Problem 7 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Given is a random variable \(X\) with probability density function \(f\) given by \(f(x)=0\) for \(x<0\), and for \(x>1\), and \(f(x)=4 x-4 x^{3}\) for \(0 \leq x \leq 1\). Determine the expectation and variance of the random variable \(2 X+3\).

Step-by-Step Solution

Answer

Expectation of \(2X+3\) is \(\frac{61}{15}\); Variance is \(\frac{44}{225}\).

Step 1: Verify Probability Distribution

First, we need to ensure that \( f(x) \) is a valid probability density function. This is done by checking that the integral of the probability density function over its entire range equals 1. We compute \( \int_0^1 (4x - 4x^3) \, dx \) to verify this. Performing the integration, we have \[ \int_0^1 (4x - 4x^3) \, dx = \left[ 2x^2 - x^4 \right]_0^1 = (2(1)^2 - (1)^4) - (2(0)^2 - (0)^4) = 2 - 1 = 1 \].This shows that \( f(x) \) is indeed a proper probability density function.

Step 2: Compute Expectation of X

The expectation \( E(X) \) for a continuous random variable is given by the integral \( E(X) = \int_{-\infty}^{\infty} x f(x) \, dx \). For our function, this simplifies to:\[ E(X) = \int_{0}^{1} x (4x - 4x^3) \, dx = \int_{0}^{1} (4x^2 - 4x^4) \, dx \].Performing the integration gives:\[ \int_{0}^{1} (4x^2 - 4x^4) \, dx = \left[ \frac{4x^3}{3} - \frac{4x^5}{5} \right]_0^1 = \frac{4}{3} - \frac{4}{5} = \frac{20}{15} - \frac{12}{15} = \frac{8}{15} \].Thus, \( E(X) = \frac{8}{15} \).

Step 3: Compute Expectation of 2X+3

Using the linearity of expectation, the expectation of a linear transformation \( aX + b \) of a random variable \( X \) is \( aE(X) + b \). Thus, we have:\[ E(2X+3) = 2E(X) + 3 = 2 \cdot \frac{8}{15} + 3 = \frac{16}{15} + 3 \].Converting \( 3 \) to a fraction: \( 3 = \frac{45}{15} \).Thus, \\[ E(2X+3) = \frac{16}{15} + \frac{45}{15} = \frac{61}{15} \].

Step 4: Compute Variance of X

Variance \( \text{Var}(X) \) is defined by \( \text{Var}(X) = E(X^2) - (E(X))^2 \). First, compute \( E(X^2) = \int_{0}^{1} x^2 (4x - 4x^3) \, dx \):\[ E(X^2) = \int_{0}^{1} (4x^3 - 4x^5) \, dx = \left[ \frac{4x^4}{4} - \frac{4x^6}{6} \right]_0^1 = \frac{1}{1} - \frac{4}{6} = 1 - \frac{2}{3} = \frac{1}{3} \].Therefore, \( \text{Var}(X) = \frac{1}{3} - \left( \frac{8}{15} \right)^2 \).Calculate \( \left( \frac{8}{15} \right)^2 = \frac{64}{225} \).\[ \text{Var}(X) = \frac{1}{3} - \frac{64}{225} = \frac{75}{225} - \frac{64}{225} = \frac{11}{225} \].

Step 5: Compute Variance of 2X+3

The variance of a linear transformation \( aX + b \) of a random variable \( X \) is \( a^2\text{Var}(X) \). Thus:\[ \text{Var}(2X+3) = 2^2 \cdot \text{Var}(X) = 4 \cdot \frac{11}{225} = \frac{44}{225} \].

Key Concepts

Expectation

The concept of expectation, often known as the expected value or mean, is a fundamental aspect of probability and statistics, especially when dealing with continuous random variables. The expectation gives a central value around which the data tends to cluster.
For a continuous random variable, the expectation is calculated using the formula:

• \( E(X) = \int_{-fty}^{fty} x f(x) \, dx \)

This integral calculates the area under the curve of the function weighted by the value of \( x \).
In the example, the expectation \( E(X) \) is computed over the interval where \( f(x) \) is defined, that is from 0 to 1. Thus, showcasing how \( X \) behaves on average according to the defined probability density function.
For transformations like \( 2X+3 \), the linearity property is useful. It indicates \( E(aX + b) = aE(X) + b \). Hence, calculating \( E(2X+3) \) becomes straightforward once the expectation of \( X \) is known.
This result means that if you were to observe \( 2X+3 \) over a long period, its average value would stabilize around this expected value.

Variance

Variance is a measure of how spread out the values of a random variable are around the mean. For continuous random variables, variance quantifies the amount of variability or dispersion in the dataset.
It's calculated using the formula:

• \( \text{Var}(X) = E(X^2) - (E(X))^2 \)

While \( E(X^2) \) provides the second moment about the origin, subtracting \( (E(X))^2 \) standardizes this to return the spread about the mean. This ensures that we talk about the variability instead of just the raw values.
In our problem, once \( \text{Var}(X) \) is determined for \( X \), finding the variance for a linear transformation \( 2X+3 \) involves the property \( \text{Var}(aX + b) = a^2 \times \text{Var}(X) \) since b shifts the mean but does not impact variability.
This calculation highlights how the variance scales with a multiplication factor squared, showing the direct effect on variability by scaling the random variable.

Continuous Random Variable

A continuous random variable is one which can take any value within a given range. Unlike discrete variables which assume distinct integers, continuous ones can take on infinite possibilities. This finer granularity allows for more precise modeling of outcomes.
Probability distributions for continuous variables are defined using Probability Density Functions (PDFs). The PDF, \( f(x) \), represents the relative likelihood of the variable taking on a particular value. For \( X \) to be a proper continuous random variable, \( \int_{-fty}^{fty} f(x) \, dx = 1 \) must hold true.
This guarantees that all probabilities within the range add up to 1, making the distribution valid.

• \( X \) being continuous implies it does not have a probability mass at each point, but rather distribution over a range.

In practice, for our random variable with \( f(x) = 4x - 4x^3 \) for \( 0 \leq x \leq 1 \), ensuring the integral equals 1 confirms it adheres to the rules of a continuous distribution. Understanding the properties of such variables aids in further calculations, such as finding expectation and variance.