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 \(\mathrm{CDF}\). $$ f(x)=\left\\{\begin{array}{ll} \frac{\pi}{8} \sin (\pi x / 4), & \text { if } 0 \leq x \leq 4 \\ 0, & \text { otherwise } \end{array}\right. $$

Short answer

(a) \( P(X \geq 2) = \frac{1}{2} \); (b) \( E(X) = 2 \); (c) \( F(x) = \frac{1}{2} \left( 1 - \cos\left(\frac{\pi x}{4}\right) \right) \) for \( 0 \leq x \leq 4 \).

Step-by-step solution

Find P(X ≥ 2)

To find the probability \( P(X \geq 2) \), we calculate the integral of the PDF from 2 to 4: \[ \int_{2}^{4} \frac{\pi}{8} \sin\left( \frac{\pi x}{4} \right) \, dx. \]Using the substitution \( u = \frac{\pi x}{4} \) and \( du = \frac{\pi}{4} dx \), we get \[ dx = \frac{4}{\pi} \, du. \]Thus, the integral becomes:\[ \int_{\frac{\pi \cdot 2}{4}}^{\pi} \frac{\pi}{8} \sin(u) \cdot \frac{4}{\pi} \, du = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \sin(u) \, du. \]Calculating the integral, we find:\[ -\frac{1}{2} \left. \cos(u) \right|_{\frac{\pi}{2}}^{\pi} = -\frac{1}{2} \left( \cos(\pi) - \cos\left(\frac{\pi}{2}\right) \right) = -\frac{1}{2} \left( -1 - 0 \right) = \frac{1}{2}. \]Thus, \( P(X \geq 2) = \frac{1}{2}. \)

Compute E(X)

The expected value \( E(X) \) is computed using the integral:\[ E(X) = \int_{0}^{4} x \cdot \frac{\pi}{8} \sin\left(\frac{\pi x}{4}\right) \, dx. \]Using integration by parts, let \( u = x \) and \( dv = \frac{\pi}{8} \sin\left( \frac{\pi x}{4} \right) \, dx \), then \( du = dx \) and integrating \( dv \) gives \( v = -\frac{4}{\pi} \cos\left(\frac{\pi x}{4}\right) \).Thus, we apply integration by parts:\[ E(X) = \left. -\frac{4x}{8} \cos\left(\frac{\pi x}{4}\right) \right|_{0}^{4} + \int_{0}^{4} \frac{4}{8} \cos\left(\frac{\pi x}{4}\right) \, dx. \]The first term becomes \( 0 \) when evaluated from 0 to 4, and the second integral evaluates using substitution, resulting in \( 2 \). Therefore, \( E(X) = 2 \).

Determine the CDF

The cumulative distribution function (CDF) \( F(x) \) is defined as:\[ F(x) = \int_{0}^{x} \frac{\pi}{8} \sin\left(\frac{\pi t}{4}\right) \, dt. \]Using the substitution \( u = \frac{\pi t}{4} \), the integral becomes:\[ F(x) = \left. -\frac{1}{2} \cos(u) \right|_{0}^{\frac{\pi x}{4}}. \]Compute:\[ F(x) = -\frac{1}{2} \cos\left(\frac{\pi x}{4}\right) + \frac{1}{2} \cos(0) = \frac{1}{2} \left( 1 - \cos\left(\frac{\pi x}{4}\right) \right). \]For \( 0 \leq x \leq 4 \), the CDF is:\[ F(x) = \frac{1}{2} \left( 1 - \cos\left(\frac{\pi x}{4}\right) \right), \] and \( F(x) = 0 \) for \( x < 0 \), and \( F(x) = 1 \) for \( x > 4 \).

Key concepts

Probability Density Function (PDF)

A Probability Density Function (PDF) is a fundamental concept in the study of continuous random variables. The PDF describes the likelihood of a random variable taking on a particular value. However, unlike in discrete cases, the probability of a continuous variable taking a specific value is zero. Instead, the probability is determined for a range of values.

For example, given a PDF like:\[f(x) = \left\{\begin{array}{ll}\frac{\pi}{8} \sin \left( \frac{\pi x}{4} \right), & \text{if } 0 \leq x \leq 4 \0, & \text{otherwise}\end{array}\right.\]This implies that the function describes how the probability is distributed over the interval from 0 to 4, while being zero outside of it. To find the probability that the random variable falls within a specific range, such as from 2 to 4, you integrate the PDF over that range. In our specific solution, this integral reveals that the probability of the random variable being greater than or equal to 2 is \( \frac{1}{2} \).

Thus, the PDF not only shows which values are more likely but also defines how probabilities are spread over the variable's possible values.

Expected Value (E(X))

The Expected Value, denoted as \( E(X) \), of a continuous random variable can be thought of as its average or mean value that we expect when repeated measurements are taken. It is determined by integrating the product of the variable's value and its PDF over its entire range.

For a function \( f(x) \) given by:\[E(X) = \int_{a}^{b} x \cdot f(x) \, dx\]This represents the balance point of the probability distribution. In our solved exercise, the integration was performed from 0 to 4 using:\[E(X) = \int_{0}^{4} x \cdot \frac{\pi}{8} \sin\left(\frac{\pi x}{4}\right) \, dx\]Applying integration by parts, where the bounds and variables are carefully managed, simplifies to a result of \( E(X) = 2 \). This value indicates the central or expected outcome of the random variable under this distribution.

Cumulative Distribution Function (CDF)

The Cumulative Distribution Function (CDF), denoted as \( F(x) \), describes the probability that a continuous random variable, say \( X \), takes a value less than or equal to \( x \). It is found by integrating the PDF from the lowest value of interest up to \( x \). This provides an aggregate probability measure.

The CDF is defined as:\[F(x) = \int_{0}^{x} f(t) \, dt\]Using the example PDF given in the exercise:\[F(x) = \int_{0}^{x} \frac{\pi}{8} \sin\left(\frac{\pi t}{4}\right) \, dt\]By substituting and computing the definite integral, we find:\[F(x) = \frac{1}{2} \left( 1 - \cos\left(\frac{\pi x}{4}\right) \right), \quad \text{for } 0 \leq x \leq 4\]The CDF starts at zero, increases steadily as \( x \) increases, and reaches 1 at the upper limit of the distribution. Thus, a CDF not only helps us understand the entire distribution but also gives the probability up to any specific point in that range.