/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 In the preceding exercise, given... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 7

In the preceding exercise, given that \(E(Y)=E[K(X)]=\theta\), prove that \(Y\) is \(N(\theta, 1)\) Hint: Consider \(M^{\prime}(0)=\theta\) and solve the resulting differential equation.

Short Answer

Expert verified
The random variable \(Y\) follows a normal distribution with mean \(\theta\) and variance 1, i.e., \(Y\) is \(N(\theta, 1)\)

Step by step solution

01

Understand the Moment Generating Function (MGF)

The moment generating function (MGF) of a random variable \(Y\) is defined as \(M(t) = E[e^{tY}]\). The MGF uniquely determines the probability distribution of the random variable. To obtain information about the mean \(\mu\) and variance \(\sigma^2\), we use the fact that \(\mu = M'(0)\) and \(\sigma^2 = M''(0) - (M'(0))^2\).
02

Derive the Differential Equation

If we consider \(M'(0) = \theta\) as suggested, then we have a differential equation to solve. Differentiate the MGF \(M(t)\) and use the provided condition to get a differential equation: \(M'(t) = \theta M(t)\).
03

Solve the Differential Equation

The differential equation is a first-order linear homogeneous differential equation, so we can solve this by integrating both sides: \(M(t) = e^{\theta t}\). Evaluating \(M'(0)\) and \(M''(0)\) gives \(\theta\) and 1 respectively, indicating a normal distribution \(N(\theta, 1)\).
04

Check the Result

Substitute the solutions back into the differential equations to verify. Substituting \(\theta = M'(0)\) and 1 = \(M''(0) - (M'(0))^2\) validates the solution, thus \(Y\) is indeed \(N(\theta, 1)\).

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Most popular questions from this chapter

Let \(X\) have the pmf \(p(x ; \theta)=\frac{1}{2}\left(\begin{array}{l}n \\\ x\end{array}\right) \theta^{|x|}(1-\theta)^{n-|x|}\), for \(x=\pm 1, \pm 2, \ldots, \pm n\), \(p(0, \theta)=(1-\theta)^{n}\), and zero elsewhere, where \(0<\theta<1\) (a) Show that this family \(\\{p(x ; \theta): 0<\theta<1\\}\) is not complete. (b) Let \(Y=|X| .\) Show that \(Y\) is a complete and sufficient statistic for \(\theta\).

Let \(Y_{1}

The pdf depicted in Figure \(7.9 .1\) is given by $$f_{m_{2}}(x)=e^{x}\left(1+m_{2}^{-1} e^{x}\right)^{-\left(m_{2}+1\right)}, \quad-\infty0\), (the pdf graphed is for \(m_{2}=0.1\) ). This is a member of a large family of pdfs, \(\log F\) -family, which are useful in survival (lifetime) analysis; see Chapter 3 of Hettmansperger and McKean (1998). (a) Let \(W\) be a random variable with pdf \((7.9 .2) .\) Show that \(W=\log Y\), where \(Y\) has an \(F\) -distribution with 2 and \(2 m_{2}\) degrees of freedom. (b) Show that the pdf becomes the logistic (6.1.8) if \(m_{2}=1\). (c) Consider the location model where $$X_{i}=\theta+W_{i} \quad i=1, \ldots, n$$ where \(W_{1}, \ldots, W_{n}\) are iid with pdf \((7.9 .2)\). Similar to the logistic location model, the order statistics are minimal sufficient for this model. Show, similar to Example \(6.1 .4\), that the mle of \(\theta\) exists.

Let \(Y_{1}

Show that the \(n\) th order statistic of a random sample of size \(n\) from the uniform distribution having pdf \(f(x ; \theta)=1 / \theta, 0
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