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Illustrative Example \(8.2 .1\) of this section dealt with a random sample of size \(n=2\) from a gamma distribution with \(\alpha=1, \beta=\theta .\) Thus the mgf of the distribution is \((1-\theta t)^{-1}, t<1 / \theta, \theta \geq 2 .\) Let \(Z=X_{1}+X_{2}\). Show that \(Z\) has a gamma distribution with \(\alpha=2, \beta=\theta .\) Express the power function \(\gamma(\theta)\) of Example \(8.2 .1\) in terms of a single integral. Generalize this for a random sample of size \(n .\)

Short Answer

Expert verified
Sum \(Z\) of two i.i.d. random variables from a gamma distribution with parameters \(\alpha=1\), \(\beta=\theta\) also has a gamma distribution with parameters \(\alpha=2\), \(\beta=\theta\). The power function can be expressed as \(\gamma(\theta)=1- \int_{c/\theta}^{\infty} u e^{-u} du\). For a random sample of size \(n\), the sum \(S\) follows a gamma distribution with \(\alpha=n\), \(\beta=\theta\), and the power function takes a similar form with the integral involving the gamma function.

Step by step solution

01

Write down the mgf of \(X_1+X_2\)

By independence, the mgf of Z, \(M_Z(t)\), is simply the product of the mgfs of \(X_1\) and \(X_2\), which are both \((1- \theta t)^{-1}\). Thus, \(M_Z(t) = (1-\theta t)^{-2}, t < 1 / \theta\) .
02

Identify the gamma distribution

This is the mgf of a gamma distribution with \(\alpha=2\) and \(\beta=\theta\). Thus \(Z=X_1+X_2\) has a gamma distribution with \(\alpha=2\), \(\beta=\theta.\)
03

Express the power function \(\gamma(\theta)\) in terms of a single integral

The power function from Example \(8.2.1\) is \(\gamma(\theta)=1- \int_{c}^{\infty} \frac{z e^{-z / \theta}}{\theta^2}dz\). We can substitute \(u=z/\theta\), which implies \(z=\theta u\) and \(dz=\theta du\). The new lower bound is \(c/\theta\) and the new upper bound is \(\infty\), so the integral becomes \(\gamma(\theta)=1- \int_{c/\theta}^{\infty} u e^{-u} du\).
04

Generalize for a random sample

If we consider a sample of size \(n\), the sum \(S=X_{1}+X_{2}+\dots+X_{n}\) will have a gamma distribution with \(\alpha=n\), \(\beta=\theta\). The power function can be expressed in terms of a single integral similar to the above procedure. It will involve the gamma function \(\Gamma(\alpha, z)\), which is a generalization of the factorial function to complex numbers. The lower bound of the integral in the power function will be \(c/\theta\), where \(c\) is the critical region depending on the null hypothesis and alternative in the testing problem.

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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, or MGF, is a powerful tool in probability theory. It helps describe the distribution of a random variable by generating moments. In simpler terms, the MGF helps us understand the characteristics of our variable, like mean and variance.

For a random variable \(X\), the MGF is defined as \(M_X(t) = \mathbb{E}[e^{tX}]\). Here, \(\mathbb{E}\) denotes the expected value. What's special about the MGF is that each coefficient of \(t^k\) in its expansion represents the \(k\)-th moment of \(X\).

In the given exercise, because the variables \(X_1\) and \(X_2\) are independent, we can find the MGF of their sum \(Z = X_1 + X_2\) as the product of their MGFs. Thus, \(M_Z(t) = (1-\theta t)^{-1} \times (1-\theta t)^{-1} = (1-\theta t)^{-2}\). This expression is crucial for identifying the distribution of \(Z\).
Power Function
The power function, often denoted as \(\gamma(\theta)\), helps in hypothesis testing by determining the power of a statistical test. The power of a test signifies the probability that the test will correctly reject a false null hypothesis.

