Problem 5
Question
Determine the expectation and variance of the \(\operatorname{Ber}(p)\) distribution.
Step-by-Step Solution
Verified Answer
The expectation is \( p \) and the variance is \( p(1-p) \).
1Step 1: Understanding Bernoulli Distribution
The Bernoulli distribution is a discrete probability distribution of a random variable which can take the value 1 with probability \( p \) and the value 0 with probability \( 1-p \). This type of distribution models the outcome of a single experiment that can result in one of two possible outcomes: success (1) or failure (0). Hence, only two parameters are involved: the probability \( p \) of success, and consequently \( 1-p \) is the probability of failure.
2Step 2: Write Down Expectation Formula
The expectation (mean) of a random variable \( X \) with a Bernoulli distribution \( \operatorname{Ber}(p) \) can be calculated using the formula: \( E(X) = \sum_{x} x \, P(X=x) \). Since the variable can only be 0 or 1, we focus on these two values.
3Step 3: Calculate Expectation
Substitute the values 0 and 1 into the expectation formula: \[E(X) = 0 \, \cdot \, P(X=0) + 1 \, \cdot \, P(X=1) = 0 \, \cdot \, (1-p) + 1 \, \cdot \, p = p\]Thus, the expectation \( E(X) \) for a \( \operatorname{Ber}(p) \) distribution is \( p \).
4Step 4: Write Down Variance Formula
The variance of a random variable \( X \) is given by the formula: \( \text{Var}(X) = E(X^2) - (E(X))^2 \). Since, for a Bernoulli distribution, \( E(X^2) = p \) because the possible values 0 and 1 satisfy \( X^2 = X \).
5Step 5: Calculate Variance
First calculate \( E(X^2) \):\[E(X^2) = 0^2 \, \cdot \, (1-p) + 1^2 \, \cdot \, p = p\]Then find the variance:\[\text{Var}(X) = E(X^2) - (E(X))^2 = p - p^2 = p(1-p)\]Thus, the variance \( \text{Var}(X) \) for a \( \operatorname{Ber}(p) \) distribution is \( p(1-p) \).
Key Concepts
Expectation of a Bernoulli DistributionVariance of a Bernoulli DistributionProbability Distribution
Expectation of a Bernoulli Distribution
The expectation, or mean, is a fundamental concept used to describe the average or central value of a random variable. For the Bernoulli distribution—denoted as \( \operatorname{Ber}(p) \)—this distribution is quite straightforward to understand. The Bernoulli distribution deals with a random event that can result in either a success or a failure. When we say success, it's simply our particular condition of interest turning out to be true, taking a value of \(1\). Any other outcome, not meeting the success condition, is a failure, taking a value of \(0\).
To compute the expectation of a Bernoulli-distributed random variable \(X\), we use the formula:
\[ E(X) = 0 \cdot (1-p) + 1 \cdot p = p \]
The expectation, \( E(X) \), is simply \( p \). It represents the probability of success across many trials. Think of it as the expected "average" value of results from a very large number of these binary trials.
Thus, if you repeat an experiment many times, with each trial being independent and having probability \( p \) of success, you'd expect, on average, successes to occur \( p \) proportion of the time.
To compute the expectation of a Bernoulli-distributed random variable \(X\), we use the formula:
\[ E(X) = 0 \cdot (1-p) + 1 \cdot p = p \]
The expectation, \( E(X) \), is simply \( p \). It represents the probability of success across many trials. Think of it as the expected "average" value of results from a very large number of these binary trials.
Thus, if you repeat an experiment many times, with each trial being independent and having probability \( p \) of success, you'd expect, on average, successes to occur \( p \) proportion of the time.
Variance of a Bernoulli Distribution
Variance is another essential concept that describes how much the values of a random variable deviate from the expectation (mean). For a Bernoulli distribution, variance helps us understand the variability or spread of the outcomes.
In the context of \( \operatorname{Ber}(p) \), the variance is derived using the formula:
\[ \text{Var}(X) = E(X^2) - (E(X))^2 \]
The calculation of \( E(X^2) \) is simple as it equals \( p \) because the possible outcomes are 0 and 1, and \( X^2 = X \). Therefore, the variance for a Bernoulli random variable \( X \) simplifies to:
\[ \text{Var}(X) = p - p^2 = p(1-p) \]
This result shows us how spread out or variable the outcomes are.
In the context of \( \operatorname{Ber}(p) \), the variance is derived using the formula:
\[ \text{Var}(X) = E(X^2) - (E(X))^2 \]
The calculation of \( E(X^2) \) is simple as it equals \( p \) because the possible outcomes are 0 and 1, and \( X^2 = X \). Therefore, the variance for a Bernoulli random variable \( X \) simplifies to:
\[ \text{Var}(X) = p - p^2 = p(1-p) \]
This result shows us how spread out or variable the outcomes are.
- If \(p\) is close to 0 or 1, the variance is low since outcomes are more predictable.
- If \(p\) is close to 0.5, there's a higher variance, representing more unpredictability in outcomes.
Probability Distribution
A probability distribution is a statistical function that describes all the possible values and their probabilities for a particular experiment. For a Bernoulli distribution, the focus is especially on simple experiments with two possible outcomes.
In the Bernoulli distribution \( \operatorname{Ber}(p) \), the probabilities are:
Each trial in this context is independent of the others and results in binary outcomes, making it a perfect model for many real-world scenarios, such as flipping a coin or a single yes/no survey question.
Additionally, knowing how to graphically represent a Bernoulli probability distribution on a bar chart can be insightful. The height of the bars represents probabilities: one bar at \( x = 0 \) with height \( 1-p \), and another at \( x = 1 \) with height \( p \).
Probability distributions, like the Bernoulli, are fundamental to understanding the likelihood of different outcomes in random scenarios, allowing us to make informed predictions and decisions based on statistical data.
In the Bernoulli distribution \( \operatorname{Ber}(p) \), the probabilities are:
- The probability of success (outcome 1) is \( p \).
- The probability of failure (outcome 0) is \( 1-p \).
Each trial in this context is independent of the others and results in binary outcomes, making it a perfect model for many real-world scenarios, such as flipping a coin or a single yes/no survey question.
Additionally, knowing how to graphically represent a Bernoulli probability distribution on a bar chart can be insightful. The height of the bars represents probabilities: one bar at \( x = 0 \) with height \( 1-p \), and another at \( x = 1 \) with height \( p \).
Probability distributions, like the Bernoulli, are fundamental to understanding the likelihood of different outcomes in random scenarios, allowing us to make informed predictions and decisions based on statistical data.
Other exercises in this chapter
Problem 3
For a certain random variable \(X\) it is known that \(\mathrm{E}[X]=2\), \(\operatorname{Var}(X)=3\). What is \(\mathrm{E}\left[X^{2}\right]\) ?
View solution Problem 4
Let \(X\) be a random variable with \(\mathrm{E}[X]=2\), \(\operatorname{Var}(X)=4\). Compute the expectation and variance of \(3-2 X\).
View solution Problem 6
The random variable \(Z\) has probability density function \(f(z)=3 z^{2} / 19\) for \(2 \leq z \leq 3\) and \(f(z)=0\) elsewhere. Determine \(\mathrm{E}[Z]\).
View solution Problem 7
Given is a random variable \(X\) with probability density function \(f\) given by \(f(x)=0\) for \(x1\), and \(f(x)=4 x-4 x^{3}\) for \(0 \leq x \leq 1\). Deter
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