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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
By using the properties of the moment generating function (MGF), the derivative of the MGF, and differential equations, it can be verified that the random variable 'Y' follows a Normal distribution with mean \(\theta\) and variance '1'.

Step by step solution

01

Establish the Moment Generating Function (MGF)

Start by establishing the moment generating function (MGF) for the normal distribution. For a normal random variable 'Y', the MGF is defined as \(M(t) = e^{\mu t + \frac{1}{2}\sigma^2t^2}\). Since variance is '1', substitute \(\mu = \theta\) and \(\sigma = 1\) into the MGF to get \(M(t) = e^{\theta t + \frac{1}{2}t^2}\).
02

Compute the Derivative of the MGF

Next, compute the derivative of the moment generating function. The derivative \(M^{\prime}(t)\) can be found by using the chain rule, which gives \(M^{\prime}(t) = (\theta + t)e^{\theta t + \frac{1}{2}t^2}\).
03

Set t=0 in the Derivative of the MGF

Set \(t=0\) in \(M^{\prime}(t)\) as per the hint. This results in \(M^{\prime}(0) = \theta\), proving that the expected value \(E(Y) = \theta\) as provided in the problem statement.
04

Formulate the Differential Equation

Then establish that \(M^{\prime}(0) = \theta\) can be formulated into a differential equation. The differential equation based on the MGF of the normal distribution is \(\frac{d}{dt}\left[M(t)\right]=M(t)(\mu+\sigma^2t)\). Set this equal to \(\theta\) and solve for 'Y' to validate that \(Y\) indeed follows a Normal distribution \(N(\theta, 1)\).
05

Verify the Solution

Finally, verify the solution by substituting \(Y\) in your discovered equation to confirm that it fulfills the conditions of the equation, thereby validating that \(Y\) follows a Normal Distribution \(N(\theta, 1)\).

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid with the distribution \(N\left(\theta, \sigma^{2}\right),-\infty<\theta<\infty\). Prove that a necessary and sufficient condition that the statistics \(Z=\sum_{1}^{n} a_{i} X_{i}\) and \(Y=\sum_{1}^{n} X_{i}\), a complete sufficient statistic for \(\theta\), are independent is that \(\sum_{1}^{n} a_{i}=0\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a normal distribution with mean zero and variance \(\theta, 0<\theta<\infty\). Show that \(\sum_{1}^{n} X_{i}^{2} / n\) is an unbiased estimator of \(\theta\) and has variance \(2 \theta^{2} / n\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(N\left(\theta_{1}, \theta_{2}\right)\) distribution. (a) Show that \(E\left[\left(X_{1}-\theta_{1}\right)^{4}\right]=3 \theta_{2}^{2}\). (b) Find the MVUE of \(3 \theta_{2}^{2}\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample of size \(n\) from a distribution with pdf \(f(x ; \theta)=\theta x^{\theta-1}, 00 .\) (a) Show that the geometric mean \(\left(X_{1} X_{2} \cdots X_{n}\right)^{1 / n}\) of the sample is a complete sufficient statistic for \(\theta\). (b) Find the maximum likelihood estimator of \(\theta\), and observe that it is a function of this geometric mean.

Show that the product of the sample observations is a sufficient statistic for \(\theta>0\) if the random sample is taken from a gamma distribution with parameters \(\alpha=\theta\) and \(\beta=6\).
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