/*! 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 9 Let \(X\) be a continuous rv wit... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) be a continuous rv with cdf $$ F(x)=\left\\{\begin{array}{cl} 0 & x \leq 0 \\ \frac{x}{4}\left[1+\ln \left(\frac{4}{x}\right)\right] & 04 \end{array}\right. $$ [This type of cdf is suggested in the article "Variability in Measured Bedload-Transport Rates" (Water 91Ó°ÊÓ Bull., 1985: 39 -48) as a model for a certain hydrologic variable.] What is a. \(P(X \leq 1)\) ? b. \(P(1 \leq X \leq 3)\) ? c. The pdf of \(X\) ?

Short Answer

Expert verified
a. 0.5966 b. 0.3692 c. \(f(x) = \frac{1}{4}(1 + \ln(\frac{x}{4}))\) for \( 0 < x \leq 4 \).

Step by step solution

01

Find P(X ≤ 1)

To find \( P(X \leq 1) \), use the cumulative distribution function (cdf) for the interval \( 0 < x \leq 4 \). Plug \( x = 1 \) into the cdf:\[F(1) = \frac{1}{4}\left[1 + \ln\left(\frac{4}{1}\right)\right] = \frac{1}{4}\left[1 + \ln(4)\right]\]Calculate this to find \( F(1) \).
02

Calculate P(X ≤ 1)

Now, substitute for \( \ln(4) = 1.3863 \) approximately:\[F(1) = \frac{1}{4}(1 + 1.3863) = \frac{1}{4} \times 2.3863 = 0.596575\]So, \( P(X \leq 1) \approx 0.5966 \).
03

Find P(1 ≤ X ≤ 3)

To find \( P(1 \leq X \leq 3) \), compute the difference \( F(3) - F(1) \). First, find \( F(3) \) using the cdf:\[F(3) = \frac{3}{4}\left[1 + \ln\left(\frac{4}{3}\right)\right]\]Calculate this part to find \( F(3) \).
04

Calculate F(3)

Calculate \( \ln\left(\frac{4}{3}\right) \) approximately as \( 0.28768 \):\[F(3) = \frac{3}{4}(1 + 0.28768) = \frac{3}{4} \times 1.28768 = 0.96576\]Now, use this to find the probability.
05

Calculate P(1 ≤ X ≤ 3)

Subtract \( F(1) \) from \( F(3) \):\[P(1 \leq X \leq 3) = F(3) - F(1) = 0.96576 - 0.596575 = 0.369185\]So, \( P(1 \leq X \leq 3) \approx 0.3692 \).
06

Find the pdf of X

The probability density function (pdf) \( f(x) \) is the derivative of the cdf \( F(x) \) where it is continuous. Compute the derivative for \( 0 < x \leq 4 \):\[f(x) = \frac{d}{dx}\left[\frac{x}{4}\left(1 + \ln\left(\frac{4}{x}\right) \right) \right]\]Expand and calculate each term.
07

Calculation of the pdf

Let's differentiate:\[f(x) = \frac{d}{dx}\left(\frac{x}{4}\right) + \frac{d}{dx}\left(\frac{x}{4} \ln\left(\frac{4}{x}\right) \right)\]Calculate the derivative of \( \frac{x}{4} \) to be \( \frac{1}{4} \), and use the product rule for the second term. Simplifying gives:\[f(x) = \frac{1}{4}(1 - \ln\left(\frac{4}{x}\right))\]
08

Simplify the pdf

Substitute \( 0 < x \leq 4 \) which confirms that the pdf is:\[f(x) = \frac{1}{4}\left(1 + \ln\left(\frac{x}{4}\right)\right)\]So the pdf of \( X \) for \( 0 < x \leq 4 \) is this expression.

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Key Concepts

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

Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) of a continuous random variable is a fundamental concept in probability and statistics. It describes the probability that a random variable is less than or equal to a certain value. For a variable \(X\), the CDF, denoted as \(F(x)\), is defined as:
  • \(F(x) = P(X \leq x)\)
The CDF is a non-decreasing function that starts at 0 and approaches 1 as \(x\) increases. It provides a complete picture of the distribution of a continuous random variable, mapping each possible value to a probability. In the exercise above, the CDF is given by different expressions based on the value of \(x\):
  • \(F(x) = 0\) for \(x \leq 0\)
  • \(F(x) = \frac{x}{4}\left[1+\ln \left(\frac{4}{x}\right)\right]\) for \(0 < x \leq 4\)
  • \(F(x) = 1\) for \(x > 4\)
These expressions indicate the behavior of the random variable across its possible range, illustrating how the probability is distributed.
Probability Density Function (PDF)
For continuous random variables, the Probability Density Function (PDF) gives a functional form that influences the likelihood of a variable taking on a particular value. However, the probability of \(X\) being exactly a single value \(x\) is technically zero. Instead, the PDF, often denoted \(f(x)\), describes the density of the probability around \(x\). Integrating the PDF over a range gives the probability that \(X\) lies within that range:
  • \(P(a < X < b) = \int_{a}^{b} f(x) \, dx\)
From the exercise, the PDF can be found by differentiating the CDF over the interval where it is continuous, specifically for \(0 < x \leq 4\):
  • First term: Derivative of \(\frac{x}{4}\) is \(\frac{1}{4}\).
  • Second term: Product rule for \(\frac{x}{4} \ln\left(\frac{4}{x}\right)\)
After simplification, this results in:
  • \(f(x) = \frac{1}{4}(1 + \ln\left(\frac{x}{4}\right))\)
This expression represents the normalized rate at which probability is assigned along the range of \(X\).
Integration in Statistics
Integration plays a crucial role in statistics, particularly in handling continuous random variables. When dealing with PDFs, integration allows us to calculate probabilities over intervals. This process sums up infinitesimal slices of probability across a specified range:
  • The definite integral of the PDF from \(a\) to \(b\) gives the probability that a continuous random variable falls within this interval: \(\int_{a}^{b} f(x) \, dx\).
  • Integrating the PDF over its entire range gives 1, confirming that the total probability is distributed.
In the provided exercise, integration is implicitly used to transition from the PDF to probability outcomes. For example, knowing \(P(1 \leq X \leq 3)\) involves the differences \(F(3) - F(1)\), which conceptually relies on integrating the PDF across this range. Differentiating the CDF to find the PDF is the reverse procedure, using calculus to articulate how probability is distributed along a continuous spectrum. This forms the backbone of probability interpretations for continuous random variables.

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