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For the proof of Theorem 4.8.1, we assumed that the cdf was strictly increasing over its support. Consider a random variable \(X\) with cdf \(F(x)\) that is not strictly increasing. Define as the inverse of \(F(x)\) the function $$ F^{-1}(u)=\inf \\{x: F(x) \geq u\\}, \quad 0

Short Answer

Expert verified
The random variable \( F^{-1}(U) \), where \( U \) is uniformly distributed between 0 and 1, has a cumulative distribution function \( F(x) \), as per the steps outlined above.

Step by step solution

01

Understand the Problem

We are to prove that the random variable \( F^{-1}(U) \), where \( U \) is uniformly distributed between 0 and 1, has a cumulative distribution function \( F(x) \). Basically, we want to show that applying the inverse CDF of \( F(x) \) to a uniform distribution results in \( F(x) \) being the CDF of the resultant random variable.
02

Define CDF of Random Variable \( F^{-1}(U) \)

Let's denote \( Z = F^{-1}(U) \). Then compute the CDF of \( Z \), \( G(z) = P(Z ≤ z) \). We need to compute this to show that it equals \( F(x) \).
03

Compute CDF of Random Variable \( F^{-1}(U) \)

We have \( G(z) = P(Z ≤ z) = P(F^{-1}(U) ≤ z) = P(U ≤ F(z)) \). Since \( U \) follows uniform distribution over (0,1), the \( P(U ≤ F(z)) = F(z) \). Thus, \( G(z) = F(z) \).
04

Statement of Conclusion

From step 3, we find that the cumulative distribution function of the random variable \( F^{-1}(U) \) equals \( F(x) \). Thus we've shown that \( Z = F^{-1}(U) \) has cumulative distribution function \( F(x) \).

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Key Concepts

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

Cumulative Distribution Function
The cumulative distribution function (CDF) is a fundamental concept in probability and statistics that describes the probability that a random variable will take a value less than or equal to a certain point.

First and foremost, to comprehend the CDF, it's crucial to recognize that it's a non-decreasing function which levels off to 1 as we move to the right along the x-axis. Its primary job is to encapsulate the distribution of probabilities in one neat formula or graph.

For a random variable (X), the CDF is represented as (F(x)), defined by the probability (P(X leq x)). This function is crucial for understanding the behavior of random variables, because it tells us the chance of the variable falling within a certain range.

In our exercise, we explored the inverse CDF, which, rather than taking a value of (X) and giving us the cumulative probability up to that point, it takes a probability (between 0 and 1) and provides us with the corresponding value of (X). The importance of the CDF extends to various applications like statistical inference and hypothesis testing.
Uniform Distribution
A uniform distribution, in the context of probability, refers to a type of probability distribution where all outcomes are equally likely to occur within the defined range.

When we speak of a continuous uniform distribution, we generally indicate a distribution over an interval, say (0,1). The key property here is that the probability is constant at any point in that interval—it's uniform.

For a uniform distribution from (0,1), any number between 0 and 1 is as likely as any other number. Its CDF is beautifully linear over the support, increasing from 0 to 1 as (x) ranges from 0 to 1, which is used in the exercise to derive results about the inverse CDF.
Random Variables
Random variables are the linchpins of probability theory. They are essentially variables whose values are determined by the outcomes of random phenomena.

There are two types of random variables: discrete, which can take a countable number of distinct outcomes, and continuous, which can take infinitely many values.

Our focus in these problems is usually on continuous random variables, since we're deriving and applying CDFs which require an interval of numbers. The random variable (Z) in the exercise is defined by applying the inverse CDF to a uniform distribution, representing a transformation from the uniform random variable (U) to a new variable with its own distribution.
Probability
Probability is a measure of the likelihood that a particular event will occur. It's a value between 0 and 1, with 0 indicating the event will never occur and 1 indicating certainty.

It all comes down to assigning numbers to events based on how likely they are, and these numbers possess properties that allow us to compute probabilities for complex events based on simpler ones.

The concept of probability is inherent in the definition of the CDF which gives us the probability that our random variable (X) is less than or equal to some value. In the given exercise, we dealt with a specific probability, namely the uniformly distributed (U), where every value between 0 and 1 had an equal chance of occurring. This becomes the basis for deriving the desired CDF of our new random variable.

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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 \(f(x)=\frac{1}{6}, x=1,2,3,4,5,6\), 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?

Consider the sample of data (data are in the file ex4.4.3data.rda): \(\begin{array}{rrrrrrrrrrr}13 & 5 & 202 & 15 & 99 & 4 & 67 & 83 & 36 & 11 & 301 \\ 23 & 213 & 40 & 66 & 106 & 78 & 69 & 166 & 84 & 64 & \end{array}\) (a) Obtain the five-number summary of these data. (b) Determine if there are any outliers. (c) Boxplot the data. Comment on the plot.

Using Exercise \(3.3 .22\), show that $$ \int_{0}^{p} \frac{n !}{(k-1) !(n-k) !} z^{k-1}(1-z)^{n-k} d z=\sum_{w=k}^{n}\left(\begin{array}{l} n \\ w \end{array}\right) p^{w}(1-p)^{n-w} $$ where \(0

Consider the problem from genetics of crossing two types of peas. The Mendelian theory states that the probabilities of the classifications (a) round and yellow, (b) wrinkled and yellow, (c) round and green, and (d) wrinkled and green are \(\frac{9}{16}, \frac{3}{16}, \frac{3}{16}\), and \(\frac{1}{16}\), respectively. If, from 160 independent observations, the observed frequencies of these respective classifications are \(86,35,26\), and 13, are these data consistent with the Mendelian theory? That is, test, with \(\alpha=0.01\), the hypothesis that the respective probabilities are \(\frac{9}{16}, \frac{3}{16}, \frac{3}{16}\), and \(\frac{1}{16}\).

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