Problem 1 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Often one is interested in the distribution of the deviation of a random variable \(X\) from its mean \(\mu=\mathrm{E}[X]\). Let \(X\) take the values \(80,90,100,110\), and 120, all with probability \(0.2 ;\) then \(\mathrm{E}[X]=\mu=100\). Determine the distribution of \(Y=|X-\mu|\). That is, specify the values \(Y\) can take and give the corresponding probabilities.

Step-by-Step Solution

Answer

Y can be 0, 10, or 20 with probabilities 0.2, 0.4, and 0.4, respectively.

Step 1: Identify the Deviations

Calculate the absolute deviations for each value of \(X\) from the mean \(\mu\). The deviations are: \(|80-100|=20\), \(|90-100|=10\), \(|100-100|=0\), \(|110-100|=10\), and \(|120-100|=20\). These give us the possible values of \(Y\).

Step 2: List Unique Values of Y

From the deviations, the unique values \(Y\) can take are \(0, 10,\) and \(20\).

Step 3: Calculate the Probability for Each Y Value

Determine the probability of each value of \(Y\). \(Y=0\) occurs when \(X=100\), so its probability is \(0.2\). \(Y=10\) occurs when \(X=90\) or \(110\), so its probability is \(0.2+0.2=0.4\). \(Y=20\) occurs when \(X=80\) or \(120\), so its probability is \(0.2+0.2=0.4\).

Step 4: Specify Distribution of Y

The distribution of \(Y\) is as follows: \(\Pr(Y=0) = 0.2\), \(\Pr(Y=10) = 0.4\), \(\Pr(Y=20) = 0.4\).

Key Concepts

Random Variable

A random variable is a fundamental concept in probability and statistics. It is a variable whose possible values are determined by the outcome of a random phenomenon. In simpler terms, it's a way to quantify the results of random events.

For instance, in the given exercise, the random variable is represented by \( X \), which can take the values 80, 90, 100, 110, and 120. The probability of each of these outcomes is \( 0.2 \). So, the probability distribution of \( X \) is defined, which shows how probable each value is.

Understanding random variables helps in predicting and understanding patterns in uncertain situations, like predicting stock prices or even the weather.

Expectation

Expectation, also known as the expected value, is a vital concept when dealing with random variables. It provides a measure of the central tendency of a probability distribution. You can think of it as the average or mean value you would get if you could repeat the random experiment many times.

Mathematically, the expectation \( \mathrm{E}[X] \) of a random variable \( X \) is calculated by multiplying each possible value of \( X \) by its probability and adding these products together. In our exercise, the expectation of \( X \) is given as 100.

This is computed because \( X \) takes every value (80, 90, 100, 110, 120) in the dataset with equal probability, which results in the expectation being equivalent to the mean of these values. Expectation plays a critical role in various statistical analyses and helps determine the expected outcomes in probabilistic scenarios.

Absolute Deviation

Absolute deviation is a measure of how much a particular value in a dataset deviates from the mean of the dataset. It focuses on how spread out the numbers are around the mean.

In the exercise, we calculated the absolute deviation of each value of \( X \) from the mean \( \mu = 100 \). This means we took the absolute value of the difference between \( X \) and \( \mu \), resulting in values 20, 10, 0, 10, and 20. These deviations tell us how far each value is from the average, without regard to direction (which is why we use absolute values).

Understanding absolute deviation helps in recognizing the extent of variability in a dataset. It is particularly useful because it gives a clear picture of the dispersion of data points around the mean.

Probability Concepts

Probability concepts form the foundation of understanding random events and their likelihoods. It helps in answering questions about how likely certain events are to happen under specific conditions.

In our exercise, after determining the absolute deviations, we looked at what probability each possible deviation could occur (represented by \( Y \)). The calculation showed that there is a 0.2 probability for \( Y = 0 \) and a 0.4 probability for both \( Y = 10 \) and \( Y = 20 \).

This helps us construct a probability distribution for \( Y \), which is a list showing each possible deviation value and its associated probability. Understanding these probability concepts is essential for analyzing and predicting how likely different outcomes are, and they are widely applied in fields such as finance, insurance, and science.