Question

Suppose a random variable \(Z\) has CDF $$ F(z)=\left\\{\begin{array}{ll} 0, & \text { if } z<0 \\ z^{2} / 9, & \text { if } 0 \leq z \leq 3 \\ 1, & \text { if } z>3 \end{array}\right. $$ Find each of the following: (a) \(P(Z>1)\) (b) \(P(1

Short answer

(a) \(\frac{8}{9}\), (b) \(\frac{1}{3}\), (c) \(f(z) = \frac{2z}{9}\) for \(0 \leq z \leq 3\), (d) \(E(Z) = 2\).

Step-by-step solution

Calculate F(1)

Use the CDF formula for the interval where 0 ≤ z ≤ 3. Therefore, when z=1, we have:\[F(1) = \frac{1^2}{9} = \frac{1}{9}\]

Calculate P(Z > 1)

Since Z is greater than 1, it's inside the interval [0, 3], so:\[P(Z > 1) = 1 - F(1) = 1 - \frac{1}{9} = \frac{8}{9}\]

Calculate F(2)

Use the CDF formula again for F(2) since 2 is also in the interval 0 ≤ z ≤ 3:\[F(2) = \frac{2^2}{9} = \frac{4}{9}\]

Calculate P(1 < Z < 2)

Calculate the probability that Z is between 1 and 2:\[P(1 < Z < 2) = F(2) - F(1) = \frac{4}{9} - \frac{1}{9} = \frac{3}{9} = \frac{1}{3}\]

Find the PDF f(z)

The PDF is the derivative of the CDF. Differentiate F(z) with respect to z for the interval 0 ≤ z ≤ 3:\[f(z) = \frac{d}{dz} \left(\frac{z^2}{9}\right) = \frac{2z}{9}\]Thus, the PDF of Z is:\[f(z) = \left\{ \begin{array}{ll} 0, & z < 0\ \frac{2z}{9}, & 0 \leq z \leq 3\ 0, & z > 3 \end{array} \right.\]

Calculate E(Z)

Use the formula for expected value by integrating z times the PDF over the interval [0, 3]:\[E(Z) = \int_0^3 z \cdot \frac{2z}{9} \, dz = \int_0^3 \frac{2z^2}{9} \, dz\]Calculate the integral:\[E(Z) = \frac{2}{9} \int_0^3 z^2 \, dz = \frac{2}{9} \left[ \frac{z^3}{3} \right]_0^3 = \frac{2}{9} \cdot \frac{27}{3} = \frac{2 \times 9}{9} = 2\]

Key concepts

Cumulative Distribution Function (CDF)

The cumulative distribution function, or CDF, is a fundamental concept in probability theory that gives the probability that a random variable takes on a value less than or equal to a specific number. For a random variable, say \(Z\), the CDF is denoted as \(F(z)\). Here's how it works:

• For \(z < 0\), \(F(z) = 0\), meaning \(Z\) cannot take any negative value.
• For \(0 \leq z \leq 3\), \(F(z) = \frac{z^2}{9}\), which describes how the probability accumulates as \(z\) increases within this interval.
• For \(z > 3\), \(F(z) = 1\), indicating that \(Z\) will take a value at or below 3.

Understanding the CDF is crucial because it provides a complete description of the probability distribution of a random variable. It is useful in calculating probabilities for different ranges of values.

Probability Density Function (PDF)

The probability density function, or PDF, describes the likelihood of a random variable to take on a specific value. Unlike the CDF, which accumulates probability, the PDF shows the rate of probability distribution across different values. The PDF is essentially the derivative of the CDF.
For the given \(Z\) in the range \(0 \leq z \leq 3\), the PDF, \(f(z)\), is derived by taking the derivative of \(F(z)\), which results in:

• \(f(z) = \frac{2z}{9}\) for \(0 \leq z \leq 3\)
• Zero for any \(z\) outside this interval.

This PDF tells us how densely or sparsely probabilities are distributed across the values \(Z\) can assume. The overall area under the PDF curve across its range sums to 1, indicating total probability.

Expected Value

The expected value, often denoted as \(E(Z)\), is a measure of the central tendency of a random variable. It provides a single average value expected from the distribution and is calculated through integration when dealing with continuous random variables.
For our random variable \(Z\) which follows a given PDF, the expected value is determined by integrating \(z\) times its PDF over its domain:

• \(E(Z) = \int_0^3 z \cdot \frac{2z}{9} \, dz\)

This calculation gives \(E(Z) = 2\), illustrating that on average, the random variable \(Z\) will have a value of 2. The expected value helps in predicting the average outcome of probabilistic experiments.

Integration in Probability

Integration is a key tool in probability, especially when dealing with continuous random variables. It's used to both transition from CDF to PDF, and for calculations such as the expected value. By integrating, we can find the total area under the PDF, representing the total probability.

• To derive the PDF, integrate the CDF over the desired range.
• To find the expected value, integrate the product of the variable and its PDF over the range of possible values.

In these contexts, integration allows us to understand and interpret continuous distributions efficiently. It translates abstract probability concepts into tangible results and insights for practical applications.