Problem 76 - Applied Statistics and Probability for Engineers

Question

Three electron emitters produce electron beams with changing kinetic energies that are uniformly distributed in the ranges \([3,7],[2,5],\) and \([4,10] .\) Let \(Y\) denote the total kinetic energy produced by these electron emitters. (a) Suppose that the three beam energies are independent. Determine the mean and variance of \(Y\). (b) Suppose that the covariance between any two beam energies is \(-0.5 .\) Determine the mean and variance of \(Y\). (c) Compare and comment on the results in parts (a) and (b).

Step-by-Step Solution

Answer

Mean of Y is 15.5 in both cases; variance is 5.083 assuming independence and 3.083 with negative covariance.

Step 1: Define Random Variables

Let \( X_1, X_2, \) and \( X_3 \) represent the kinetic energies produced by the three electron emitters, each following a uniform distribution: \( X_1 \sim U(3,7) \), \( X_2 \sim U(2,5) \), and \( X_3 \sim U(4,10) \). The total energy \( Y \) is given by \( Y = X_1 + X_2 + X_3 \).

Step 2: Calculate Means of Uniform Distributions

The mean of a uniform distribution \( U(a, b) \) is \( \mu = \frac{a+b}{2} \). So, calculate:- \( \mu_1 = \frac{3+7}{2} = 5 \) for \( X_1 \).- \( \mu_2 = \frac{2+5}{2} = 3.5 \) for \( X_2 \).- \( \mu_3 = \frac{4+10}{2} = 7 \) for \( X_3 \).

Step 3: Calculate Variances of Uniform Distributions

The variance of a uniform distribution \( U(a, b) \) is \( \sigma^2 = \frac{(b-a)^2}{12} \). Calculate:- \( \sigma_1^2 = \frac{(7-3)^2}{12} = \frac{16}{12} = \frac{4}{3} \) for \( X_1 \).- \( \sigma_2^2 = \frac{(5-2)^2}{12} = \frac{9}{12} = \frac{3}{4} \) for \( X_2 \).- \( \sigma_3^2 = \frac{(10-4)^2}{12} = \frac{36}{12} = 3 \) for \( X_3 \).

Step 4: Mean of Total Energy for Part (a)

Assuming independence, the mean of \( Y \) is the sum of the means of \( X_1, X_2, \) and \( X_3 \):\[ \mu_Y = \mu_1 + \mu_2 + \mu_3 = 5 + 3.5 + 7 = 15.5 \]

Step 5: Variance of Total Energy for Part (a)

The variance of \( Y \) assuming independence is the sum of the variances of \( X_1, X_2, \) and \( X_3 \):\[ \sigma_Y^2 = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 = \frac{4}{3} + \frac{3}{4} + 3 = \frac{16}{12} + \frac{9}{12} + \frac{36}{12} = \frac{61}{12} \approx 5.083 \]

Step 6: Mean of Total Energy for Part (b)

The mean of \( Y \) remains the same: \( \mu_Y = 15.5 \), because the mean is not affected by the covariance.

Step 7: Variance of Total Energy for Part (b)

If \( \, \text{Cov}(X_i, X_j) = -0.5 \, \) for any two different \( i, j \), then:\[ \sigma_Y^2 = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 + 2(\text{Cov}(X_1, X_2) + \text{Cov}(X_1, X_3) + \text{Cov}(X_2, X_3)) \] Calculate:\[ \sigma_Y^2 = \frac{4}{3} + \frac{3}{4} + 3 + 2(-0.5 - 0.5 - 0.5) = \frac{61}{12} - 3 = \frac{37}{12} \approx 3.083 \]

Step 8: Compare and Comment (Part c)

In part (a), assuming independence, the variance of \( Y \) is higher at approximately \( 5.083 \) compared to part (b), where the variance is reduced to \( 3.083 \) due to negative covariance, indicating some cancellation effect among the energy fluctuations of the emitters, while the mean remains unchanged.

Key Concepts

Random Variables

In probability and statistics, a **random variable** is a variable whose possible values are numerical outcomes of a random phenomenon. Think of it as a box containing a bunch of numbers, and during an experiment, you pick one number out of the box.
In our exercise, the random variables are \( X_1, X_2, \) and \( X_3 \), representing the kinetic energies from the three different electron emitters. These values are not set but can vary according to specific conditions defined by their probability distributions.
Random variables can be either discrete, taking on a finite set of values (like rolling a die), or continuous, having values over a continuous range (like our kinetic energies). Understanding random variables helps us make sense of experiments and processes with inherent randomness.

Uniform Distribution

A **uniform distribution** describes an equal likelihood of all outcomes within a specified range. Imagine spinning a wheel divided into equal sections — each section is just as likely to land under the spinner as any other.
For a continuous uniform distribution, defined by an interval \([a, b]\), any point within the interval is equally likely, and the probability density is constant. This is the case with each of our random variables \(X_1, X_2, \) and \(X_3\).

• \(X_1 \sim U(3,7)\): Uniformly distributed energies between 3 and 7.
• \(X_2 \sim U(2,5)\): Uniformly distributed energies between 2 and 5.
• \(X_3 \sim U(4,10)\): Uniformly distributed energies between 4 and 10.

The key takeaway is that for a uniform distribution, the mean \( \mu \) is the midpoint, calculated as \( \mu = \frac{a+b}{2} \), and provides a central measure for this evenly spread range.

Covariance

**Covariance** measures how much two random variables change together. It tells us if the variables have a tendency to increase or decrease simultaneously.
In our problem, the covariance between any two electron emitters is given as \( -0.5 \), indicating a negative relationship.
A negative covariance means that as one variable increases, the other tends to decrease. For our emitters, if one emitter's energy is higher than average, the others might be lower. This negative relationship impacts the overall variability of their combined energies.

• Positive covariance: Variables move in the same direction.
• Negative covariance: Variables move in opposite directions.
• Zero covariance: No consistent linear relationship.

Covariance is essential in determining total variance when combining multiple variables, especially when they are not independent.

Mean and Variance

**Mean** and **variance** are fundamental concepts in statistics. The mean, often referred to as the average, provides a central value for a set of data. The variance, on the other hand, tells us how much the data is spread out.
For our exercise:

• The mean of each uniform distribution is found by \(\mu = \frac{a+b}{2}\), which tells us the expected value of energy for each emitter.
• The variance of a uniform distribution is \( \sigma^2 = \frac{(b-a)^2}{12} \), showing us the expected spread from the mean.

Calculating the mean and variance helps in understanding the behavior of the total kinetic energy \(Y\). In statistical calculations, the mean of combined independent variables is simply the sum of individual means, while the variance is the sum of individual variances when variables are independent. However, when negative covariance is involved, it reduces the total variance, as we've seen in the exercise outcomes.

Independent Events

In probability, **independent events** are scenarios where the occurrence of one event does not affect the probability of another. This is like tossing a coin; the result of one toss doesn't change the odds for the next.
For the emitters, assuming that the "beam energies are independent" means the energy produced by one emitter doesn't influence the others. In terms of our calculations, independence implies that the total variance of \(Y\) is merely the sum of the variances of \(X_1, X_2,\) and \(X_3\), without any additional terms for covariance affecting the result.
However, in situations where the events are not independent, such as when a covariance is involved, this assumption changes. Independency allows for elegant calculations and simplifies understanding complex systems by reducing them to their constituent parts with no interference between them.