Problem 17 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(U\) be a discrete random variable taking the values \(a_{1}, \ldots, a_{r}\) with probabilities \(p_{1}, \ldots, p_{r} .\) a. Suppose all \(a_{i} \geq 0\), but that \(\mathrm{E}[U]=0\). Show then $$ a_{1}=a_{2}=\cdots=a_{r}=0 . $$ In other words; \(\mathrm{P}(U=0)=1\).

Step-by-Step Solution

Answer

All \( a_{i} = 0 \), so \( \mathrm{P}(U=0) = 1 \).

Step 1: Understanding the Expectation

The expected value \( \mathrm{E}[U] \) of a discrete random variable \( U \) can be calculated as \( \mathrm{E}[U] = \sum_{i=1}^{r} a_{i} p_{i} \). We are given that \( \mathrm{E}[U] = 0 \). So, \( \sum_{i=1}^{r} a_{i} p_{i} = 0 \).

Step 2: Applying Condition on Values

We are given that all \( a_{i} \geq 0 \). If all \( a_{i} \) are non-negative, then each term \( a_{i} p_{i} \) in the sum must also be non-negative because \( p_{i} \geq 0 \) for all probabilities.

Step 3: Analyzing the Sum Condition

Since the sum \( \sum_{i=1}^{r} a_{i} p_{i} = 0 \) and each term \( a_{i} p_{i} \geq 0 \), the only way this sum can equal zero is if each term is zero. That means \( a_{i} p_{i} = 0 \) for every \( i \).

Step 4: Examining Each Term

For \( a_{i} p_{i} = 0 \) and \( p_{i} > 0 \) (since they are probabilities and sum to 1), it must be that \( a_{i} = 0 \). This must hold for every \( i \), thus \( a_{1}=a_{2}=\cdots=a_{r}=0 \).

Step 5: Conclusion

Since each \( a_{i} = 0 \), the only value \( U \) can take is 0, with probability 1. Thus, \( \mathrm{P}(U=0)=1 \), confirming the statement that \( a_{1}=a_{2}=\cdots=a_{r}=0 \).

Key Concepts

Discrete Random Variable

In the world of probability, a **discrete random variable** is a type of variable that takes on a countable number of distinct values. It is often used to represent outcomes of random phenomena such as the roll of a die or the flip of a coin.
For instance, if you are dealing with a six-sided die, the discrete random variable can take values 1 through 6. Each value corresponds to an outcome of the die roll. When we talk about discrete random variables, each possible outcome has a probability assigned to it. Let's denote the outcomes of a random variable as \( a_1, a_2, \ldots, a_r \). These could be any distinct numbers depending on the context of the problem at hand. Importantly, the total probability for all possible outcomes must sum up to one, reflecting that one of these outcomes will definitely occur when the experiment is carried out. Discrete random variables heavily rely on their probability distributions, which help in determining the expected value, variance, and many other statistical measures.

Probability

**Probability** is a fundamental concept in statistics and is used to quantify the uncertainty of events. For any given event, probability values range from 0 to 1, where 0 indicates impossibility, and 1 indicates certainty.
When we consider a discrete random variable, each possible outcome is associated with a probability. This probability distribution essentially "weights" each outcome according to its likelihood.For a discrete random variable \( U \) with outcomes \( a_1, a_2, \ldots, a_r \), the probabilities \( p_1, p_2, \ldots, p_r \) fulfill the condition:

• Each probability \( p_i \geq 0 \)
• The sum of all probabilities is 1: \( \sum_{i=1}^{r} p_i = 1 \)

Understanding probability is crucial since it underpins the computation of the expected value. Moreover, if each term \( a_i p_i \) equals zero, as demonstrated in our problem, it further confirms the calculated expectation.

Expectation

The **expectation** or expected value is a measure of the "central tendency" of a random variable, representing the average outcome one would expect if an experiment could be repeated an infinite number of times. In a sense, it is similar to the concept of the mean.
For a discrete random variable \( U \), its expected value \( \mathrm{E}[U] \) is calculated as the sum of each outcome multiplied by its corresponding probability:\[\mathrm{E}[U] = \sum_{i=1}^{r} a_i p_i\]In our exercise, the expected value is zero: \( \mathrm{E}[U] = 0 \). Since all \( a_i \)'s are non-negative, this means every term in the sum \( a_i p_i \) must also be non-negative due to probabilities being non-negative. To achieve a total sum of zero, each term \( a_i p_i \) must be zero, which occurs only when each \( a_i = 0 \).
This leads to the conclusion that the only possibility is \( a_1 = a_2 = \ldots = a_r = 0 \), meaning \( U \) consistently takes the value 0, and thus \( \mathrm{P}(U=0) = 1 \). Understanding expectation is pivotal, as it serves as a crucial stepping stone in verifying either regular or exceptional outcomes for random variables.