Chapter 3: Problem 15
Let \(X\) be a random variable such that \(E\left(X^{2 m}\right)=(2 m) ! /\left(2^{m} m !\right), m=\) \(1,2,3, \ldots\) and \(E\left(X^{2 m-1}\right)=0, m=1,2,3, \ldots\) Find the mgf and the pdf of \(X\).
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Chapter 3: Problem 15
Let \(X\) be a random variable such that \(E\left(X^{2 m}\right)=(2 m) ! /\left(2^{m} m !\right), m=\) \(1,2,3, \ldots\) and \(E\left(X^{2 m-1}\right)=0, m=1,2,3, \ldots\) Find the mgf and the pdf of \(X\).
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Show that the failure rate (hazard function) of the Pareto distribution is $$\frac{h(x)}{1-H(x)}=\frac{\alpha}{\beta^{-1}+x}$$ Find the failure rate (hazard function) of the Burr distribution with cdf $$G(y)=1-\left(\frac{1}{1+\beta y^{\top}}\right)^{\alpha}, \quad 0 \leq y<\infty$$ In each of these two failure rates, note what happens as the value of the variable increases.
Let the number \(X\) of accidents have a Poisson distribution with mean \(\lambda \theta\). Suppose \(\lambda\), the liability to have an accident, has, given \(\theta\), a gamma pdf with parameters \(\alpha=h\) and \(\beta=h^{-1}\); and \(\theta\), an accident proneness factor, has a generalized Pareto pdf with parameters \(\alpha, \lambda=h\), and \(k\). Show that the unconditional pdf of \(X\) is $$\frac{\Gamma(\alpha+k) \Gamma(\alpha+h) \Gamma(\alpha+h+k) \Gamma(h+k) \Gamma(k+x)}{\Gamma(\alpha) \Gamma(\alpha+k+h) \Gamma(h) \Gamma(k) \Gamma(\alpha+h+k+x) x !}, \quad x=0,1,2, \ldots$$ sometimes called the generalized Waring pmf.
Let \(X\) and \(Y\) have a bivariate normal distribution with parameters
\(\mu_{1}=\) \(3, \mu_{2}=1, \sigma_{1}^{2}=16, \sigma_{2}^{2}=25\), and
\(\rho=\frac{3}{5} .\) Determine the following probabilities:
(a) \(P(3
Let \(T=W / \sqrt{V / r}\), where the independent variables \(W\) and \(V\) are, respectively, normal with mean zero and variance 1 and chi-square with \(r\) degrees of freedom. Show that \(T^{2}\) has an \(F\) -distribution with parameters \(r_{1}=1\) and \(r_{2}=r\). Hint: What is the distribution of the numerator of \(T^{2} ?\)
Evaluate \(\int_{2}^{3} \exp \left[-2(x-3)^{2}\right] d x\).
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