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Let \(f(x)=\frac{1}{6}, x=1,2,3,4,5,6\), zero elsewhere, be the pmf of a distribution of the discrete type. Show that the pmf of the smallest observation of a random sample of size 5 from this distribution is $$ g_{1}\left(y_{1}\right)=\left(\frac{7-y_{1}}{6}\right)^{5}-\left(\frac{6-y_{1}}{6}\right)^{5}, \quad y_{1}=1,2, \ldots, 6 $$ zero elsewhere. Note that in this exercise the random sample is from a distribution of the discrete type. All formulas in the text were derived under the assumption that the random sample is from a distribution of the continuous type and are not applicable. Why?

Short Answer

Expert verified
The given function, \(g_{1}\left(y_{1}\right)=\left(\frac{7-y_{1}}{6}\right)^{5}-\left(\frac{6-y_{1}}{6}\right)^{5}\), for \(y_{1}=1,2, \ldots, 6\), is indeed the probability mass function of the smallest observation in a random sample of size 5 from the Discrete Uniform Distribution \(f(x)\). It can be derived by calculating the number of outcomes where the smallest observation is \(y_1\). The formulas for a continuous type distribution cannot be used here because the distribution is discrete, meaning the outcomes can only take the specific values 1-6.

Step by step solution

01

Understand the given distribution

The given discrete distribution function, \(f(x)\), is defined as \(\frac{1}{6}\) for \(x\) equals 1 through 6, and zero elsewhere. This represents a Discrete Uniform Distribution, where all outcomes are equally probable within the given range (1 to 6).
02

Understand the question

The task is to find the probability mass function (pmf) of the smallest observation in a random sample of size 5 from this distribution. You are given this pmf \(g_1(y_1)\) and are required to show that it indeed represents the pmf of the smallest observation of a randomly drawn sample of size 5.
03

Explanation of the pmf of the smallest observation

For pmf of the smallest observation \(y_1\) in a sample of size 5, the smallest value can be from 1 to 6. Each observation in the sample space is greater or equal to \(y_1\) and less than \(y_1+1\). Because all outcomes are equally likely in the given distribution, the probability is the total number of outcomes where the smallest observation is \(y_1\). Which means, all these outcomes divided by the total number of all outcomes in sample space. The pmf is then \(g_{1}(y_{1})= \frac{(7-y_{1})^5 - (6-y_{1})^5}{6^5}\), for \(y_1 = 1, 2, 3, 4, 5, 6\) and zero elsewhere.
04

Understanding the relationship between the given distribution and the computed pmf

The computed result \(g_{1}(y_{1})= \frac{(7-y_{1})^5 - (6-y_{1})^5}{6^5}\) simplifies to \(g_{1}(y_{1})= (1 - y_{1}/6)^5 - (1 - (y_{1}+1)/6)^5\), for \(y_1 =1,2,3,4,5,6\), which matches the given pmf. The formula derived in Step 3 is the pmf of the smallest observation from a random sample of size 5 from the given Discrete Uniform Distribution.
05

Explaining why a different formula cannot be used

The difference between a discrete and a continuous type distribution lies in their possible outcomes. Discrete distributions represent situations where the outcomes can only take specific, separate values (like throwing dice), while continuous distributions can take any value in a given range. Therefore, formulas derived for continuous type distributions, involve integrals and cannot be applied here because our outcomes cannot take any value within a range; they can only take the specific six values.

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Key Concepts

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

Discrete Uniform Distribution
When we flip a fair coin or roll a fair six-sided die, each outcome has the same chance of occurring. This is the essence of a Discrete Uniform Distribution, a core concept in probability and statistics that deals with events where all possible outcomes are equally likely.

In a Discrete Uniform Distribution, the probability mass function (pmf) assigns the same probability to each of the finite number of outcomes. When rolling a fair die, the outcomes are discrete, meaning they can only be one of the six whole numbers: 1, 2, 3, 4, 5, or 6. In our example, with a die roll, the pmf assigns a probability of \(\frac{1}{6}\) to each outcome since there are six possible outcomes and each is equally likely.

The uniformity of this distribution simplifies finding probabilities for different events because each outcome has an identical weight. However, understanding this concept is crucial as it sets the foundation for more complex probability problems, including finding the distribution of a sample statistic like the smallest observation in a random sample.
Random Sample
Taking a random sample is akin to drawing names from a hat; it's a fundamental method in statistics to ensure that every possible subset of the population has an equal chance of being selected. When we refer to taking a 'random sample' of size 5 from a distribution, it means we randomly choose 5 elements, independently and with equal probability, from the set of outcomes.

For our six-sided die, a random sample of size 5 could look like this: 3, 1, 4, 6, 2. It's important to note that each sample drawn is not affected by the previous one, meaning each draw is independent of others, thus maintaining the discrete uniform nature of the distribution.

A random sample helps in making inferences about the entire population without bias. For instance, we can estimate the likelihood of the smallest value in our sample, which is what the exercise is centered around. Understanding how to construct a pmf for a random sample is a vital skill in any statistics exercise.
Statistics Exercise
In a statistics exercise, the objective is to apply probabilistic concepts to solve real-world scenarios. Through step-by-step analysis, one can dissect complex problems into manageable parts. To illustrate, the task might be to find the pmf of the smallest observation in a random sample from a Discrete Uniform Distribution. The exercise demands comprehension of the distribution, sample selection, and underlying probability principles.

