \) are more likely.
  • For \( Y = \log(U+1) \): Similarly, the density function is found to be \( e^y \) for \( -\infty < y \leq \log(2) \). This function shows that as \( Y \) grows, the probability of observing such values diminishes.
Density functions illustrate how probabilities "densify" over intervals, providing a continuous alternative to discrete probability techniques. They're indispensable in fields such as statistics, machine learning, and natural simulations.

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Most popular questions from this chapter

There are an unknown number of moose on Isle Royale (a National Park in Lake Superior). To estimate the number of moose, 50 moose are captured and tagged. Six months later 200 moose are captured and it is found that 8 of these were tagged. Estimate the number of moose on Isle Royale from these data, and then verify your guess by computer program (see Exercise 36 ).

Suppose that \(X\) is a random variable which represents the number of calls coming in to a police station in a one-minute interval. In the text, we showed that \(X\) could be modelled using a Poisson distribution with parameter \(\lambda\), where this parameter represents the average number of incoming calls per minute. Now suppose that \(Y\) is a random variable which represents the number of incoming calls in an interval of length \(t\). Show that the distribution of \(Y\) is given by $$P(Y=k)=e^{-\lambda t} \frac{(\lambda t)^{k}}{k !}i.e., \(Y\) is Poisson with parameter \(\lambda t\). Hint: Suppose a Martian were to observe the police station. Let us also assume that the basic time interval used on Mars is exactly \(t\) Earth minutes. Finally, we will assume that the Martian understands the derivation of the Poisson distribution in the text. What would she write down for the distribution of $Y ?$$$

It is often assumed that the auto traffic that arrives at the intersection during a unit time period has a Poisson distribution with expected value \(m\). Assume that the number of cars \(X\) that arrive at an intersection from the north in unit time has a Poisson distribution with parameter \(\lambda=m\) and the number \(Y\) that arrive from the west in unit time has a Poisson distribution with parameter \(\lambda=\bar{m} .\) If \(X\) and \(Y\) are independent, show that the total number \(X+Y\) that arrive at the intersection in unit time has a Poisson distribution with parameter \(\lambda=m+\bar{m}\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be \(n\) mutually independent random variables, each of which is uniformly distributed on the integers from 1 to \(k\). Let \(Y\) denote the minimum of the \(X_{i}\) 's. Find the distribution of \(Y\).

(a) Compute the leading digits of the first 100 powers of 2 , and see how well these data fit the Benford distribution. (b) Multiply each number in the data set of part (a) by \(3,\) and compare the distribution of the leading digits with the Benford distribution.

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