In Example 8.2.1, the power function is initially presented as an integral. We have \(\gamma(\theta) = 1 - \int_{c}^{\infty} \frac{ze^{-z/\theta}}{\theta^2} dz\). By transforming variables, this can be simplified to an integral of the form \(\gamma(\theta) = 1 - \int_{c/\theta}^{\infty} ue^{-u} du\). This transformation simplifies evaluating the power function because it uses standard integral forms that can be easily computed or looked up.
Random Sample
When dealing with probability, a random sample is a crucial concept that refers to a collection of random variables each drawn independently from the same probability distribution.For the given problem, we're looking at the case where the random sample consists of \(n\) variables drawn from a gamma distribution.

When considering such a sample, the sum \(S = X_1 + X_2 + \dots + X_n\) becomes a very interesting quantity. When all \(X_i's\) are from a gamma distribution with parameters \(\alpha = 1\) and \(\beta = \theta\), the sum \(S\) also follows a gamma distribution, except with parameters \(\alpha = n\) and the same \(\beta = \theta\). This result is quite significant because it allows us to extend our insights from a smaller sample to larger samples efficiently.
Gamma Function
The Gamma Function is a cornerstone of probability and statistics, especially when dealing with distributions like the gamma distribution. It is denoted by \(\Gamma(n)\) and is a generalization of the factorial function for real and complex numbers. For a positive integer \(n\), \(\Gamma(n) = (n-1)!\).

More generally, for any \(z > 0\), the Gamma function is defined as \(\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\). This integral form is quite useful in evaluating the MGFs and in solving problems involving continuous probability distributions.

In the exercise, the power function's integral involves a similar form to the Gamma function, illuminating its use in simplifying and solving complex statistical problems. Knowing properties of the Gamma function helps greatly in these calculations, making it easier to find solutions and verify hypotheses.

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Most popular questions from this chapter

Let \(X\) have the pmf \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=0,1\), zero elsewhere. \(W_{\text {e }}\) test the simple hypothesis \(H_{0}: \theta=\frac{1}{4}\) against the alternative composite hypothesis \(H_{1}: \theta<\frac{1}{4}\) by taking a random sample of size 10 and rejecting \(H_{0}: \theta=\frac{1}{4}\) if aulul only if the observed values \(x_{1}, x_{2}, \ldots, x_{10}\) of the sample observations are such that \(\sum_{1}^{10} x_{i} \leq 1 .\) Find the power function \(\gamma(\theta), 0<\theta \leq \frac{1}{4}\), of this test.

Let \(X\) be \(N(0, \theta)\) and, in the notation of this section, let \(\theta^{\prime}=4, \theta^{\prime \prime}=9\), \(\alpha_{a}=0.05\), and \(\beta_{a}=0.10 .\) Show that the sequential probability ratio test can be based upon the statistic \(\sum_{1}^{n} X_{i}^{2} .\) Determine \(c_{0}(n)\) and \(c_{1}(n)\).

. Let \(X\) and \(Y\) have a joint bivariate normal distribution. An observation \((x, y)\) arises from the joint distribution with parameters equal to either $$\mu_{1}^{\prime}=\mu_{2}^{\prime}=0,\quad\left(\sigma_{1}^{2}\right)^{\prime}=\left(\sigma_{2}^{2}\right)^{\prime}=1, \quad \rho^{\prime}=\frac{1}{2}$$ or $$\mu_{1}^{\prime \prime}=\mu_{2}^{\prime \prime}=1,\left(\sigma_{1}^{2}\right)^{\prime \prime}=4, \quad\left(\sigma_{2}^{2}\right)^{\prime \prime}=9, \rho^{\prime \prime}=\frac{1}{2}$$ Show that the classification rule involves a second-degree polynomial in \(x\) and \(y\).

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample of size 10 from a normal distribution \(N\left(0, \sigma^{2}\right) .\) Find a best critical region of size \(\alpha=0.05\) for testing \(H_{0}: \sigma^{2}=1\) against \(H_{1}: \sigma^{2}=2 .\) Is this a best critical region of size \(0.05\) for testing \(H_{0}: \sigma^{2}=1\) against \(H_{1}: \sigma^{2}=4 ?\) Against \(H_{1}: \sigma^{2}=\sigma_{1}^{2}>1 ?\)

Show that the likelihood ratio principle leads to the same test when testing a simple hypothesis \(H_{0}\) against an alternative simple hypothesis \(H_{1}\), as that given by the Neyman-Pearson theorem. Note that there are only two points in \(\Omega\).

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