Question

Let \(\mu\) be the average of the random variable \(x\). Then the quantities \(\left(x_{i}-\mu\right)\) are the deviations of \(x\) from its average. Show that the average of these deviations is zero. Hint: Remember that the sum of all the \(p_{i}\) must equal 1.

Short answer

The average of the deviations \( (x_{i} - \textbacklash mu) \) is zero.

Step-by-step solution

Define the average \(\textbackslash mu\)

The average of the random variable \( x \) is given by \(\textbackslash mu = E[x] = \textbackslash sum_{i} p_{i} x_{i}\), where \( p_{i}\) are the probabilities associated with \(x_{i}\).

Write down the deviations from the average

The deviations of \( x \) from its average \( \textbackslash mu \) are given by \((x_{i} - \textbackslash mu)\).

Find the expectation of the deviations

The average of the deviations can be written as the expected value of \((x_{i} - \textbackslash mu)\): \[ E[(x - \textbackslash mu)] = \textbackslash sum_{i} p_{i} (x_{i} - \textbackslash mu) \]

Distribute and rearrange terms inside the summation

Distribute \( p_{i} \) to get: \[ E[(x - \textbackslash mu)] = \textbackslash sum_{i} p_{i} x_{i} - \textbackslash mu \textbackslash sum_{i} p_{i} \]

Recognize that the sum of probabilities \((\textbackslash sum_{i} p_{i}) \) is 1

Since \(\textbackslash sum_{i} p_{i} = 1\), substitute this into the equation to get: \[ E[(x - \textbackslash mu)] = \textbacklash sum_{i} p_{i} x_{i} - \textbackslash mu \textbackslash cdot 1 \]

Substitute the definition of \( \textbackslash mu \) into the equation

Recall that \(\textbackslash mu = \textbacklash sum_{i} p_{i} x_{i}\). Substitute \( \textbacklash mu \) in to get: \[ E[(x - \textbackslash mu)] = \textbacklash mu - \textbacklash mu \]

Simplify the equation

Simplify to show: \( E[(x - \textbacklash mu)] = 0 \).

Key concepts

Random Variable

In probability and statistics, a **random variable** is a numerical outcome of a random phenomenon. Think of it as a variable whose possible values are numerical outcomes of a random process. For example, when you roll a die, the outcome can be any number from 1 to 6. Each outcome corresponds to a certain probability.
There are two main types of random variables:

• **Discrete Random Variables**: These take on a countable number of distinct values. An example is the roll of a die, where the possible outcomes are 1, 2, 3, 4, 5, and 6.


• **Continuous Random Variables**: These can take on an infinite number of possible values. An example would be the amount of rainfall in a day, which could be any value within a range.

For any random variable, we can calculate an average value, also called the **expectation** or **mean**. If the random variable is denoted by **x**, the expectation is usually denoted by **E[x]** or **μ** (mu). If we have a set of probabilities **p_i** associated with outcomes **x_i**, the mean is computed as
\( \textbackslash mu = E[x] = \textbackslash sum_{i} p_{i} x_{i} \)

Deviation

The **deviation** of a random variable is the difference between each outcome and the mean of the random variable. This shows how much each value differs from the average value. Mathematically, for a random variable **x_i** with a mean **μ**, the deviation is given by
\( (x_{i} - \textbackslash mu) \).
Understanding deviation is essential to grasp the dispersion or spread of values around the mean. It's a foundational concept in statistics because it helps quantify variability in data. If we sum the deviations for all possible outcomes of a random variable weighed by their probabilities, surprisingly, the result is always zero. This happens because the positive and negative deviations balance each other out. Mathematically, we want to show that
\( \textbackslash sum_{i} p_{i} (x_{i} - \textbackslash mu) = 0 \)
Here's why this works:

• We start with the expectation of deviations, \( E[(x - \textbackslash mu)] \).
• Distribute the probability term, so it becomes \( \textbacklash sum_{i} p_{i} x_{i} - \textbackslash mu \textbackslash sum_{i} p_{i} \).
• Since the sum of all probabilities \( \textbackslash sum_{i} p_{i} \) is 1, it simplifies to \( \textbacklash sum_{i} p_{i} x_{i} - \textbackslash mu \).
• Replace \( \textbacklash sum_{i} p_{i} x_{i} \) with the mean \( \textbacklash mu \).
• The final expression is \( \textbacklash mu - \textbacklash mu = 0 \), showing that the expected value of the deviations is zero.

Probabilities

Probabilities are a measure of the likelihood that a particular event will occur. They are fundamental to understanding random variables and deviations. Here's why:

• **Probability Values**: Probabilities are numbers between 0 and 1, where 0 means an event will not happen and 1 means it will certainly happen. In between, they quantify the event's likelihood.


• **Sum of Probabilities**: The sum of the probabilities of all possible outcomes of a random variable must equal 1. This is because one of the outcomes must happen.


• **Expected Value**: By associating each outcome of a random variable with its probability, you can find the expected value, which provides a measure of the center of the distribution of the variable.

In our exercise, we used the probabilities **p_i** to calculate the mean **μ** and then examined how deviations from this mean behave. It's crucial to understand that probabilities serve as weights, ensuring that the mean reflects the central tendency accurately, even when outcomes have different likelihoods.