Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

A continuous random variable \(X\) is said to have a uniform distribution on the interval \([a, b]\) if the PDF has the form $$ f(x)= \begin{cases}\frac{1}{b-a}, & \text { if } a \leq x \leq b \\ 0, & \text { otherwise }\end{cases} $$ (a) Find the probability that the value of \(X\) is closer to \(a\) than it is to \(b\). (b) Find the expected value of \(X\). (c) Find the CDF of \(X\).

Step-by-Step Solution

Answer

(a) \(\frac{1}{2}\), (b) \(\frac{a+b}{2}\), (c) \(F(x) = \frac{x-a}{b-a}\) for \(a \leq x \leq b\).

Step 1: Understanding the Problem

We need to solve a series of problems involving a uniform distribution for a random variable \(X\) over the interval \([a, b]\).

Step 1: Find Probability that \(X\) is Closer to \(a\) than \(b\)

To find the probability that \(X\) is closer to \(a\) than \(b\), we need to determine the values of \(x\) such that \(|x-a| < |x-b|\). Solving this inequality gives the region where \(a < x < \frac{a+b}{2}\), since at the midpoint \(x = \frac{a+b}{2}\), \(x\) is equidistant from both \(a\) and \(b\). The length of this interval is \(\frac{b-a}{2}\). The probability is then \(\frac{\text{length of interval}}{b-a} = \frac{1}{2}\). So, \(P(X < \frac{a+b}{2}) = \frac{1}{2}\).

Step 2: Find the Expected Value of \(X\)

The expected value \(E(X)\) of a uniform distribution over \([a, b]\) is calculated using the formula \(E(X) = \frac{a+b}{2}\). This makes sense because the uniform distribution implies symmetry, which results in the mean being the midpoint of the interval.

Step 3: Find the CDF of \(X\)

The cumulative distribution function (CDF) \(F(x)\) for \(a \leq x \leq b\) is given by the formula \(F(x) = \int_{a}^{x} f(t) \, dt\). Substituting \(f(x) = \frac{1}{b-a}\), we get:\[ F(x) = \int_{a}^{x} \frac{1}{b-a} \, dt = \frac{x-a}{b-a} \]For \(x < a\), \(F(x) = 0\), and for \(x > b\), \(F(x) = 1\). Thus, the CDF is:\[ F(x) = \begin{cases} 0 & \text{if } x < a \ \frac{x-a}{b-a} & \text{if } a \leq x \leq b \ 1 & \text{if } x > b \end{cases} \]

Step 5: Conclusion

We have solved the problems using the properties of a uniform distribution. The specific results are for the uniform distribution over an interval.

Key Concepts

Probability Density Function

The probability density function (PDF) for a uniform distribution is quite straightforward, making it an ideal starting point for learning. When a continuous random variable \(X\) follows a uniform distribution over an interval \([a, b]\), the PDF is defined as a constant function on the interval. This is because each outcome within the interval is equally likely. Mathematically, the PDF is given by:

• \(f(x) = \frac{1}{b-a}\) for \(a \leq x \leq b\)
• \(f(x) = 0\) for \(x < a\) or \(x > b\)

This function tells us that the probability of \(X\) taking any specific value in \([a, b]\) is uniformly distributed. In practical terms, this means the height of the PDF is constant at \(\frac{1}{b-a}\), resulting in a rectangular shape when graphed. This consistent probability height introduces the concept of a *flat* distribution between \(a\) and \(b\).
What's important to remember is that, although the PDF gives the likelihood of \(X\) being near a specific point, the probability of \(X\) being exactly any single point is actually zero due to the continuous nature of \(X\). Instead, the PDF is primarily used to determine probabilities over intervals.

Expected Value

The expected value of a uniform distribution is crucial because it gives us insight into the central tendency of the distribution. For a random variable \(X\) uniformly distributed over \([a, b]\), the expected value \(E(X)\) is the midpoint of the interval. This is expressed mathematically as:

• \(E(X) = \frac{a+b}{2}\)

The reason the expected value is the midpoint is due to the symmetry of the uniform distribution. If you visualize the distribution, the outcomes spread evenly across the interval, with equal probability height. Therefore, the center, or the average location, of these outcomes is naturally the midpoint. This concept is intuitive but details the fundamental property of uniform distributions as fair and balanced across the range.
Understanding the expected value can also aid in context-based applications, like determining the average outcome expected over an interval. Remember that the expected value is a concept that applies to all continuous random variables, but its computation for the uniform distribution is highly simplified due to its symmetric nature.

Cumulative Distribution Function

The cumulative distribution function (CDF) indicates the probability that a continuous random variable \(X\) will take a value less than or equal to \(x\). For a uniform distribution over \([a, b]\), the CDF is as follows:

• \(F(x) = 0\) for \(x < a\)
• \(F(x) = \frac{x-a}{b-a}\) for \(a \leq x \leq b\)
• \(F(x) = 1\) for \(x > b\)

The logic here is simple: before reaching \(a\), no value has been chosen yet, hence the probability is 0. Between \(a\) and \(b\), probabilities scale linearly—it increases from 0 at \(a\) to 1 at \(b\). So the CDF forms an upward-sloping line from \(a\) to \(b\). This line precisely represents the area under the PDF curve from \(a\) to \(x\).
The CDF is fundamental for understanding how probabilities accumulate over an interval, transitioning gradually from 0 to 1. By integrating the PDF over \([a, b]\), this accumulation explains why \(F(b) = 1\)—reaching the end of the interval means everything possible has already occurred.

Continuous Random Variable

A continuous random variable, like \(X\) in our uniform distribution example, can take on an infinite number of possible values within any given interval. The hallmark of a continuous random variable is that it is not restricted to discrete (countable) values, but rather can vary continuously across any range.In the context of the uniform distribution on \([a, b]\), \(X\) can assume any real number value between \(a\) and \(b\). This makes such variables distinct from discrete random variables, which have specific, countable outcomes.Some features of continuous random variables include:

• They are described using probability density functions (PDFs) since individual probabilities at exact points are zero.
• They are characterized by cumulative distribution functions (CDFs) that compute probabilities over intervals.
• Expected values for these variables signify average outcomes when observations from the distribution are made.
• The probability of \(X\) being in any specific sub-interval of \([a, b]\) is given by the area under the PDF curve over that sub-interval.

Understanding the nature of continuous random variables helps in distinguishing how exact probabilities and averages are computed differently than for discrete variables. Fields such as physics, engineering, and finance often need models based on continuous variables to handle real-world measurements and analyses.