Chapter 12: Problem 10
Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) are independent random variables with \(P\left(X_{i}>x\right)=e^{-2 x} .\) What can you say about \(\frac{1}{n} \sum_{i=1}^{n} X_{i}\) as \(n \rightarrow \infty ?\)
Short Answer
Expert verified
The sample mean \( \frac{1}{n} \sum_{i=1}^{n} X_{i} \) converges to \( \frac{1}{2} \) as \( n \to \infty \).
Step by step solution
01
Identify Distribution of Xi
The given probability statement \( P(X_i > x) = e^{-2x} \) suggests that the variable \( X_i \) follows an exponential distribution. Based on this, identify the parameter of the exponential distribution. For an exponential random variable \( X \) with rate \( \lambda \), \( P(X > x) = e^{-\lambda x} \), which implies \( \lambda = 2 \). Therefore, \( X_i \sim \text{Exponential}(2) \).
02
Calculate Expected Value and Variance
For an exponential distribution \( X \sim \text{Exponential}(\lambda) \), the mean (expected value) \( E[X] \) is \( \frac{1}{\lambda} \) and the variance \( \text{Var}(X) \) is \( \frac{1}{\lambda^2} \). With \( \lambda = 2 \), this gives \( E[X_i] = \frac{1}{2} \) and \( \text{Var}(X_i) = \frac{1}{4} \).
03
Identify the Sample Mean's Behavior
We are interested in the behavior of the sample mean \( \frac{1}{n} \sum_{i=1}^{n} X_{i} \) as \( n \to \infty \). The expected value of the sample mean is \( E\left[\frac{1}{n} \sum_{i=1}^{n} X_i\right] = \frac{1}{n} \sum_{i=1}^{n} E[X_i] = \frac{1}{2} \).
04
Use the Law of Large Numbers
According to the law of large numbers, the sample mean \( \frac{1}{n} \sum_{i=1}^{n} X_{i} \) will converge in probability to the expected value \( E[X_i] = \frac{1}{2} \) as \( n \to \infty \). This means that as the number of observations \( n \) increases, the sample mean will get arbitrarily close to \( \frac{1}{2} \).
05
Conclusion: Behavior of the Sample Mean
As \( n \to \infty \), the sample mean \( \frac{1}{n} \sum_{i=1}^{n} X_{i} \) converges in probability to \( \frac{1}{2} \). This is a result of each \( X_i \) being an exponential random variable with mean \( \frac{1}{2} \) and by the application of the law of large numbers.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Distribution
An exponential distribution is a continuous probability distribution used to model the time or space between events in a process where events occur continuously and independently at a constant rate.
- For a random variable following an Exponential distribution with rate parameter \( \lambda \), the probability that it takes a value greater than \( x \) is given by: \[ P(X > x) = e^{-\lambda x} \]
- This suggests the random variable \( X \) has an exponential distribution with rate \( \lambda \).
Expected Value
The expected value, often called the mean, of a random variable gives a measure of the central tendency or "average" of the probability distribution.
- For exponential distribution with rate \( \lambda \), the expected value \( E[X] \) is calculated as:\[ E[X] = \frac{1}{\lambda} \]
- In the given exercise, since \( \lambda = 2 \), the expected value of each \( X_i \) is:\[ E[X_i] = \frac{1}{2} \]
Variance
Variance is a key concept in probability and statistics, providing insight into the spread or dispersion of a set of values. It is calculated for a random variable to describe how much the values differ from the expected value.
- For an exponentially distributed random variable \( X \) with rate \( \lambda \), the variance is given by:\[ \text{Var}(X) = \frac{1}{\lambda^2} \]
- Given \( \lambda = 2 \) in our exercise, the variance of each \( X_i \) is:\[ \text{Var}(X_i) = \frac{1}{4} \]