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Given a continuous uniform distribution, show that (a) \(\mu=\frac{A+B}{2},\) and (b) \(\sigma^{2}-\frac{(B-\lambda)^{2}}{12}\)

Short Answer

Expert verified
The mean of a continuous uniform distribution on the interval [A, B] is \(\mu=\frac{A+B}{2}\) and the variance is \(\sigma^{2}=\frac{(B-A)^{2}}{12}\).

Step by step solution

01

Analyze the Mean (μ)

A continuous uniform distribution defined on the interval [A, B] has a mean (or expected value) defined as \(\mu =\frac{A+B}{2}\). This is derived from the basic formula for expected value, \(\mu = \int xf(x)dx\), where f(x) is the probability density function of the distribution, in this case is \(\frac{1}{B-A}\) for A ≤ x ≤ B, else 0. We would integrate this expression over the range of possible x values (A to B), if we simplify we can see that \(\mu =\frac{A+B}{2}\).
02

Analyze the Variance (σ²)

The variance of a continuous uniform distribution defined on the interval [A, B] is given by the formula \(\sigma^{2}=\frac{(B-A)^{2}}{12}\). This can be derived from the formula for variance, \(\sigma^{2} = \int (x-\mu)^2 f(x)dx\), integrating this equation and simplifications, we arrive at the formula \(\sigma^{2}=\frac{(B-A)^{2}}{12}\). This calculation provides the dispersion measure around the mean.
03

Finalize the results

As a result of the defining properties of the uniform distribution, we have shown that (a) the mean of a continuous uniform distribution on the interval [A, B] is \(\mu=\frac{A+B}{2}\), and (b) the variance is \(\sigma^{2}=\frac{(B-A)^{2}}{12}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The expected value, often denoted as μ or E(X), is a fundamental concept in probability and statistics, particularly when working with continuous distributions. In a discrete context, the expected value is calculated as the sum of all possible outcomes weighted by their respective probabilities. However, for continuous distributions, we determine it using integration.

In the case of a continuous uniform distribution, which is defined over an interval [A, B], the expected value is the midpoint of A and B, mathematically expressed as \(\mu=\frac{A+B}{2}\). This intuitively makes sense because, in a uniform distribution, all outcomes are equally likely, and hence the 'center of mass' of the distribution is right in the middle of the interval. The calculation follows from integrating the product of the value x and its probability density function f(x) across the range of possible values.

Understanding the expected value is critical for predicting the long-run average of results over many trials or occurrences of a random process. It essentially gives us a measure of the 'center' of the distribution. In practical terms, if we were to repeatedly sample from a continuous uniform distribution, the average of the results would tend to the expected value as the number of samples increases.
Probability Density Function
The probability density function (PDF) is a function that describes the relative likelihood for a continuous random variable to occur at a given point. For the continuous uniform distribution, the PDF is exceptionally simple. It is defined by a constant function over the interval [A, B] and is zero elsewhere.

The formula for the PDF of a continuous uniform distribution looks like this: \(f(x) = \frac{1}{B-A}\) for \(A \leq x \leq B\), and zero for other values of x. This constant value ensures that the total area under the curve of the PDF over the interval [A, B] is equal to one, satisfying the basic requirement for any probability distribution.

The uniform PDF is distinctive because it allocates equal probability to all values within the specified interval; it illustrates one of the simplest cases of probability distributions. Knowing this function is essential as it is used to compute other vital statistical measures, such as the mean and variance, and for understanding how probabilities are allocated across different outcomes.
Variance
Variance is a measure of how dispersed the values in a distribution are around the mean, or expected value. It captures the degree of spread in the possible outcomes. In the context of a continuous uniform distribution, the variance is mathematically derived as \(\sigma^{2}=\frac{(B-A)^{2}}{12}\).

To calculate this, one would typically square the distance of each possible value from the mean, multiply by the probability of that value (here constant due to the uniform nature of the distribution), and integrate that over the interval [A, B]. The result, in this case, is a function of the range of the distribution, (B - A), which makes sense because a broader interval implies a higher dispersion of values and, therefore, a greater variance.

Variance is useful when comparing the spread of different distributions, or when we want to understand the volatility or risk associated with the outcomes of a random process. A low variance indicates that data points are generally close to the mean and hence to each other, while a high variance indicates that the data points are spread out over a wider range of values.

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