Chapter 10: Problem 10
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be an independent trials process with
density
$$
f(x)=\frac{1}{2} e^{-|x|}, \quad-\infty
Short Answer
Expert verified
Mean = 0, Variance = 2; As \(n \to \infty\), \(S_n^*\) tends to normal distribution \(N(0,1)\), \(A_n\) converges to 0.
Step by step solution
01
Identify the probability distribution
The given density function \( f(x) = \frac{1}{2} e^{-|x|} \) is the probability density function (PDF) of a Laplace distribution, specifically \( X \sim \text{Laplace}(0, 1) \) where 0 is the location parameter (mean) and 1 is the scale parameter.
02
Find the mean of the Laplace distribution
The mean \( \mu \) of a Laplace distribution \( \text{Laplace}(0, b) \) is given by the location parameter. Hence, the mean of \( f(x) = \frac{1}{2} e^{-|x|} \) is 0.
03
Find the variance of the Laplace distribution
The variance \( \sigma^2 \) of a Laplace distribution \( \text{Laplace}(0, b) \) is \( 2b^2 \). For \( b=1 \), the variance is \( 2 \cdot 1^2 = 2 \).
04
Define and find the moment generating function (MGF) for \(X_1\)
For an \(X \sim \text{Laplace}(0, 1)\), the moment generating function is given by \( M_X(t) = \frac{1}{1-t^2} \), for \(-1 < t < 1\).
05
Find the moment generating function for \(S_n\)
Since \(S_n = X_1 + X_2 + \cdots + X_n\), and \(X_i\) are i.i.d., we have \(M_{S_n}(t) = (M_X(t))^n = \left(\frac{1}{1-t^2}\right)^n\) for \(-1 < t < 1\).
06
Find the moment generating function for \(A_n\)
\(A_n = \frac{S_n}{n}\), therefore \(M_{A_n}(t) = M_{S_n}(\frac{t}{n}) = \left(\frac{1}{1-\left(\frac{t}{n}\right)^2}\right)^n\).
07
Find the moment generating function for \(S_n^*\)
Since \(S_n^* = \frac{S_n - n\mu}{\sigma\sqrt{n}}\), where \(\mu = 0\) and \(\sigma^2 = 2\), we have \(M_{S_n^*}(t) = M_{S_n}\left(\frac{t}{\sqrt{2n}}\right) = \left(\frac{1}{1-\left(\frac{t}{\sqrt{2n}}\right)^2}\right)^n\).
08
Analyze the MGF of \(S_n^*\) as \(n \to \infty\)
As \(n \to \infty\), \(\frac{t}{\sqrt{2n}} \to 0\), so \(M_{S_n^*}(t) = e^{\frac{t^2}{2}}\), which implies \(S_n^*\) converges in distribution to a standard normal distribution by the Central Limit Theorem.
09
Analyze the MGF of \(A_n\) as \(n \to \infty\)
As \(n \to \infty\), \(M_{A_n}(t) \to e^{\frac{t^2}{2n}}\), which approximates to 1 for large \(n\). This thinly implies \(A_n\) converges to a point mass at the mean, 0, by the Law of Large Numbers.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Moment Generating Function
The moment generating function (MGF) helps us understand the characteristics of a probability distribution, such as the Laplace distribution we're working with. It's defined as a function that provides the moments of the distribution. For a random variable \(X\), the MGF is given by the expected value: \[ M_X(t) = E(e^{tX}) \]. When dealing with a Laplace distribution, specifically \(X \sim \text{Laplace}(0, 1)\), the moment generating function becomes: \[ M_X(t) = \frac{1}{1-t^2} \] for \(-1 < t < 1\). This function is crucial because it not only gives us insights into various moments like the mean and variance, but also helps connect to other distributions using transformations.
- The MGF exists for all \(t\) within the range \(-1 < t < 1\).
- It dramatically simplifies the approach to understanding the behavior of sums of random variables.
Variance
Variance is a measure of how much the values of a random variable spread out from the mean. For a distribution like the Laplace distribution, variance gives us an understanding of its dispersion. The Laplace distribution \( \text{Laplace}(0, b) \) has a variance given by: \[ \sigma^2 = 2b^2 \].
In our case, for a Laplace distribution \( \text{Laplace}(0, 1) \), the variance becomes \( 2 \cdot 1^2 = 2 \). Variance is key to understanding the volatility or risk associated with random variables in the distribution.
In our case, for a Laplace distribution \( \text{Laplace}(0, 1) \), the variance becomes \( 2 \cdot 1^2 = 2 \). Variance is key to understanding the volatility or risk associated with random variables in the distribution.
- It is always non-negative, as it represents spread.
- Variance is used in determining suitable models for prediction and analysis.
Mean
The mean is the 'average' of a probability distribution and provides a central value around which data points tend to cluster. In the context of a Laplace distribution, the mean is equivalent to the location parameter. For our specific Laplace distribution \( \text{Laplace}(0, 1) \), the mean is: \[ \mu = 0 \]. The mean serves as a balancing point of the distribution.
However, it's important to remember that while the mean gives a central tendency, it doesn't reflect the distribution's spread or other characteristics like skewness.
However, it's important to remember that while the mean gives a central tendency, it doesn't reflect the distribution's spread or other characteristics like skewness.
- The mean can sometimes be referred to as the expected value or average.
- It helps in comparing different distributions when evaluating performance or characteristics.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental statistical principle that describes how the distribution of the sum (or average) of a large number of independent, identically distributed random variables tends towards a normal distribution, regardless of the original distribution. In our exercise, for \(S_n^*\) (a scaled and centered version of \(S_n\)), we find as \(n \to \infty\): \[ M_{S_n^*}(t) = e^{\frac{t^2}{2}} \]. This aligns with the MGF of a standard normal distribution, implying \(S_n^*\) converges to a normal distribution due to the CLT.
- The CLT is powerful: it allows for approximations even when the original distribution isn't normal.
- CLT relies on the independence and large number of samples to be applicable.