Problem 58 - Calculus Single Variable

Question

Calculate the variance \(\operatorname{Var}(X)\) of a random variable \(X\) whose probability density function is the given function \(f\). $$ f(x)=x / 4 \quad 1 \leq x \leq 3 $$

Step-by-Step Solution

Answer

The variance is \( \frac{11}{36} \).

Step 1: Understand the Variance Formula

To find the variance of a continuous random variable with a given probability density function (pdf) \( f(x) \), we use the formula: \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 \] where \( E(X) \) is the expected value of \( X \) and \( E(X^2) \) is the expected value of \( X^2 \).

Step 2: Calculate Expected Value, E(X)

The expected value \( E(X) \) is calculated as \[ E(X) = \int_{1}^{3} x \cdot f(x) \, dx = \int_{1}^{3} x \cdot \left( \frac{x}{4} \right) \, dx = \frac{1}{4} \int_{1}^{3} x^2 \, dx \]Evaluating this integral gives: \[ E(X) = \frac{1}{4} \cdot \left[ \frac{x^3}{3} \right]_{1}^{3} = \frac{1}{4} \cdot \left( \frac{27}{3} - \frac{1}{3} \right) = \frac{1}{4} \cdot \frac{26}{3} = \frac{13}{6} \]

Step 3: Calculate Expected Value of X^2, E(X^2)

The expected value \( E(X^2) \) is calculated as \[ E(X^2) = \int_{1}^{3} x^2 \cdot f(x) \, dx = \int_{1}^{3} x^2 \cdot \left( \frac{x}{4} \right) \, dx = \frac{1}{4} \int_{1}^{3} x^3 \, dx \]Evaluating the integral gives: \[ E(X^2) = \frac{1}{4} \cdot \left[ \frac{x^4}{4} \right]_{1}^{3} = \frac{1}{4} \cdot \left( \frac{81}{4} - \frac{1}{4} \right) = \frac{1}{4} \cdot \frac{80}{4} = \frac{20}{4} = 5 \]

Step 4: Substitute Values into Variance Formula

Now, substitute \( E(X) = \frac{13}{6} \) and \( E(X^2) = 5 \) into the variance formula: \[ \operatorname{Var}(X) = E(X^2) - (E(X))^2 = 5 - \left( \frac{13}{6} \right)^2 \]Calculate \( \left( \frac{13}{6} \right)^2 = \frac{169}{36} \). Then:\[ \operatorname{Var}(X) = 5 - \frac{169}{36} = \frac{180}{36} - \frac{169}{36} = \frac{11}{36} \]

Step 5: Final Result

The variance of the random variable \( X \) with the given pdf is \( \operatorname{Var}(X) = \frac{11}{36} \).

Key Concepts

Probability Density Function

A probability density function, often abbreviated as pdf, is a crucial concept in understanding continuous random variables. The pdf describes how the probability of the random variable is distributed over a range of values.
Unlike discrete random variables that have specific probabilities for each outcome, continuous random variables have a range of outcomes, and probabilities are assigned to intervals rather than specific points.
In mathematical terms, for a continuous random variable \(X\) with pdf \(f(x)\), the probability that \(X\) lies within a certain interval \([a, b]\) is given by the integral \(\int_{a}^{b} f(x) \, dx\).
A key property of a pdf is that when you integrate the function over its entire range, it must equal 1. This reflects the fact that the probability of all possible outcomes must be 100%.
In the given exercise, the pdf is defined as \(f(x) = \frac{x}{4}\) for \(1 \leq x \leq 3\). To ensure this is a valid pdf, the integral over the range \([1, 3]\) should equal 1, confirming it is correctly distributed across the interval.

Expected Value

The expected value, often referred to as the mean, is a measure of the center of a probability distribution. It gives you a kind of "average" value that the random variable is expected to take.
For continuous random variables, the expected value \(E(X)\) is determined using the formula \(E(X) = \int_{a}^{b} x \cdot f(x) \, dx\), where \(f(x)\) is the probability density function.
This involves integrating the product of \(x\) and the pdf over the entire range where the random variable is defined. The result is the expected value \(E(X)\), providing insight into the central tendency of the distribution.
In our exercise, this was calculated by integrating \(x \cdot \frac{x}{4}\) from 1 to 3, resulting in an expected value of \(\frac{13}{6}\).
The expected value plays a significant role in calculating variance, as it is used to determine how much the random variable deviates from this central value.

Continuous Random Variable

A continuous random variable is a type of random variable that can take on an infinite number of possible values within a given range. This distinguishes it from a discrete random variable, which has a finite or countable number of outcomes.
Continuous random variables require a probability density function to describe the likelihood of the variable falling within a particular interval. They are capable of representing a vast array of real-world phenomena such as heights, weights, time durations, and more.
The key characteristics of continuous random variables are:

• The probability of the variable taking a specific value is zero. Instead, probabilities are associated with intervals of values.
• The cumulative distribution function (CDF) is often used to describe probabilities for continuous variables by showing the probability that a variable takes a value less than or equal to a certain number.

In the given problem, \(X\) is a continuous random variable defined over the interval from 1 to 3, and the probability density function \(f(x) = \frac{x}{4}\) helps to determine probabilities across this range.