Question

1. The Weibull distribution has many applications in reliability engineering, survival analysis, and general insurance. function Let p> 0, δ > 0. Consider the probability density x>0 zero otherwise. Find the probability distribution of w-x6 a) Determine the probability distribution of W by finding the c.d.f. of W, Fw(w). Find the cd.f. of X, Fx(x) = P(X x). “Hint', 1: u-substitution: u "Hint" 2: There is no such thing as a negative cumulative distribution function "Hint" 3: Should be Fx(0)-0, Fx(o)1 i) Find the c.d.f. of W, Fw(w) - P(Wsw) P(x8s w). iii) What is the name of the probability distribution of W? What are its parameters? b) Determine the probability distribution of W by finding the p.d.f. of W, Jw(w), using the change-of-variable technique. DFind the pdf. of W. w(w) -Ax(e (w)) a ii) What is the name of the probability distribution of W? What are its parameters? dx

Answer 1

$1.~(a)~CDF~of~W~is\\ F_W(w)=P(X^\delta\leq w)=P(X\leq w^{\frac{1}{\delta}})=\int_0^{w^{\frac{1}{\delta}}}\beta \delta x^{\delta-1}\exp\left(-\beta x^\delta \right )dx\\ Take~\beta x^\delta=y~then~ \beta \delta x^{\delta-1}dx=dy\\ then~F_W(w)=\int_0^w \exp(-y)dy=1-\exp(-w),~w>0\\ Hence~F_W(w)=0~if~w\leq 0\\ =1-\exp(-w)~if~w>0\\ (b)~(i)~The~pdf~of~W~is\\ f_W(w)=\frac{\mathrm{d} F_W(w)}{\mathrm{d} w}=\exp(-w),~w>0\\ =0~otherwise\\ (ii)~Hence~W\sim Exponential~distribution~with~parameter=1.\\ (c)~MGF~of~W~is\\ M_W(t)=E(e^{Wt})=\int_0^\infty e^{tw}e^{-w}dw=\int_0^\infty e^{-(1-t)w}dw\\ =\frac{1}{1-t}~provided~t<1$