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Chapter 12: Problem 2

Show that $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{2} & \text { for } 0

Short Answer

Expert verified
The function is a valid density function and the distribution function is \( F(x) = \begin{cases} 0 & \text{if } x < 0, \\ \frac{1}{2}x & \text{if } 0 \leq x < 2, \\ 1 & \text{if } x \geq 2. \end{cases} \)

Step by step solution

01

Verify Non-Negativity

For a function to be a density function, it must be non-negative for all real numbers. Here, \( f(x) = \frac{1}{2} \) for \( 0 < x < 2 \) and \( f(x) = 0 \) otherwise. Since both these values are non-negative, the function fulfills this criterion.
02

Check that Total Area Integrates to 1

The integral of the density function over its entire range should equal 1. Therefore, we need to evaluate \[ \int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{2} \frac{1}{2} \, dx. \] Evaluate the integral: \( \int_{0}^{2} \frac{1}{2} \, dx = \left[ \frac{1}{2}x \right]_0^2 = \frac{1}{2} \times 2 - \frac{1}{2} \times 0 = 1. \) Thus, the total area under the curve \( f(x) \) equals 1, confirming it as a density function.
03

Define the Distribution Function

The distribution function \( F(x) \) is defined as the integral of \( f(x) \) from \(-\infty\) to \( x \). We calculate \( F(x) \) in parts: 1. **For** \( x < 0 \), \( F(x) = 0 \).2. **For** \( 0 \leq x < 2 \), evaluate \( F(x) = \int_{0}^{x} \frac{1}{2} \, dt = \frac{1}{2}x. \)3. **For** \( x \geq 2 \), \( F(x) = 1 \). Hence, \( F(x) \) is given by: \[ F(x) = \begin{cases} 0 & \text{if } x < 0, \ \frac{1}{2}x & \text{if } 0 \leq x < 2, \ 1 & \text{if } x \geq 2. \end{cases} \]

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Key Concepts

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

Distribution Function
In the realm of probability, the distribution function is crucial for understanding how probabilities distribute over outcomes. For a density function like our example, the distribution function, denoted as \( F(x) \), is established by integrating the density from \(-\infty\) to \( x \). This process accumulates the probabilities up to \( x \).

To articulate \( F(x) \) in different scenarios, we express it in a piecewise function:
  • \( F(x) = 0 \) when \( x < 0 \). Before reaching the starting point (0), there's no accumulated probability.
  • For \( 0 \leq x < 2 \), the integral calculates as \( F(x) = \frac{1}{2}x \), representing a linear increase in probability as \( x \) grows from 0 to just below 2.
  • When \( x \geq 2 \), \( F(x) = 1 \), showing that the entire probability is accumulated beyond 2, as our density does not extend past this bound.
The shape of \( F(x) \) generally rises from 0 to 1, aligning with the total probability of the sample space.
Non-Negativity
A fundamental property of a probability density function (pdf) is its non-negativity over all possible values. This ensures that probabilities are never less than zero. After all, a probability can't be negative. For the given function, \( f(x) = \frac{1}{2} \) when \( 0 < x < 2 \) and \( f(x) = 0 \) otherwise. Both values conform to non-negativity as they are either zero or positive.

When verifying non-negativity:
  • Check that for any \( x \), \( f(x) \ge 0 \).
  • Within the defined range, maintain positivity or zero values.
This stipulation is essential for a valid probability density function, ensuring that the model aligns with real-world interpretations of probability.
Integration over Range
Integration over the function's range is pivotal to ensure it sums to 1. This condition signifies that all probability is accounted for, indicating no missing or excess probability.

We compute
  • \( \int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{2} \frac{1}{2} \, dx \).
  • This integral evaluates to \( \left[ \frac{1}{2}x \right]_0^2 = 1 \), confirming the total probability is complete and equals 1.
A valid density must integrate to 1, defining the total probability across its range. Calculating the integration not only confirms completeness but also validates that the pdf is properly normalized over its domain.
Calculus
Calculus plays a vital role in understanding and working with probability functions, particularly through the use of integration and differentiation. The integral helps in accumulating probabilities as seen in distribution functions, while derivatives might be used in more advanced topics like finding expectations or other statistical measures.

In our context,
  • Integration converts the density function into a distribution function, \( F(x) \), showing probabilities accrued up to \( x \).
  • Calculus assures that transformations between density and distribution conserve properties like total area (or probability).
Mastering the calculus involved in probability concepts allows us to better model and predict outcomes, addressing complex problems with mathematical precision.

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Most popular questions from this chapter

Suppose the waiting time for the first success in an experiment is geometrically distributed with mean \(1 / p\). (a) Find the probability that the first success occurs on the \(k\) th trial. (b) The experiment is repeated after the first success. Assume that the waiting time for the second success has the same distribution as the waiting time for the first success. Find the probability mass function for the distribution of the second success.

Let \(X\) and \(Y\) be two independent random variables with probability mass function described by the following table: $$\begin{array}{rcc} \hline {\boldsymbol{k}} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{k}) & \boldsymbol{P}(\boldsymbol{Y}=\boldsymbol{k}) \\ \hline-3 & 0.1 & 0.1 \\ -1 & 0.1 & 0.2 \\ 0 & 0.2 & 0.1 \\ 0.5 & 0.3 & 0.3 \\ 2 & 0.15 & 0.1 \\ 2.5 & 0.15 & 0.2 \\ \hline \end{array}$$ (a) Find \(E(X)\) and \(E(Y)\). (b) Find \(E(X+Y)\). (c) Find \(\operatorname{var}(X)\) and \(\operatorname{var}(Y)\). (d) Find \(\operatorname{var}(X+Y)\).

The median lifetime is defined as the age \(x_{m}\) at which the probability of not having failed by age \(x_{m}\) is \(0.5\). Use a graphing calculator to numerically approximate the median lifetime if the hazard-rate function is $$ \lambda(x)=1.2+0.3 e^{0.5 x}, \quad x \geq 0 $$

Use the Poisson approximation. For a certain vaccine, 1 in 500 individuals experiences some side effects. Find the probability that, in a group of 200 people, at least 1 person experiences side effects.

The median lifetime is defined as the age \(x_{m}\) at which the probability of not having failed by age \(x_{m}\) is \(0.5\). Use a graphing calculator to numerically approximate the median lifetime if the hazard-rate function is $$ \lambda(x)=0.5+0.1 e^{0.2 x}, \quad x \geq 0 $$
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