Chapter 4: Problem 25
Let \(Y_{1}
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Let \(y_{1}
Let \(\bar{X}\) and \(\bar{Y}\) be the means of two independent random samples, each of size \(n\), from the respective distributions \(N\left(\mu_{1}, \sigma^{2}\right)\) and \(N\left(\mu_{2}, \sigma^{2}\right)\), where the common variance is known. Find \(n\) such that $$ P\left(\bar{X}-\bar{Y}-\sigma / 5<\mu_{1}-\mu_{2}<\bar{X}-\bar{Y}+\sigma / 5\right)=0.90 $$
Assume that \(Y_{1}\) has a \(\Gamma(\alpha+1,1)\) -distribution, \(Y_{2}\) has a
uniform \((0,1)\) distribution, and \(Y_{1}\) and \(Y_{2}\) are independent.
Consider the transformation \(X_{1}=\) \(Y_{1} Y_{2}^{1 / \alpha}\) and
\(X_{2}=Y_{2}\)
(a) Show that the inverse transformation is: \(y_{1}=x_{1} / x_{2}^{1 /
\alpha}\) and \(y_{2}=x_{2}\) with support \(0
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?
Let \(X\) have a pdf of the form \(f(x ; \theta)=\theta x^{\theta-1}, 0
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