Question

The simplest continuous random variable is the one whose distribution is constant over some interval \((a, b)\) and zero elsewhere. This is the uniform distribution. $$ \begin{array}{ll} \mathrm{f}(\mathrm{x})=[1 /(\mathrm{b}-\mathrm{a})], & \mathrm{a} \leq \mathrm{X} \leq \mathrm{b} \\ \text { and }=0, & \text { elsewhere } \end{array} $$ Find the mean and variance of this distribution.

Short answer

The mean of the uniform distribution is \(E[X] = \frac{1}{2}(b+a)\) and the variance is \(Var(X) = \frac{1}{12}(b - a)^2\).

Step-by-step solution

Find the Mean (Expected Value) E[X]

First, let's calculate the expected value of the random variable X using the given probability density function: $$ E[X] = \int_{-\infty}^{\infty} xf(x) dx = \int_{a}^{b} x \frac{1}{b-a} dx $$ Compute the integral: $$ E[X] = \frac{1}{b - a} \int_{a}^{b} x dx = \frac{1}{b - a} \left[ \frac{1}{2}x^2 \right]_a^b $$ Substitute the interval limits a and b: $$ E[X] = \frac{1}{b - a} \left[ \frac{1}{2}b^2 - \frac{1}{2}a^2 \right] = \frac{1}{2}(b+a) $$ So, the mean (expected value) of the distribution is \(E[X] = \frac{1}{2}(b+a)\).

Find E[X^2]

To compute the variance, we need to find E[X^2] first. Using the probability density function again, we have: $$ E[X^2] = \int_{-\infty}^{\infty} x^2f(x) dx = \int_{a}^{b} x^2 \frac{1}{b-a} dx $$ Compute the integral: $$ E[X^2] = \frac{1}{b - a} \int_{a}^{b} x^2 dx = \frac{1}{b - a} \left[ \frac{1}{3}x^3 \right]_a^b $$ Substitute the interval limits a and b: $$ E[X^2] = \frac{1}{b - a} \left[ \frac{1}{3}b^3 - \frac{1}{3}a^3 \right] = \frac{1}{3}(b^3 - a^3)/(b - a) $$ So, the value of E[X^2] for the distribution is \(E[X^2] = \frac{1}{3}(b^3 - a^3)/(b - a)\).

Find the Variance Var(X)

Now, we'll find the variance by using the formula: $$ Var(X) = E[X^2] - E[X]^2 $$ Substitute the values of E[X] and E[X^2]: $$ Var(X) = \frac{1}{3}(b^3 - a^3)/(b - a) - \left( \frac{1}{2}(b+a) \right)^2 $$ Simplify further: $$ Var(X) = \frac{1}{12}(b^2 - 2ab + a^2) = \frac{1}{12}(b - a)^2 $$ Hence, the variance of the given distribution is \(Var(X) = \frac{1}{12}(b - a)^2\). So, the mean of the uniform distribution is \(E[X] = \frac{1}{2}(b+a)\), and the variance is \(Var(X) = \frac{1}{12}(b - a)^2\).

Key concepts

Continuous Random Variable

Understanding the concept of a continuous random variable is essential for diving into topics like probability and statistics. Unlike a discrete random variable, which can only take on a countable number of values (like the roll of a dice), a continuous random variable can assume any value within a certain range or interval.

Consider the scenario of measuring the time it takes to complete a task. The amount of time, theoretically, can be any number within an interval, say between 0 and 2 hours, accounting for every possible fraction of a second. This is a continuous random variable, because you aren't just counting whole seconds or minutes—you're considering every possible moment within that range.

Importantly, you cannot find the probability of the exact value of a continuous random variable; it is always zero since there are infinite possibilities. Instead, you calculate the likelihood that the variable falls within an interval, a concept central to the continuous uniform distribution.

Probability Density Function

The probability density function (PDF) is a tool to describe the likelihood of a continuous random variable falling within a specific range. Think of it as a curve that shows how the probabilities are distributed across the values the variable can take. For the uniform distribution, the PDF is particularly straightforward.

The formula provided in the exercise, \( f(x) = \frac{1}{b - a} \) for \( a \leq x \leq b \) and 0 elsewhere, represents a PDF. The function's value is the same across the interval \( (a, b) \) and drops to zero outside it. This means that the variable is equally likely to take any value within \( (a, b) \), truly defining a 'uniform' distribution.

Knowing the PDF allows us to compute important characteristics of the distribution, such as the mean and variance, which are key to understanding the behavior of the random variable.

Mean of Distribution

The mean of the distribution, often referred to as the expected value or average, is a crucial measure in probability and statistics. For a uniform distribution, it represents the midpoint between the lower and upper bounds of the range over which the variable is distributed.

In the case of our uniform distribution, the formula \( E[X] = \frac{1}{2}(b+a) \) tells you the exact center of the range \( (a, b) \). This makes sense because, with each value equally likely, the mean sits smack in the middle. You can visualize it as the balancing point of the distribution—if you were to balance the PDF on the tip of your finger, the mean would be the point where it stays level.

Variance of Distribution

While the mean tells us where the center of the distribution lies, the variance of the distribution gives us an idea of how spread out the values are around the mean. For our uniform distribution, the variance formula \( Var(X) = \frac{1}{12}(b-a)^2 \) calculates the degree of dispersion.

A smaller variance indicates that the values are clustered more closely around the mean, while a larger variance signifies a wider spread. The factor of \( \frac{1}{12} \) in the formula is due to the specific properties of the uniform distribution, resulting in a spread that is symmetric around the mean. In more complex distributions, the formula for variance can vary considerably, but the concept remains the same: variance is a measure of distribution 'width' or 'spread'.