Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

A PDF for a continuous random variable \(X\) is given. Use the PDF to find (a) \(P(X \geq 2),(b) E(X)\), and \((c)\) the \(\mathrm{CDF}\). $$ f(x)=\left\\{\begin{array}{ll} \frac{4}{3} x^{-2}, & \text { if } 1 \leq x \leq 4 \\ 0, & \text { otherwise } \end{array}\right. $$

Step-by-Step Solution

Answer

(a) \(P(X \geq 2) = \frac{1}{3}\); (b) \(E(X) = \frac{8}{3} \ln(2)\); (c) \(F(x) = 1 - \frac{4}{3x}\), for \(1 \leq x \leq 4\).

Step 1: Find the Probability P(X ≥ 2)

To find \(P(X \geq 2)\), we need to calculate the integral of the PDF from 2 to 4, as we are given that the variable is defined from 1 to 4.\[P(X \geq 2) = \int_{2}^{4} \frac{4}{3} x^{-2} \, dx\]Now, compute the integral:\[\int \frac{4}{3} x^{-2} \, dx = -\frac{4}{3x}\]Evaluate from 2 to 4:\[P(X \geq 2) = \left[-\frac{4}{3x} \right]_{2}^{4} = \left(-\frac{4}{12} \right) - \left(-\frac{4}{6} \right) = \frac{1}{3}\]

Step 2: Find the Expectation E(X)

To find the expectation, calculate the integral of \(x\) multiplied by the PDF over the interval from 1 to 4:\[E(X) = \int_{1}^{4} x \cdot \frac{4}{3} x^{-2} \, dx = \int_{1}^{4} \frac{4}{3} x^{-1} \, dx\]The integral of \(x^{-1}\) is the natural logarithm, \(\ln|x|\):\[\int \frac{4}{3} x^{-1} \, dx = \frac{4}{3} \ln|x|\]Evaluate from 1 to 4:\[E(X) = \left[\frac{4}{3} \ln|x| \right]_{1}^{4} = \frac{4}{3} \ln(4) - \frac{4}{3} \ln(1) = \frac{8}{3} \ln(2)\]

Step 3: Find the Cumulative Distribution Function (CDF)

The cumulative distribution function \(F(x)\) is obtained by integrating the PDF from the lower bound up to \(x\), for 1 \(\leq x \leq 4\):\[F(x) = \int_{1}^{x} \frac{4}{3} t^{-2} \, dt\]Using the integral already calculated:\[F(x) = \left[-\frac{4}{3t} \right]_{1}^{x} = \left(-\frac{4}{3x} \right) - \left(-\frac{4}{3} \right) = 1 - \frac{4}{3x}\]Thus, the CDF is:\[\ F(x) = \begin{cases} 0, & x < 1 \ 1 - \frac{4}{3x}, & 1 \leq x \leq 4 \ 1, & x > 4 \end{cases}\]

Key Concepts

Probability Density Function

A probability density function (PDF) for a continuous random variable gives us the likelihood of the variable taking on a specific value. Unlike a discrete probability distribution, the PDF does not give probabilities directly. Instead, it provides a curve where the area under the curve corresponds to probability. For a given range of the variable, the area under the PDF curve over that interval represents the probability that the variable falls within the interval.

In our example, we're working with a continuous random variable \( X \) that follows the PDF:

• \( f(x) = \frac{4}{3} x^{-2} \) for \( 1 \leq x \leq 4 \)
• \( f(x) = 0 \), otherwise.

To find \( P(X \geq 2) \), we calculated the integral of the PDF from 2 to 4, which equaled \( \frac{1}{3} \). This tells us there's a 1/3 probability of \( X \) being greater than or equal to 2.

Cumulative Distribution Function

The cumulative distribution function (CDF) of a continuous random variable provides the probability that the variable will take a value less than or equal to a certain number. It is a significant concept because it captures the entire probability distribution of a random variable. The CDF increases from 0 to 1 as \( x \) moves from the smallest possible value to the largest possible value.

For the PDF provided, the CDF \( F(x) \) can be determined by integrating the PDF from the lower limit up to a particular value \( x \). Through integration, we found

• \( F(x) = 1 - \frac{4}{3x} \) for \( 1 \leq x \leq 4 \)
• and specific conditions: \( F(x) = 0 \) for \( x < 1 \) and \( F(x) = 1 \) for \( x > 4 \).

This CDF tells us how probabilities accumulate as the variable \( x \) increases.

Expectation Value

The expectation value, or the expected value, of a continuous random variable \( X \) represents the theoretical mean of \( X \). It is calculated as the integral of \( x \) times the PDF over the entire range where the PDF is defined. It gives us a sense of the 'center' of the distribution or the average outcome you would expect if you sampled many times.

In our problem, we found the expectation value \( E(X) \) by computing the integral from 1 to 4:

• \( E(X) = \int_{1}^{4} x \cdot \frac{4}{3} x^{-2} \, dx = \int_{1}^{4} \frac{4}{3} x^{-1} \, dx \)
• This evaluates to \( \frac{8}{3} \ln(2) \).

The result, \( \frac{8}{3} \ln(2) \), gives the expected average outcome of the random variable if the process is repeated many times.