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Chapter 1: Problem 12

Find the cdf \(F(x)\) associated with each of the following probability density functions. Sketch the graphs of \(f(x)\) and \(F(x)\). (a) \(f(x)=3(1-x)^{2}, 0

Short Answer

Expert verified
The CDFs are \(F(x) = 1 - (1-x)^3\) for (a), \(F(x) = 1 - 1 / x\) for (b), and \(F(x) = x / 3\) for \(0 < x < 1\) and \(F(x) = (x - 2) / 3 + 1 /3\) for \(2 < x < 4\) in (c). The median and the 25th percentile for each distribution need to be calculated by solving the respective CDF equal to 0.5 and 0.25.

Step by step solution

01

Part a

Given the PDF \(f(x) = 3(1-x)^{2}\) for \(0 < x < 1\). To find the CDF, integrate from the lower limit to \(x\): \(F(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} 3(1-t)^{2} dt = [-(1-t)^3]|_{0}^{x} = 1 - (1-x)^3\). The graph of \(f(x)\) is a downward parabola while \(F(x)\) starts from 0 at \(x = 0\) and ends at 1 at \(x = 1\). Solve \(1 - (1-x)^3 = 0.5\) for the median and \(1 - (1-x)^3 = 0.25\) for the 25th percentile.
02

Part b

Given the PDF \(f(x) = 1 / x^{2}\) for \(1 < x < \infty\). The CDF is \(F(x) = \int_{1}^{x} f(t) dt =\int_{1}^{x} 1 / t^{2} dt = [-1/t] |_{1}^{x} = 1 - 1 / x\). The graph of \(f(x)\) is a downward hyperbola and \(F(x)\) starts at 0 at \(x = 1\) and ends at 1 as \(x -> \infty\). Solve \(1 - 1 / x = 0.5\) for the median and \(1 - 1 / x = 0.25\) for the 25th percentile.
03

Part c

Given the PDF \(f(x) = 1 / 3\) for \(0 < x < 1\) and \(2 < x < 4\). The CDF is \(F(x) = \int_{0}^{x} f(t) dt\) for \(0 < x < 1\) and \(F(x) = \int_{2}^{x} f(t) dt + 1 / 3\) for \(2 < x < 4\). The graphs of \(f(x)\) are straight lines at height 1/3 while \(F(x)\) raises linearly to 1/3 for \(x < 1\) and raises linearly to 1 for \(2 < x < 4\). Solve \(x/3 = 0.5\) for the median and \(x/3 = 0.25\) for the 25th percentile. Since the solutions are outside the range of \(x\), take \(\min(\max(x, 0), 1)\) where \(x\) is the solution.

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Key Concepts

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

Cumulative Distribution Function (CDF)
A Cumulative Distribution Function (CDF) provides a complete description of the probability distribution of a random variable. The CDF, denoted as \( F(x) \), expresses the probability that a random variable \( X \) will take a value less than or equal to \( x \). For continuous random variables, the CDF continuously increases from 0 to 1. To find the CDF from a Probability Density Function (PDF), we integrate the PDF over the range from the lower limit to \( x \).

Consider a PDF \( f(x) = 3(1-x)^2 \) for \( 0 < x < 1 \). The CDF is obtained by integrating the PDF:
  • \( F(x) = \int_{0}^{x} 3(1-t)^2 \, dt \)
  • This integration result is \( F(x) = 1 - (1-x)^3 \)
Evaluating and sketching, we see that \( f(x) \) is a downward parabola while \( F(x) \) rises from 0 to 1 as \( x \) increases from 0 to 1.
Median of a Distribution
The median of a distribution is the value that divides the probability distribution into two equal parts. It's the point \( x \) where the CDF is equal to 0.5. It tells us that there is a 50% chance that the random variable is less than or equal to the median value.

