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 \(\mathrm{CDF}\). $$ f(x)=\left\\{\begin{array}{ll} (8-x) / 32, & \text { if } 0 \leq x \leq 8 \\ 0, & \text { otherwise } \end{array}\right. $$

Short answer

(a) \(P(X \geq 2) = \frac{3}{4}\), (b) \(E(X) = \frac{16}{3}\), (c) CDF is \(F(x) = \frac{x}{4} - \frac{x^2}{64}\) for \(0 \leq x \leq 8\).

Step-by-step solution

Check PDF Validity

The probability density function (PDF) must integrate to 1 over its entire range. Let's verify this for the given PDF \(f(x) = \frac{8-x}{32}\) for \(0 \leq x \leq 8\).\[\int_0^8 \frac{8-x}{32} \, dx \]Find the integral and confirm it equals 1.

Calculate the Integral

Integrate the PDF over the range from 0 to 8:\[\int_0^8 \frac{8-x}{32} \, dx = \left[ \frac{8x}{32} - \frac{x^2}{64} \right]_0^8\]Substitute the limits and simplify:\[= \left( \frac{64}{32} - \frac{64}{64} \right) - \left( 0 - 0 \right) = 1\]This confirms that the PDF is valid as its integral equals 1.

Find P(X ≥ 2)

To find \(P(X \geq 2)\), calculate the integral of \(f(x)\) from 2 to 8:\[P(X \geq 2) = \int_2^8 \frac{8-x}{32} \, dx\]Calculate this integral:\[\left[ \frac{8x}{32} - \frac{x^2}{64} \right]_2^8\]Substitute the values:\[= \left( \frac{64}{32} - \frac{64}{64} \right) - \left( \frac{16}{32} - \frac{4}{64} \right) = 1 - \frac{1}{4} = \frac{3}{4}\]

Calculate Expected Value E(X)

The expected value \(E(X)\) is given by\[E(X) = \int_{0}^8 x \cdot \frac{8-x}{32} \, dx\]First, expand the integrand:\[E(X) = \int_{0}^8 \left( \frac{8x}{32} - \frac{x^2}{32} \right) \, dx\]Then integrate each term separately:\[= \left[ \frac{8x^2}{64} - \frac{x^3}{96} \right]_0^8\]Calculate the integral:\[= \left( 8 - \frac{64}{12} \right) = \frac{16}{3}\]

Find the Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF), \(F(x)\), is the integral of the PDF from its minimum value up to \(x\):\[F(x) = \int_{0}^{x} \frac{8-t}{32} \, dt\]Calculate this integral and find the expression for \(F(x)\):\[= \left[ \frac{8t}{32} - \frac{t^2}{64} \right]_0^x\]\[= \frac{8x}{32} - \frac{x^2}{64}\]Thus, the CDF for \(0 \leq x \leq 8\) is \(F(x) = \frac{x}{4} - \frac{x^2}{64}\).For \(x < 0\), \(F(x) = 0\) and for \(x > 8\), \(F(x) = 1\).

Key concepts

Continuous Random Variable

A continuous random variable, like the one involving the Probability Density Function (PDF) in this exercise, represents a variable that can take an uncountable number of values. Imagine it as a smooth curve on a graph where any point along the horizontal axis is a possible outcome. Unlike discrete random variables, which are countable and have distinct separate values, continuous variables can take any value within a range.

The PDF is crucial when dealing with continuous random variables. It provides the probability that the random variable falls within a specific range of values. The critical property of a PDF is that the area under the curve represents the probability, and this area must be equal to 1 when calculated over the entire range the variable can take. This ensures that all possible outcomes are considered, meaning the total probability is 100%.

In our exercise, verifying that the integral of the PDF equals 1 confirms the validity of the PDF. A correct and valid PDF, such as the one described, allows us to move forward with calculations like determining probabilities and expected values.

Expected Value

The expected value, often denoted by \( E(X) \), represents the average or mean value of a random variable when an experiment is repeated many times. It is a crucial concept in probability and statistics because it provides a summary measure that describes the center or typical value one can expect from a distribution.

For continuous random variables, calculating the expected value involves integrating the product of the variable and its PDF over the entire range. Specifically, in our exercise, we have the integral \( \int_{0}^8 x \cdot \frac{8-x}{32} \, dx \) that results in \( E(X) = \frac{16}{3} \).

This result gives insight into the average outcome you might expect from the random variable \( X \) if it were to be observed repeatedly. Despite \( X \) taking values between 0 and 8, the expected value provides a single number that effectively summarizes the distribution of \( X \).

The expected value is instrumental in real-world applications where decisions need to be made based on the probability of future events, such as in economics, risk management, and various fields of engineering.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a fundamental concept in probability, especially for continuous random variables. It provides a way to calculate the probability that a random variable will take a value less than or equal to a particular number. This is done by integrating the PDF from the smallest possible value up to any value \( x \).

In our solution, the CDF is expressed as \( F(x) = \int_{0}^{x} \frac{8-t}{32} \, dt \), resulting in \( F(x) = \frac{x}{4} - \frac{x^2}{64} \) for the range \( 0 \leq x \leq 8 \). The CDF provides values which range from 0 to 1, with 0 before the minimum value (where no event has occurred) and 1 after the maximum (where the event is certain to have occurred).

The CDF is valuable because it allows for straightforward calculation of probabilities and facilitates comparison between different distributions. It gives a complete picture of the distribution of a random variable by showing not just how likely individual outcomes are (as the PDF does) but how likely it is that outcomes will fall below any given point. This makes the CDF particularly useful for statistical hypothesis testing and the creation of confidence intervals.

Understanding the CDF is essential for anyone working with statistical data as it provides a clearer view of the probability landscape across the entire spectrum of possible outcomes.