Problem 4 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(X\) and \(Y\) be two independent random variables, where \(X\) has an \(N(2,5)\) distribution and \(Y\) has an \(N(5,9)\) distribution. Define \(Z=3 X-2 Y+1\). a. Compute \(E[Z]\) and \(\operatorname{Var}(Z)\). b. What is the distribution of \(Z ?\) c. Compute \(\mathrm{P}(Z \leq 6)\).

Step-by-Step Solution

Answer

a. \(E[Z] = -3\), \(\operatorname{Var}(Z) = 81\). b. \(Z \sim N(-3, 81)\). c. \(P(Z \leq 6) \approx 0.8413\).

Step 1: Understand the Problem

We need to find the expected value and variance for a new random variable, \(Z = 3X - 2Y + 1\), where \(X\) and \(Y\) are independent and normally distributed. Additionally, we should determine the distribution of \(Z\) and calculate the probability \(P(Z \leq 6)\).

Step 2: Compute Expected Value of Z

The expected value of a linear combination of independent random variables is a linear combination of the expected values. Thus, \(E[Z] = E[3X - 2Y + 1] = 3E[X] - 2E[Y] + 1\).We know: \(E[X] = 2\) and \(E[Y] = 5\).Therefore, \(E[Z] = 3(2) - 2(5) + 1 = 6 - 10 +1 = -3\).

Step 3: Compute Variance of Z

The variance of a linear combination of independent random variables is the sum of the variances of each term. For \(Z = 3X - 2Y + 1\), the variance is: \(\operatorname{Var}(Z) = (3^2)\operatorname{Var}(X) + (-2)^2\operatorname{Var}(Y)\).Given \(\operatorname{Var}(X) = 5\) and \(\operatorname{Var}(Y) = 9\), \(\operatorname{Var}(Z) = 3^2 \times 5 + (-2)^2 \times 9 = 45 + 36 = 81\).

Step 4: Determine Distribution of Z

\(X\) is distributed as \(N(2, 5)\) and \(Y\) as \(N(5, 9)\). Since \(Z = 3X - 2Y + 1\), which is a linear transformation of normal variables, \(Z\) is also normally distributed. It will have mean \(E[Z] = -3\) and variance \(\operatorname{Var}(Z) = 81\). Thus, \(Z \sim N(-3, 81)\).

Step 5: Compute Probability P(Z ≤ 6)

To compute \(P(Z \leq 6)\), we standardize \(Z\) using the formula: \(Z_{std} = \frac{Z - \mu}{\sigma}\), where \(\mu = -3\) and \(\sigma = \sqrt{81} = 9\).This gives us: \(Z_{std} = \frac{6 - (-3)}{9} = \frac{9}{9} = 1\).Thus, \(P(Z \leq 6) = P(Z_{std} \leq 1)\). Using the standard normal distribution table, \(P(Z_{std} \leq 1) \approx 0.8413\).

Key Concepts

Normal Distribution

The normal distribution is a key concept in probability theory, often referred to as a Gaussian distribution. This distribution is symmetric and follows a bell-shaped curve, indicating that data near the mean are more frequent in occurrence than data far from the mean.

Here are some crucial points about normal distribution:

• A normal distribution is characterized by its mean \(\mu\) and variance \(\sigma^2\).
• The mean determines the center of the distribution.
• The variance indicates how spread out the data is around the mean.

In our exercise, random variables \(X\) and \(Y\) show normal distributions with different parameters. \(X\) follows a distribution \((N(2,5)\) with mean 2 and variance 5, and \(Y\) follows \(N(5,9)\) with mean 5 and variance 9. Understanding these parameters is crucial for calculating any linear combination of these variables, like the variable \(Z\).

Expected Value

The expected value is a fundamental concept in probability, often referred to as the mean or average. It helps provide the center point of a random variable's distribution.

Here's how to interpret the expected value:

• It indicates the long-term average if an experiment is repeated many times.
• It's helpful in predicting the future outcome of random processes.

To compute the expected value of \(Z\) in the problem, we use the properties of linear combinations for independent random variables:\[E[Z] = E[3X - 2Y + 1] = 3E[X] - 2E[Y] + 1\]Given \(E[X] = 2\) and \(E[Y] = 5\), the expected value of \(Z\) calculates to \(-3\). This final answer captures where the center of \(Z\)'s distribution lies.

Variance

Variance is a measure that quantifies the spread of random variables from their expected value. A higher variance indicates that data points are spread out more widely from the mean.

These points help in understanding variance:

• The variance is always non-negative.
• The square root of the variance is called the standard deviation, which is in the same unit as the data.

In the provided problem, the calculation of the variance for \(Z\) uses the rule for linear combinations of independent random variables:\[\text{Var}(Z) = 3^2 \cdot \text{Var}(X) + (-2)^2 \cdot \text{Var}(Y)\]Substituting \(\text{Var}(X) = 5 \) and \(\text{Var}(Y) = 9\), \(\text{Var}(Z) \) results in 81. This value shows how much \(Z\) is spread around its mean, \(-3\).

Independent Random Variables

Independent random variables are those whose occurrence is not affected by each other. When dealing with probability, independence simplifies calculation considerably, especially with operations involving multiple random variables.

Key aspects of independent random variables include:

• The probability of their joint occurrence is the product of their individual probabilities.
• For independent variables, events related to one provide no information about events related to the other.

Since \(X\) and \(Y\) in the exercise are independent, calculating the variance and expected value for the new variable \(Z\) becomes straightforward. Independence justifies the separate computations for each component of \(Z\), leading us to effective conclusions about its distribution.