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Let \(A\) be an event. Then \(I_{A}\) , the indicator random variable of \(A\) , equals 1 if \(A\) occurs and equals 0 otherwise. Show that the expectation of the indicator random variable of \(A\) equals the probability of \(A,\) that is, \(E\left(I_{A}\right)=p(A)\)

Short Answer

Expert verified
E(I_A) = P(A)

Step by step solution

01

Define the Indicator Random Variable

The indicator random variable of event A, denoted as I_A, is defined such that I_A equals 1 if event A occurs, and 0 if event A does not occur.
02

Establish the Definition of Expectation

The expectation of a random variable X, E(X), is defined as the sum of all possible values of X, each multiplied by its probability. For a random variable Y with discrete values,\[E(Y) = \sum_{k} y_k \, P(Y = y_k)\]
03

Apply the Expectation Formula to the Indicator Variable

For the indicator random variable I_A, there are two possible values: 0 and 1. We need to calculate their contributions to the expectation. Therefore,\[E(I_A) = 0 \, \cdot \, P(I_A = 0) + 1 \, \cdot \, P(I_A = 1)\]
04

Substitute the Probabilities

Given that P(A) is the probability that event A occurs and P(A^c) is the probability that event A does not occur, we have:\[P(I_A = 1) = P(A)\]\[P(I_A = 0) = P(A^c)\]
05

Final Calculation

Substituting these values back into the expectation formula gives:\[E(I_A) = 0 \, \cdot \, P(A^c) + 1 \, \cdot \, P(A) = P(A)\]

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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 is a fundamental concept in probability and statistics. It represents the average outcome of a random variable over a large number of trials.
Formally, the expected value of a random variable X, denoted as E(X), is calculated by summing the products of each possible value of X and its corresponding probability:
\[ E(X) = \sum_{k} x_k \, P(X = x_k) \]

This formula tells us that for each outcome x_k, we multiply it by the probability (P) that X takes on the value x_k, and sum these products.

In the case of the indicator random variable I_A, which equals 1 if event A occurs and 0 otherwise, the expected value is particularly straightforward to calculate.
First, recall that for I_A:
  • I_A = 1 if A occurs
  • I_A = 0 if A does not occur

Using the expectation definition, the expected value for I_A becomes:\[E(I_A) = 0 \, P(I_A = 0) + 1 \, P(I_A = 1)\]
This expression simplifies to the probability that event A occurs: \(E(I_A) = P(A)\).
This is a powerful result because it links the probability of an event directly to its indicator random variable's expected value.
probability

Probability quantifies the likelihood of an event occurring. It is a value between 0 and 1, where 0 indicates that the event will not happen and 1 indicates certainty.

Several foundational principles help us understand probability better:
  • Law of Large Numbers: As the number of trials increases, the experimental probability will converge to the theoretical probability.
  • Complementary Events: The sum of the probabilities of an event and its complement is always 1. If P(A) is the probability of event A, then \(P(A^c) = 1 - P(A)\), where A^c is the complement of A.
  • Additive Rule: For any two mutually exclusive events A and B, \(P(A \cup B) = P(A) + P(B)\).

When dealing with indicator random variables, we focus on the probability of binary outcomes. For instance, with I_A:
  • Probability of I_A being 1: This is the probability that event A occurs, denoted as P(A).
  • Probability of I_A being 0: This is the probability that event A does not occur, denoted as P(A^c).

Through our earlier calculation, we observed that the expected value of the indicator random variable I_A equals the probability of the event A.
discrete random variables

Discrete random variables take on a countable number of distinct values. Examples include rolling a dice, where outcomes range from 1 to 6, or the number of heads in a series of coin tosses.

In probability theory, the key elements related to discrete random variables are:
  • Probability Mass Function (PMF): This function describes the probability of each possible value of the discrete random variable. If X is a discrete random variable taking on values x1, x2, ..., xk, then its PMF is P(X = xk).
  • Cumulative Distribution Function (CDF): The CDF gives the probability that the random variable X is less than or equal to some value x. It is defined as F(x) = P(X ≤ x).

The indicator random variable, I_A, is a type of discrete random variable with only two possible values: 0 and 1.

Consider how this impacts the expectation calculation:
  • When I_A = 1: This happens if event A occurs, and its probability is P(A).
  • When I_A = 0: This happens if event A does not occur, and its probability is P(A^c) = 1 - P(A).

So, the expected value of I_A becomes:\[E(I_A) = 0 \, P(I_A = 0) + 1 \, P(I_A = 1) = P(A)\].

In this way, understanding discrete random variables helps us see why the expectation of the indicator variable I_A is the probability of event A happening.

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Most popular questions from this chapter

In roulette, a wheel with 38 numbers is spun. Of these, 18 are red, and 18 are black. The other two numbers, which are neither black nor red, are 0 and \(00 .\) The probability that when the wheel is spun it lands on any particular number is 1\(/ 38\) . a) What is the probability that the wheel lands on a red number? b) What is the probability that the wheel lands on a black number twice in a row? c) What is the probability that the wheel lands on 0 or 00\(?\) do? d) What is the probability that in five spins the wheel never lands on either 0 or 00\(?\) e) What is the probability that the wheel lands on one of the first six integers on one spin, but does not land on any of them on the next spin?

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Use Chebyshev's inequality to find an upper bound on the probability that the number of tails that come up when a biased coin with probability of heads equal to 0.6 is tossed \(n\) times deviates from the mean by more than \(\sqrt{n}\) .

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