To find the median for our examples, we solve \( F(x) = 0.5 \) for each given distribution:
  • For \( f(x) = 3(1-x)^2 \): Solve \( 1 - (1-x)^3 = 0.5 \) gives the median \( x = 0.7937 \).
  • For \( f(x) = 1/x^2 \): Solve \( 1 - 1/x = 0.5 \) gives the median \( x = 2 \).
  • For \( f(x) = 1/3 \) over limited ranges: If the solution is outside specified ranges, adjustments are made to fit \( x \) within its valid range.
Percentiles
Percentiles provide another way to understand how data is distributed. Specifically, a 25th percentile, also known as the first quartile, represents the point below which 25% of the data fall. Similar to the median, we use the CDF to find the 25th percentile by solving \( F(x) = 0.25 \).

Here's how it's found in our examples:
  • For \( f(x) = 3(1-x)^2 \): Solve \( 1 - (1-x)^3 = 0.25 \) to obtain \( x = 0.6823 \).
  • For \( f(x) = 1/x^2 \): Solve \( 1 - 1/x = 0.25 \) to receive \( x = 4/3 \).
  • For the piecewise distribution: Considerations for valid \( x \) values need to be made since solutions might initially fall outside the specified range.
Mathematical Integration
Mathematical integration is a fundamental tool needed for finding the CDF from a given PDF. Integration essentially sums up infinite small areas under the curve \( f(x) \) to calculate the total area up to the point \( x \), which represents the CDF \( F(x) \).

In the context of probability:
  • Integration transforms \( f(x) \) from rate of change (PDF) into cumulative probability (CDF).
  • For example, integrating \( f(x) = 3(1-x)^2 \) gives a new function \( F(x) = 1 - (1-x)^3 \) indicating how the total probability accumulates as \( x \) increases.
Becoming comfortable with integration allows one to effectively extract valuable insights from analytic probability functions.

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Most popular questions from this chapter

Let \(X\) be a random variable with space \(\mathcal{D}\). For \(D \subset \mathcal{D}\), recall that the probability induced by \(X\) is \(P_{X}(D)=P[\\{c: X(c) \in D\\}] .\) Show that \(P_{X}(D)\) is a probability by showing the following: (a) \(P_{X}(\mathcal{D})=1\). (b) \(P_{X}(D) \geq 0\). (c) For a sequence of sets \(\left\\{D_{n}\right\\}\) in \(\mathcal{D}\), show that $$ \left\\{c: X(c) \in \cup_{n} D_{n}\right\\}=\cup_{n}\left\\{c: X(c) \in D_{n}\right\\} $$ (d) Use part (c) to show that if \(\left\\{D_{n}\right\\}\) is sequence of mutually exclusive events, then $$ P_{X}\left(\cup_{n=1}^{\infty} D_{n}\right)=\sum_{n=1}^{\infty} P_{X}\left(D_{n}\right) $$

Hunters A and B shoot at a target; the probabilities of hitting the target are \(p_{1}\) and \(p_{2}\), respectively. Assuming independence, can \(p_{1}\) and \(p_{2}\) be selected so that \(P(\) zero hits \()=P(\) one hit \()=P(\) two hits \() ?\)

Let \(\mathcal{C}=R\), where \(R\) is the set of all real numbers. Let \(\mathcal{I}\) be the set of all open intervals in \(R\). The Borel \(\sigma\) -field on the real line is given by $$ \mathcal{B}_{0}=\cap\\{\mathcal{E}: \mathcal{I} \subset \mathcal{E} \text { and } \mathcal{E} \text { is a } \sigma \text { -field }\\} $$ By definition, \(\mathcal{B}_{0}\) contains the open intervals. Because \([a, \infty)=(-\infty, a)^{c}\) and \(\mathcal{B}_{0}\) is closed under complements, it contains all intervals of the form \([a, \infty)\), for \(a \in R\). Continue in this way and show that \(\mathcal{B}_{0}\) contains all the closed and half- open intervals of real numbers.

A pair of dice is cast until either the sum of seven or eight appears. (a) Show that the probability of a seven before an eight is \(6 / 11\). (b) Next, this pair of dice is cast until a seven appears twice or until each of a six and eight has appeared at least once. Show that the probability of the six and eight occurring before two sevens is \(0.546\).

Find the 25 th percentile of the distribution having pdf \(f(x)=|x| / 4\), where \(-2
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