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 $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{b-a}, & \text { if } a \leq x \leq b \\ 0, & \text { otherwise } \end{array}\right. $$ (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) CDF: \(F(x) = \frac{x-a}{b-a}\) for \(a \leq x \leq b\).

Step 1: Understanding the Problem

We need to address three parts of the problem related to a uniform distribution on the interval \([a, b]\): (a) finding the probability that the value of \(X\) is closer to \(a\) than it is to \(b\); (b) finding the expected value of \(X\); and (c) finding the cumulative distribution function (CDF) of \(X\).

Step 2: Probability Closer to a than to b

To solve part (a), we find where \(X\) is closer to \(a\) than to \(b\). This occurs when \(|X - a| < |X - b|\). Solving \(|X - a| < |X - b|\) for \(X\) gives the condition \(X < \frac{a + b}{2}\). Thus, we want the probability \(P(X < \frac{a+b}{2})\). Using the PDF of a uniform distribution, we calculate this as: \[P(X < \frac{a+b}{2}) = \int_a^{\frac{a+b}{2}} \frac{1}{b-a} \, dx = \frac{\frac{a+b}{2} - a}{b-a} = \frac{b-a}{2(b-a)} = \frac{1}{2}\]

Step 3: Calculating Expected Value

For part (b), the expected value of a uniform distribution \(X\) on \([a, b]\) is given by \(E(X) = \frac{a+b}{2}\). This result uses the definition of the expected value:\[E(X) = \int_a^b x \cdot \frac{1}{b-a} \, dx = \left[ \frac{x^2}{2(b-a)} \right]_a^b = \frac{b^2-a^2}{2(b-a)} = \frac{(b-a)(b+a)}{2(b-a)} = \frac{a+b}{2}\]

Step 4: Deriving the CDF

To find the cumulative distribution function (CDF) part (c), we integrate the PDF from \(a\) to \(x\) for \(a \leq x \leq b\):\[F(x) = \int_a^x \frac{1}{b-a} \, dt = \left[ \frac{t}{b-a} \right]_a^x = \frac{x-a}{b-a}\]Thus, the CDF is given by:\[F(x) = \left\{ \begin{array}{ll} 0, & x < a \ \frac{x-a}{b-a}, & a \leq x \leq b \ 1, & x > b \end{array} \right.\]

Step 5: Final Answer Verification

We verified all steps and results are consistent with properties of uniform distribution: (a) Probability \(X\) is closer to \(a\) than \(b\) is \(\frac{1}{2}\), (b) Expected value \(E(X) = \frac{a+b}{2}\), and (c) the CDF is as derived. Results align with theoretical expectations for uniform distributions.

Key Concepts

Probability Density Function

Whenever we talk about continuous random variables, it's crucial to understand the Probability Density Function (PDF). The uniform distribution is one of the simplest types, characterized by its constant probability across a specified interval. For a continuous random variable \( X \) uniformly distributed over the interval \([a, b]\), the PDF is given as:

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

This implies that the likelihood of \( X \) taking on any value within \( [a, b] \) is equally likely.

The PDF helps in calculating probabilities over intervals by integrating over the desired range. For example, to find out the probability of \( X \) being close to the start of the interval \( [a, \frac{a+b}{2}] \): \[P(X < \frac{a+b}{2}) = \int_a^{\frac{a+b}{2}} \frac{1}{b-a} \, dx\]which simplifies to \( \frac{1}{2}\), signaling that there's an equal chance of \( X \) being closer to end \( a \) or \( b \) of the interval.

Expected Value

The expected value of a distribution gives a measure of the "center" of the distribution or the "average" value expected from a set of repeated trials. For a uniform distribution over \([a, b]\), the expected value is straightforwardly found at the midpoint of the interval:

• \( E(X) = \frac{a+b}{2} \)

The derivation comes from the definition of expected value for continuous distributions:\[E(X) = \int_a^b x \cdot \frac{1}{b-a} \, dx = \frac{a+b}{2}\]This expected value, or mean, tells us intuitively that if \( X \) is uniformly distributed, it is "equally away" from \( a \) and \( b \). This symmetry explains why the mean is at the middle, \( \frac{a+b}{2} \).

Students often find it helpful to visualize this: Think of it as balancing a seesaw – the midpoint will always be where balance is achieved when the weight distribution is uniform.

Cumulative Distribution Function

While the PDF gives us a "density" of probabilities, the Cumulative Distribution Function (CDF) gives the probability that \( X \) will be less than or equal to a certain value. For a uniform distribution, the CDF \( F(x) \) is derived by integrating the PDF from \( a \) to \( x \):

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

The CDF effectively "accumulates" the probability as \( x \) moves from \( a \) to \( b \). At \( x = a \), the probability is \( 0 \) because \( X \) hasn't reached or passed \( a \). As \( x \) increases and reaches \( x = b \), the probability cumulatively adds up to \( 1 \).

Hence, the CDF is a step function, starting at 0, climbing linearly, and reaching 1 at upper limit \( b \), reflecting the complete certainty that \( X \) has taken a value within \([a, b]\). This linear increase mirrors the uniform spread of values in the interval.