Improving your approach to these exercises involves recognizing the type of distribution you are working with, whether it is discrete or continuous, as the strategies for each can differ greatly. Emphasizing critical thinking, you must challenge assumptions and verify whether the standard formulas apply or if the situation requires a more tailored equation, as seen in our example where continuous distributions formulas are inapplicable.

Such exercises not only enhance your understanding of statistical methodologies but also refine your problem-solving skills, which can extend far beyond the realm of statistics.

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Most popular questions from this chapter

Consider the following permutation test for the two-sample problem with hypotheses \((4.9 .7) .\) Let \(\mathbf{x}^{\prime}=\left(x_{1}, x_{2}, \ldots, x_{n_{1}}\right)\) and \(\mathbf{y}^{\prime}=\left(y_{1}, y_{2}, \ldots, y_{n_{2}}\right)\) be the realizations of the two random samples. The test statistic is the difference in sample means \(\bar{y}-\bar{x} .\) The estimated \(p\) -value of the test is calculated as follows: 1\. Combine the data into one sample \(\mathbf{z}^{\prime}=\left(\mathbf{x}^{\prime}, \mathbf{y}^{\prime}\right)\). 2\. Obtain all possible samples of size \(n_{1}\) drawn without replacement from \(\mathrm{z}\). Each such sample automatically gives another sample of size \(n_{2}\), i.e., all elements of \(\mathbf{z}\) not in the sample of size \(n_{1}\). There are \(M=\left(\begin{array}{c}n_{1}+n_{2} \\ n_{1}\end{array}\right)\) such samples. 3\. For each such sample \(j\) : (a) Label the sample of size \(n_{1}\) by \(\mathbf{x}^{*}\) and label the sample of size \(n_{2}\) by \(\mathbf{y}^{*}\). (b) Calculate \(v_{j}^{*}=\bar{y}^{*}-\bar{x}^{*}\). 4\. The estimated \(p\) -value is \(\hat{p}^{*}=\\#\left\\{v_{j}^{*} \geq \bar{y}-\bar{x}\right\\} / M\). (a) Suppose we have two samples each of size 3 which result in the realizations: \(\mathbf{x}^{\prime}=(10,15,21)\) and \(\mathbf{y}^{\prime}=(20,25,30)\). Determine the test statistic and the permutation test described above along with the \(p\) -value. (b) If we ignore distinct samples, then we can approximate the permutation test by using the bootstrap algorithm with resampling performed at random and without replacement. Modify the bootstrap program boottesttwo.s to do this and obtain this approximate permutation test based on 3000 resamples for the data of Example \(4.9 .2 .\) (c) In general, what is the probability of having distinct samples in the approximate permutation test described in the last part? Assume that the original data are distinct values.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(\Gamma(1, \beta)\) distribution. (a) Show that the confidence interval \(\left(2 n \bar{X} /\left(\chi_{2 n}^{2}\right)^{(1-(\alpha / 2))}, 2 n \bar{X} /\left(\chi_{2 n}^{2}\right)^{(\alpha / 2)}\right)\) is an exact \((1-\alpha) 100 \%\) confidence interval for \(\beta\). (b) Using part (a), show that the \(90 \%\) confidence interval for the data of Example \(4.9 .1\) is \((64.99,136.69)\)

Verzani (2014), page 323 , presented a data set concerning the effect that different dosages of the drug AZT have on patients with HIV. The responses we consider are the p24 antigen levels of HIV patients after their treatment with AZT. Of the \(20 \mathrm{HIV}\) patients in the study, 10 were randomly assign the dosage of \(300 \mathrm{mg}\) of AZT while the other 10 were assigned \(600 \mathrm{mg}\). The hypotheses of interest are \(H_{0}: \Delta=0\) versus \(H_{1}: \Delta \neq 0\) where \(\Delta=\mu_{600}-\mu_{300}\) and \(\mu_{600}\) and \(\mu_{300}\) are the true mean p24 antigen levels under dosages of \(600 \mathrm{mg}\) and \(300 \mathrm{mg}\) of AZT, respectively. The data are given below but are also available in the file aztdoses. rda. \begin{tabular}{|l|llllllllll|} \hline \(300 \mathrm{mg}\) & 284 & 279 & 289 & 292 & 287 & 295 & 285 & 279 & 306 & 298 \\ \hline \(600 \mathrm{mg}\) & 298 & 307 & 297 & 279 & 291 & 335 & 299 & 300 & 306 & 291 \\ \hline \end{tabular} (a) Obtain comparison boxplots of the data. Identify outliers by patient. Comment on the comparison plots. (b) Compute the two-sample \(t\) -test and obtain the \(p\) -value. Are the data significant at the \(5 \%\) level of significance? (c) Obtain a point estimate of \(\Delta\) and a \(95 \%\) confidence interval for it. (d) Conclude in terms of the problem.

Let \(Y_{1}

It is known that a random variable \(X\) has a Poisson distribution with parameter \(\mu\). A sample of 200 observations from this distribution has a mean equal to \(3.4\). Construct an approximate \(90 \%\) confidence interval for \(\mu\).

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