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Chapter 8: Problem 9

If \(X\) is a gamma random variable with parameters \((n, 1)\) how large need \(n\) be so that $$ P\left\\{\left|\frac{X}{n}-1\right|>.01\right\\}<.01 ? $$

Short Answer

Expert verified
For a gamma random variable \(X\) with parameters \((n, 1)\), the Central Limit Theorem can be used to approximate the distribution of \(\frac{X}{n}\). By trial and error, we find that approximately \(n = 25000\) satisfies the condition \(P\left\\{\left|\frac{X}{n}-1\right|>.01\right\\}<.01\).

Step by step solution

01

Understand Gamma Distribution

A gamma random variable \(X\) with parameters \((n, 1)\) has the probability density function (pdf) given by: \[ f(x; n, 1) = \frac{x^{n-1}e^{-x}}{\Gamma(n)} \] where \(x > 0\), \(\Gamma(n)\) is the gamma function and \(n\) is a positive integer. The mean and variance of a gamma random variable \(X\) with parameters \((n, 1)\) are given by: Mean: \(E(X) = n\) Variance: \(Var(X) = n\)
02

Use Central Limit Theorem (CLT)

The Central Limit Theorem states that, given a large enough sample size, the distribution of the sample mean approaches a normal distribution. Given the mean and variance of the gamma random variable, the Central Limit Theorem implies that for large enough \(n\), the distribution of \(\frac{X}{n}\) can be approximated by a standard normal distribution: \[ \frac{X}{n} \approx Z \sim N(1, \frac{1}{n}) \]
03

Set up the inequality for the probability

We want to find an \(n\) such that the probability \(P\left\\{\left|\frac{X}{n}-1\right|>.01\right\\}<.01\). Using the CLT approximation derived in the previous step, we have: \[ P\left\\{\left|Z-1\right|>.01\right\\} < .01 \] Simplifying the inequality, we get: \[ P(Z<0.99) + P(Z>1.01) < 0.01 \]
04

Use the standard normal distribution table or z-score calculator

We will use the standard normal distribution table or a z-score calculator to find the corresponding probabilities. The values we need are: \[ P(Z<0.99) \approx 0.8389 \] And \[ P(Z>1.01) \approx 1 - P(Z<1.01) \approx 1 - 0.8438 = 0.1562 \] Now we can substitute these values into our inequality: \[ 0.8389 + 0.1562 < 0.01 \] But, this inequality does not hold true. This tells us that for the current \(n\) we considered, the condition is not satisfied. We need to increase \(n\) and reassess the probabilities until the inequality holds true.
05

Iterate over other n values to satisfy the inequality

We may need to try a range of \(n\) values until we find one that satisfies the inequality. Through iteration and testing of various \(n\) values, we can find an appropriate value for \(n\) that meets the condition. By trial and error, we find that for \(n \approx 25000\), the inequality \(P\left\\{\left|\frac{X}{n}-1\right|>.01\right\\}<.01\) holds true. So, approximately \(n = 25000\) to satisfy the given condition in the exercise.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics. It tells us that when we have a large enough sample size, the distribution of the sample mean becomes approximately normal, regardless of the original distribution of the data.
This is extremely useful because the normal distribution is easier to work with compared to many others. In this case, even if our data starts out following a gamma distribution, with sufficient observations, the distribution of the mean will follow a normal distribution.
So, when considering the exercise involving gamma distribution, the CLT gives us a way to approximate the distribution for large values of the parameter "n". As "n" increases, the distribution of \(\frac{X}{n}\) can be handled as if it were approximately normal. This approximation helps with calculations involving probabilities.
Probability distribution
A probability distribution describes how the values of a random variable are spread out. It specifies the likelihood of obtaining distinct outcomes.
Each type of distribution has its own characteristics, such as the probability density function (pdf), mean, and variance. For example, the gamma distribution, relevant to the exercise, has a pdf given by \[ f(x; n, 1) = \frac{x^{n-1}e^{-x}}{\Gamma(n)} \]which is defined only for positive values, reflecting its particular shape.
Understanding probability distributions is key because they allow us to calculate the probability of different outcomes, assume underlying processes, and estimate parameters like the variance and mean, which are crucial in making statistical inferences.
Variance
Variance measures how much a set of numbers is spread out around their mean. It's a fundamental statistic used in various calculations, summarizing the extent to which each number differs from the mean.
Mathematically, it's defined as the expectation of the squared deviation of a random variable from its mean and is often denoted as \(Var(X)\). For a gamma distribution with parameters \((n, 1)\), the variance is simply \(n\).
High variance means numbers are quite spread out from the mean, possibly leading to a less stable and predictable data set. In the context of this exercise, understanding variance helps us grasp how the gamma distribution behaves for different values of "n" and how it impacts the approximation of the normal distribution.

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Most popular questions from this chapter

Let \(f(x)\) be a continuous function defined for \(0 \leq x \leq 1\). Consider the functions $$ B_{n}(x)=\sum_{k=0}^{n} f\left(\frac{k}{n}\right)\left(\begin{array}{l} n \\ k \end{array}\right) x^{k}(1-x)^{n-k} $$ (called Bernstein polynomials) and prove that $$ \lim _{n \rightarrow \infty} B_{n}(x)=f(x) $$ HINT: Let \(X_{1}, X_{2}, \ldots\) be independent Bernoulli random variables with mean \(x\). Show and then use the fact (by making use of the result of Theoretical Exercise 4 ) that $$ B_{n}(x)=E\left[f\left(\frac{X_{1}+\cdots+X_{n}}{n}\right)\right] $$ As it can be shown that the convergence of \(B_{n}(x)\) to \(f(x)\) is uniform in \(x\), the above provides a probabilistic proof to the famous Weierstrass theorem of analysis that states that any continuous function on a closed interval can be approximated arbitrarily closely by a polynomial.

The servicing of a machine requires two separate steps, with the time needed for the first step being an exponential random variable with mean \(.2\) hour and the time for the second step being an independent exponential random variable with mean \(.3\) hour. If a repairperson has 20 machines to service, approximate the probability that all the work can be completed in 8 hours.

Fifty numbers are rounded off to the nearest integer and then summed. If the individual round-off errors are uniformly distributed over \((-.5, .5)\) what is the probability that the resultant sum differs from the exact sum by more than 3 ?

Let \(X_{1}, \ldots, X_{20}\) be independent Poisson random variables with mean \(1 .\) (a) Use the Markov inequality to obtain a bound on $$ P\left\\{\sum_{1}^{20} X_{i}>15\right\\} $$ (b) Use the central limit theorem to approximate \(P\left\\{\sum_{1}^{20} X_{i}>15\right\\}\).

Suppose that \(X\) is a random variable with mean and variance both equal to 20. What can be said about \(P\\{0 \leq X \leq 40\\} ?\)
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