Chapter 5: Problem 17
Let \(\bar{X}_{n}\) denote the mean of a random sample of size \(n\) from a
distribution that has pdf \(f(x)=e^{-x}, 0
Short Answer
Expert verified
The limiting distribution of \(Y_n\) as \(n\) tends to infinity is a standard normal distribution i.e. \(N(0,1)\).
Step by step solution
01
Compute \(\bar{X}_{n}\)'s mgf
For an exponentially distributed random variable \(X\) with parameter 1, the moment generating function \(M_X(t)\) is given by: \[M_X(t) = \int_0^{\infty} e^{tx} f(x) dx = \int_0^{\infty} e^{tx} e^{-x} dx = \int_0^{\infty} e^{x(t-1)} dx, t<1\] Now, for \(\bar{X}_{n}\), it's moment generating function \(M_{\bar{X}_n}(t)\) can be derived from the MGF of the exponential distribution as \(M_{\bar{X}_n}(t) = (M_X(t/n))^n = (1/(1-t/n))^n = (1/(1/n -t))^n =[(1-t/n)^{-1}]^n\)
02
Compute \(Y_n\)'s mgf
The given random variable is \(Y_{n}=\sqrt{n}\left(\bar{X}_{n}-1\right)\). The moment generating function for a linear transformation of a random variable, say \(aX+b\), is \(M_{aX+b}(t) = e^{bt}M_X(at)\). So, for \(Y_n\), It's moment generating function \(M_{Y_n}(t)\) will be \(M_{Y_n}(t) = e^{-t\sqrt{n}}M_{\bar{X}_n}(t\sqrt{n}) = e^{-t\sqrt{n}}[(1-t/\sqrt{n})^{-1}]^n = \left[e^{t / \sqrt{n}}-(t / \sqrt{n}) e^{t / \sqrt{n}}\right]^{-n}\)
03
Find Limiting Distribution
As \(n\) approaches infinity, the function \(M_{Y_n}(t)\) will converge to a certain value given \(-\sqrt{n}\infty} M_{Y_n}(t) = \lim_{n->\infty}e^{-t\sqrt{n}}[(1-t/\sqrt{n})^{-1}]^n = \lim_{n->\infty} \left[e^{t / \sqrt{n}}-(t / \sqrt{n}) e^{t / \sqrt{n}}\right]^{-n}\] After careful observation, we see that the limit is in the form of 0/0 when \(n->\infty\), which is undefined. We can use L'Hopital's rule here to get an appropriate value for the limit. Upon applying L'Hopital's rule and simplifying, we see that as \(n->\infty\) the limiting distribution of \(Y_n\) is a standard normal distribution i.e. \(N(0,1)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Moment Generating Function (mgf)
The moment generating function, commonly known as mgf, is a powerful tool in probability theory. It's a function that helps to characterize all the moments of a random variable. In simple terms, it acts like a "fingerprint" for the distribution.
For an exponential distribution with a rate parameter of 1, the mgf is given by \( M_X(t) = \int_0^{\infty} e^{tx} e^{-x} dx \). Simplifying it, you get \( M_X(t) = \frac{1}{1-t} \), valid for \( t < 1 \). The mgf provides us with easy access to obtain moments by simply differentiating the function with respect to \( t \).
When dealing with sample means, the mgf for \( \bar{X}_n \), the mean of a sample of \( n \) independent and identically distributed exponential random variables, is \((M_X(t/n))^n\). It gives \((1 - t/n)^{-n}\), because the mgf of each random variable is scaled by \(1/n\) for taking the average of the whole sample.
For an exponential distribution with a rate parameter of 1, the mgf is given by \( M_X(t) = \int_0^{\infty} e^{tx} e^{-x} dx \). Simplifying it, you get \( M_X(t) = \frac{1}{1-t} \), valid for \( t < 1 \). The mgf provides us with easy access to obtain moments by simply differentiating the function with respect to \( t \).
When dealing with sample means, the mgf for \( \bar{X}_n \), the mean of a sample of \( n \) independent and identically distributed exponential random variables, is \((M_X(t/n))^n\). It gives \((1 - t/n)^{-n}\), because the mgf of each random variable is scaled by \(1/n\) for taking the average of the whole sample.
Limiting Distribution
The limiting distribution describes the behavior of a sequence of random variables as the number of samples, \( n \), approaches infinity. It's a crucial concept in probability and statistics because it allows us to understand what happens to sample-based measures when the sample size is very large.
In this exercise, you see the role of limiting distribution when computing \( Y_n = \sqrt{n}(\bar{X}_n - 1) \). For large \( n \), the mgf of \( Y_n \) approaches the mgf of another well-known distribution. By evaluating the behavior of \( M_{Y_n}(t) \) as \( n \) goes to infinity, we find that it converges to a specific form, indicating the distribution \( Y_n \) starts to resemble.
Thus, the limiting distribution essentially forms the foundation for the Central Limit Theorem, which states that sample means will tend toward a normal distribution, regardless of the original distribution of the data, given a sufficiently large \( n \).
In this exercise, you see the role of limiting distribution when computing \( Y_n = \sqrt{n}(\bar{X}_n - 1) \). For large \( n \), the mgf of \( Y_n \) approaches the mgf of another well-known distribution. By evaluating the behavior of \( M_{Y_n}(t) \) as \( n \) goes to infinity, we find that it converges to a specific form, indicating the distribution \( Y_n \) starts to resemble.
Thus, the limiting distribution essentially forms the foundation for the Central Limit Theorem, which states that sample means will tend toward a normal distribution, regardless of the original distribution of the data, given a sufficiently large \( n \).
Standard Normal Distribution
The standard normal distribution is a probability distribution with a mean of 0 and a standard deviation of 1. This is denoted as \( N(0,1) \).
It's a cornerstone of statistics because it simplifies the process of computing probabilities and estimating parameters. When random variables are converted to this format through **standardization**, comparisons are made easier across different datasets.
Within this particular problem, as \( n \) approaches infinity, the random variable \( Y_n \) converges to a standard normal distribution. Essentially, the expression \( \sqrt{n}(\bar{X}_n - 1) \) symbolizes how far the sample mean is from the true mean in standardized form. It's this standardization which makes it culminate into \( N(0,1) \) when \( n \) is large enough, by virtue of the Central Limit Theorem.
It's a cornerstone of statistics because it simplifies the process of computing probabilities and estimating parameters. When random variables are converted to this format through **standardization**, comparisons are made easier across different datasets.
Within this particular problem, as \( n \) approaches infinity, the random variable \( Y_n \) converges to a standard normal distribution. Essentially, the expression \( \sqrt{n}(\bar{X}_n - 1) \) symbolizes how far the sample mean is from the true mean in standardized form. It's this standardization which makes it culminate into \( N(0,1) \) when \( n \) is large enough, by virtue of the Central Limit Theorem.